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$f(x)=\\simplify[all,!collectNumbers]{(x -{a})/((x -{b})^2 + {c}) }$

$f'(x)=$ [[2]]

Find when $f'(x)=0$, hence find:

$x$-coordinate of the stationary point giving a minimum $=$ [[0]]

$x$-coordinate of the stationary point giving a maximum $=$ [[1]]

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Finding the stationary points of a rational function with specific features.

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Locate the stationary points of the function.

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$f(x)=\\simplify[all,!collectNumbers,!noleadingminus]{{a}x^3+{b}x^2+{c}x+{d}}$

$f'(x)=$ [[2]]

$f''(x)=$ [[3]]

Find when $f'(x)=0$, hence find:

$x$-coordinate of the stationary point giving a minimum $=$ [[0]]

$x$-coordinate of the stationary point giving a maximum $=$ [[1]]

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On differentiating we get $\\displaystyle \\frac{df}{dx}=\\simplify[std]{{3*a}x^2+{2*b}x+{c}}$.

To find the stationary points we have to solve $\\displaystyle \\frac{df}{dx}=0$ for $x$.

So we have to solve $\\simplify[std]{{3*a}x^2+{2*b}x+{c}=0}$.

Note that the quadratic factorises and the equation becomes $\\simplify[std]{({3a}x-{r1})(x-{r2})=0}$.

Hence we have two stationary points: $x=\\simplify[std]{{r1}/{3a}}$ and $x=\\var{r2}$.

To find out the types of these stationary points we look at the sign of $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6a}*x+{2*b}}$ at the stationary points.

If $\\displaystyle \\frac{d^2f}{dx^2} \\lt 0 $ at a stationary point then it is a MAXIMUM.

If $\\displaystyle \\frac{d^2f}{dx^2} \\gt 0 $ at a stationary point then it is a MINIMUM.

If $\\displaystyle \\frac{d^2f}{dx^2} = 0 $ at a stationary point then we have to do more work!

At $x=\\var{r2}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{6*a*r2+2*b}}${lg1}$0$ hence is a {type1}.

At $\\displaystyle x=\\simplify[std]{{r1}/{3a}}$ we have $\\displaystyle \\frac{d^2f}{dx^2} = \\simplify{{2*r1+2*b}}${lg2}$0$ hence is a {type2}.

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Finding the stationary points of a cubic with two turning points

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Find the coordinates of the stationary points of the function.

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The length of the box is \$\\var{l}-2x\$, the width is \$\\var{w}-2x\$ and the height is \$x\$.

The volume is then given by

\$V=(\\var{l}-2x).(\\var{w}-2x).x\$

\$V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\$

\$\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\$

This is a quadratic equation.

\$x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\$

\$x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\$

\$\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\$

when \$x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\$ \$\\frac{d^2V}{dx^2}<0\$ and therefore is the value that gives a maximum.

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Input the value for \$x\$ correct to one decimal place.

\$x = \$ [[0]]

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Maximising the volume of a rectangular box

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A rectangular sheet of metal of length = \$\\var{l}cm\$ and width = \$\\var{w}cm\$ has a square of side \$x\\,cm\$ cut from each corner. The ends and sides will be folded upwards to form an open box.

Determine the value of \$x\$ that will maximise the volume of this box.

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Another practical application of differentiation

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An open metal tank of square base has a volume of $\\var{v}\\text{ m}^3$

Given that the square base has sides of length $x$ metres, find expressions, in terms of $x$, for the following.

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This section draws on the skills learnt the previous parts of the 'Differentiation' series of questions, and some geometry knowledge.

The hint under the steps should be all the extra information you need.

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Describe the height of the tank in terms of $x$

$h=$ [[0]] m

Describe the surface area of the tank in terms of $x$

$S=$ [[0]] $\\text{m}^2$

Find the first derivative

$S'(x)=$ [[1]]

and given that the surface area is a minimum, find the value of $x$. Therefore,

$x=$ [[0]] (give your answer to 2 decimal places)

Find the second derivative

$S''(x)=$ [[2]]

Check this is a minimum.

Substitute your value for $x$ into $S''(x)$ and determine whether is it a minimum.

Type '$Y$' for yes, '$N$' for no, or '$U$' for undefined.

[[3]]

Hence, calculate the minimum area of metal used

$A_{min}=$ [[4]] (give your answer to 2 decimal places)

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Hint: find $x$ at stationary point

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The square-based box will have volume

\\[V=x^2h,\\]

where $x$ is the base side length and $h$ is the box height. Further, the surface area $S$, of the box is

\\[S=2x^2+4xh.\\]

For this particular problem, $V={volume}$, so we have

\\[\\var{volume}=x^2h \\Rightarrow h=\\frac{\\var{volume}}{x^2},\\]

providing a function for $h$ in terms of $x$. Substituting this into the surface area function we then have

\\[S(x)=2x^2+4x\\left(\\frac{\\var{volume}}{x^2}\\right)=2x^2+4\\frac{\\var{volume}}{x}.\\]

Minima and maxima of $S(x)$ occur when $dS/dx=0$. We have

\\[\\frac{dS}{dx}=4x-4\\frac{\\var{volume}}{x^2},\\]

and so we must solve

\\[0=4x-4\\frac{\\var{volume}}{x^2}.\\]

Rearranging, we find that

\\[4x=4\\frac{\\var{volume}}{x^2}\\]

\\[\\Rightarrow x^3=\\var{volume}\\]

\\[\\Rightarrow x=\\var{volume}^{1/3}=\\var{xsol}.\\]

Finally, substituting this value back into the $h$ equation, we see that

\\[h=\\frac{\\var{volume}}{\\var{volume}^{2/3}}=\\var{hsol}.\\]

We should confirm that this set of dimensions does in fact produce a minimum surface area (rather than a maximum). This can be carried out by noting that the 2nd derivative of $S$ with respect to $x$ is

\\[\\frac{d^2S}{dx^2}=4+8\\frac{\\var{volume}}{x^3},\\]

which is positive for all positive $x$, and hence our proposed side length will produce a minimised box surface area.

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The equation for box volume can be used to express box height in terms of the side length of the base of the box. If the side length is denoted $x$, enter the function here

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Use your result from part (a) to write a function for the surface area of cardboard used in the construction of the box. Your function should be entirely in terms of the side length of the base of the box, $x$.

To minimise the surface area of cardboard, we need the derivative of the surface area function with respect to $x$. This derivative is $\\displaystyle \\frac{dS}{dx}=$

The minimum surface area will occur when the length of the sides of the box base is $x=$

Finally, the height of the box must then be $h=$

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The box volume is fixed for each student. Here it lies between 0.3 and 3$m^3$.

"}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

This question guides students through the process of determining the dimensions of a box to minimise its surface area whilst meeting a specified volume.

"}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "statement": "

A square based box is to be constructed such that its volume is exactly {volume}m$^3$. The following questions guide you through the process of determining the base side length and box height such that the minimum amount of cardboard is used to construct the box.

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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]

Firstly, differentiate.

$\\displaystyle \\frac{dy}{dx}=$ [[1]]

Gradient at $x=\\var{d}\\;$ is [[0]]

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You have not given your answer to the correct precision.

Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

$y=\\simplify {{f}x^2+{g}x+{h}}$

Firstly, find the first and second derivatives $y$.

$\\displaystyle \\frac{dy}{dx}=$ [[2]]

$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

$x$-coordinate of the turning point $=$ [[0]]

$y$-coordinate of the turning point $=$ [[1]]

The turning point is a [[4]]

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You have not given your answer to the correct precision.

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maximum

", "

minimum

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An unpowered missile is launched vertically from the ground.

At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

\\[ y=\\var{z}t-\\var{w}t^2. \\]

Calculate the maximum height reached by the missile.

Firstly, differentiate.

$\\displaystyle \\frac{dy}{dt}=$ [[0]]

Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.

$y=$ [[1]]

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Parts A and B

Here, the question takes you throught the stages needed to find the solution. The reason we differentiate is that the derivative of a function tells us its gradient at a given point, and we want to find where the function has gradient zero because when the gradient is zero we either have a maximum or a minimum point.

Part C

The first part of this question is similar to parts A and B. The tricky bit is the second part! You need to work out the value of $t$ that produces the maximum piont but that is not the final answer - you need to use that value of $t$ to find the maximum height, which you do by substituting $t$ into the original function to find $y$.

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Is the stationary point a maximum?

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Problem on a closed cylindrical tank having minimum surface area

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A closed cylindrical tank is to be built having a volume of \$\\var{v}\$ cm³.

Determine the required height, \$h\$, and radius, \$r\$, if the total surface area is to be a minimum.

", "advice": "

\$\\pi r^2h=\\var{v}\$

\$h=\\frac{\\var{v}}{\\pi r^2}\$

The total surface area is to be a minimum.

Lid + curved surface area + base

\$A=\\pi r^2+2\\pi rh+\\pi r^2\$

\$A=2\\pi r^2+2\\pi r\\left(\\frac{\\var{v}}{\\pi r^2}\\right)\$

\$A=2\\pi r^2+\\simplify{2*{v}}r^{-1}\$

\$\\frac{dA}{dr}=4\\pi r-\\simplify{2{v}}r^{-2}=0\$

\$4\\pi r=\\simplify{2*{v}}/{r^2}\$

\$r^3=\\frac{\\var{v}}{2\\pi}\$

\$r=\\simplify{({v}/(2*pi))^(1/3)}\$

From the second line we have the relation \$h=\\frac{\\var{v}}{\\pi r^2}\$ to get

\$h=2*\\simplify{({v}/(2*pi))^(1/3)}\$

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Input the cyinder height, correct to two decimal places.

\$h = \$ [[0]]

Input the required cylinder radius, correct to two decimal places.

\$r = \$ [[1]]

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Rate of change problem involving velocity & acceleration

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Calculate the speed of the missile (m/s) \$\\var{b}\$ seconds after launch. Give your answer correct to one decimal place.

\$v = \$ [[0]]m/s

What is the maximum height achieved by this missile? Give your answer correct to one decimal place.

\$h = \$ [[1]]m

"}], "advice": "

\$h=\\var{a}t-4.9t^2\$

Recall that speed is the rate of change of position with respect to time i.e. \$v=\\frac{dh}{dt}\$

\$v=\\frac{dh}{dt}=\\var{a}-2*4.9t\$

when \$t=\\var{b}\$

\$v=\\var{a}-2*4.9*\\var{b}\$

\$v=\\simplify{{a}-9.8*{b}}m/s\$

The missile will reach its maximum height when its speed = 0. i.e. \$v=\\frac{dh}{dt}=\\var{a}-2*4.9t=0\$

\$\\var{a}=9.8t\$

\$t=\\var{a}/9.8\$

The maximum height reached will occur when \$t=\\simplify{{a}/9.8}\$

\$h=\\var{a}*\\left(\\simplify{{a}/9.8}\\right)-4.9*\\left(\\simplify{{a}/9.8}\\right)^2\$

\$h=\\simplify{{a}^2/19.6}\$

\$h=\\simplify{{{a}/{19.6}^0.5}^2}\$

", "variable_groups": [], "statement": "

A missile is launched straight up in the air. The height of the missile, \$h\$ metres, above the ground \$t\$ seconds after the launch button is pressed is given by:

\$h=\\var{a}t-4.9t^2\$

", "preamble": {"css": "", "js": ""}, "type": "question"}, {"name": "SFY0004 Implicit 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "calculus", "derivative", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "first derivative using implicit differentiation", "implicit differentiation", "implicit relation"], "advice": "

On differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) + {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[(\\var{b} + 2y) \\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x}\\] and hence on further rearranging:
\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x) / ({b} + (2 * y))}\\]

", "rulesets": {}, "parts": [{"prompt": "\n

Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

Input your answer here:

$\\displaystyle \\frac{dy}{dx}= $ [[0]]

\n \n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "all,!collectNumbers", "marks": 2.0, "answer": "(({( - a)} + ( - (2 * x))) / ({b} + (2 * y)))", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "statement": "

Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{a}x+{b}y}=\\var{c}\\]
answer the following question.

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20/06/2012:

\n \t\t \t\t \t\t

Added tags.

\n \t\t \t\t \t\t

Improved display using \\displaystyle where appropriate.

\n \t\t \t\t \t\t

Changed marks to 2.

\n \t\t \t\t \t\t

3/07/2012:

\n \t\t \t\t \t\t

Added tags.

\n \t\t \t\t \n \t\t \n \t\t", "description": "\n \t\t \t\t \t\t

Implicit differentiation.

\n \t\t \t\t \t\t

Given $x^2+y^2+ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

\n \t\t \t\t \t\t

\n \t\t \t\t \n \t\t \n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "SFY0004 Implicit 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "ungrouped_variables": ["a", "c", "b", "d"], "tags": ["Calculus", "Differentiation", "calculus", "derivative", "deriving an implicit relation", "differentiate", "differentiate implicitly", "differentiation", "equation of tangent", "first derivative using implicit differentiation", "gradient", "implicit differentiation", "implicit relation", "tangent at a point"], "preamble": {"css": "", "js": ""}, "advice": "

a)

On differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\] and hence on further rearranging:

\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]

b)

On putting $x=0$ in the relation we get:

\\[\\simplify{y^2+{b}y={c}} \\Rightarrow \\simplify{y^2+{b}y-{c}=0 }\\Rightarrow (y+\\var{c})(y-1)=0\\]

Hence $a=-\\var{c}$ and $b=1$.

c)

First we find the tangent at the point $(0,-\\var{c})$.

We find using the formula we found for $\\frac{dy}{dx}$ in part a) that the gradient at $(0,-\\var{c})$ is:

\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}+{d*c}}}{\\var{b}-\\var{2*c}}=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}\\]

As the tangent goes through the point $(0,\\var{-c})$ i.e. at $x=0,\\;\\;y=-\\var{c}$ we see that the equation of the tangent is:

\\[y=\\simplify[all,fractionNumbers]{{a-d*c}/{c+1}}x-\\var{c}\\]

Next we find that the gradient at $(0,1)$ is:

\\[\\frac{dy}{dx}=\\frac{\\simplify[all,!collectnumbers]{{-a}-{d}}}{\\var{b}+2}=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}\\]

As the tangent goes through the point $(0,1)$ i.e. at $x=0,\\;\\;y=1$ we see that the equation of the tangent is:

\\[y=\\simplify[all,fractionNumbers]{-{a+d}/{c+1}}x+1\\]

", "rulesets": {"std": ["all", "fractionNumbers"]}, "parts": [{"stepsPenalty": 0, "prompt": "\n

Using implicit differentiation find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

Input your answer here:

$\\displaystyle \\frac{dy}{dx}= $ [[0]]

Input all numbers as integers not as decimals.

If you want more help click on Show steps - you will not lose any marks.

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Input all numbers as integers or as fractions, not as decimals.

", "showStrings": false, "strings": ["."], "partialCredit": 0}, "expectedvariablenames": [], "checkingaccuracy": 0.001, "vsetrange": [0, 1], "showpreview": true, "vsetrangepoints": 5, "showCorrectAnswer": true, "answersimplification": "all,!collectNumbers", "scripts": {}, "answer": "(({( - a)} + ( - (2 * x))-{d}y) / ({b} + (2 * y)+{d}x))", "marks": 2, "checkvariablenames": false, "checkingtype": "absdiff", "type": "jme"}], "showCorrectAnswer": true, "scripts": {}, "steps": [{"type": "information", "showCorrectAnswer": true, "scripts": {}, "prompt": "\n

Note that we regard $y$ as a function of $x$.

Hence we have (using the chain rule):

$\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$

And , using the product rule:

$\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

Now differentiate both sides of the relation with respect to $x$.

\n ", "marks": 0}], "type": "gapfill"}, {"prompt": "\n

Find the two points $(0,a),\\;\\;(0,b),\\;\\;a \\lt b$ which lie on the curve given by the relation.

$a=\\;$[[0]]

$b=\\;$[[1]]

(remember that $a \\lt b$).

\n ", "marks": 0, "gaps": [{"allowFractions": false, "marks": 1, "maxValue": "-c", "minValue": "-c", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "marks": 1, "maxValue": "1", "minValue": "1", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "type": "gapfill"}, {"prompt": "\n

Hence find the equations of the tangents at the points $(0,a)$ and $(0,b)$.

Input all numbers as integers or as fractions, not as decimals.

Equation of tangent at $(0,a)$:

Find the gradient of the tangent at $(0,a)$.

Gradient=[[0]].

Hence the equation of the tangent at $(0,a)$ is:

$y = \\;$[[1]]

Equation of tangent at $(0,b)$:

Find the gradient of the tangent at $(0,b)$.

Gradient=[[2]].

Hence the equation of the tangent at $(0,b)$ is:

$y = \\;$[[3]]

\n ", "marks": 0, "gaps": [{"notallowed": {"message": "

Input as an integer or as a fraction, not as a decimal.

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Input all numbers as integers or as fractions, not as decimals.

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Input as an integer or as a fraction, not as a decimal.

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Input all numbers as integers or as fractions, not as decimals.

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Given the following relation between $x$ and $y$
\\[\\simplify[all,!collectNumbers]{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]
answer the following questions.

", "type": "question", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"a": {"definition": "-random(1..9)", "templateType": "anything", "group": "Ungrouped variables", "name": "a", "description": ""}, "c": {"definition": "random(2..9 except -a+1)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "c-1", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "random(-3..3 except 0)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}}, "metadata": {"notes": "\n \t\t

30/04/2013

\n \t\t

Created new question for SFY0004 out of 1041 CBA2_5.

\n \t\t

Added more tags.

\n \t\t", "description": "\n \t\t

Implicit differentiation.

\n \t\t

Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

\n \t\t

Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.

\n \t\t

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Maria's copy of Differentiation: Exponential and Logarithmic functions - Gradient of Hill", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}, {"name": "Ben McGovern", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4872/"}, {"name": "Evi Papadaki", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/18113/"}], "tags": [], "metadata": {"description": "

Using basic derivatives to calculate the gradient function of a hill $y=-e^{x}+b\\ln{\\left(x\\right)+c$, and then substituting values to find the gradient at specific distance from the sea.

", "licence": "None specified"}, "statement": "

The cross-section of a hill with a steep cliff face is modelled by the equation:
\\[y=-e^{x}+\\var{b}\\ln{\\left(x\\right)+\\var{c}}\\]
where $y$ is the height of the ground above sea level, measured in decametres ($dam$), and $x$ is the horizontal distance from the sea, also measured in decametres.

The model applies for values of $x$ in the range $0.01\\le\\ x\\le3$.

", "advice": "

a) To find the gradient of the hill at any given horizontal distance from the sea, $x$, we need to differentiate the equation

\\[y=-e^{x}+\\var{b}\\ln{\\left(x\\right)+\\var{c}}\\]

with respect to x:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\var{b}\\times\\frac{1}{x}+0 \\\\&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\frac{1}{x}+0 \\\\&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

Therefore, the formula that describes the gradient of the hill at any horizontal distance from the sea is:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^x+\\frac{\\var{b}}{x} \\end{split}\\]

b) We can use the formula from part (a) to calculate the gradient at specific points.

Remember that the distance in this problem is measured in decametres ($dam$) and $1 ~~dam$ = $10~~ m$.

So, when $x=1$:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^1+\\frac{\\var{b}}{1} \\\\&=-e+\\var{b} \\\\ &= \\var{ansb1} ~ \\text{[2 d.p.]}\\end{split}\\]

Thus, the gradient of the hill at a horizontal distance of $1~~dam$ from the sea is $\\frac{dy}{dx}=\\var{ansb1} ~~ dam$.

When $x=0.1$:

\\[ \\begin{split} \\frac{dy}{dx}&=-e^{0.1}+\\frac{\\var{b}}{0.1} \\\\ &= \\var{ansb2} ~ \\text{[2 d.p.]}\\end{split}\\]

Thus, the gradient of the hill at a horizontal distance of $0.1~~dam$ from the sea is $\\frac{dy}{dx}=\\var{ansb2} ~~ dam$.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"b": {"name": "b", "group": "Ungrouped variables", "definition": "random (1..5)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(4..10)", "description": "", "templateType": "anything", "can_override": false}, "ansB2": {"name": "ansB2", "group": "Ungrouped variables", "definition": "precround(-e^(0.1)+10*{b},2)", "description": "", "templateType": "anything", "can_override": false}, "ansB1": {"name": "ansB1", "group": "Ungrouped variables", "definition": "precround(-e^1+{b},2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["b", "c", "ansB2", "ansB1"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Find the formula that describes the gradient of the hill at any given horizontal distance from the sea.

$\\frac{dy}{dx}=$ [[0]].

What is the gradient of the hill at a horizontal distance of:

$10$ metres ($1 ~~dam$) from the sea?

[[0]] $dam$

$1$ metre ($0.1~~ dam$) from the sea?

[[1]] $dam$

Give your answers to 2 decimal places when necessary.

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Calculating the rate of change of the temperature during a chemical reaction using the chain rule in a function of the form $T=ate^{-t}$, and finding the maximum temperature of the reaction.

", "licence": "None specified"}, "statement": "

Iron is mixed with an acid in an insulated container. The temperature, $T$°C, above room temperature, at time $t$ minutes after the metal is added to the acid is given by:
\\[T=\\var{a}te^{-t}\\]

", "advice": "

a) To find the rate of change at a specific time, we need to first differentiate the function

\\[ T=\\var{a}te^{-t} \\]

with respect to $t$.

To differentiate the function, we need to notice that $T$ is of the form $T=u(t) \\times v(t)$. In other words $T$ is the product of the functions $u=\\var{a}t$ and $v=e^{-t}$.

Therefore, we can use the product rule:

\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}\\times v+u\\times \\frac{dv}{dt} \\end{split}\\]

First, we need to calculate the derivatives $\\frac{du}{dt}$ and $\\frac{dv}{dt}$:

\\[ \\begin{split} u(t)&=\\var{a}t\\qquad \\text{and} \\qquad v(t)&=e^{-t} \\\\ \\frac{du}{dt}&=\\var{a}~~~~~\\qquad \\qquad \\frac{dv}{dt}&=-e^{-t}\\end{split} \\]

Substituting these results into the product rule formula, we can obtain the derivative of $T$:

\\[ \\begin{split} \\frac{dT}{dt}&= \\frac{du}{dt}v+u\\frac{dv}{dt} \\\\ &=\\var{a}e^{-t}+\\var{a}t \\left(-e^{-t} \\right) \\\\ &=\\var{a}e^{-t}-\\var{a}te^{-t}\\end{split}\\]

Factorising,

\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-t}(1-t)\\end{split}\\]

Now, we can substitute $t=\\var{t1}$ and calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added:

\\[ \\begin{split} \\frac{dT}{dt}&= \\var{a}e^{-\\var{t1}}(1-\\var{t1}) \\\\&= \\simplify{{a}e^{-{t1}}(1-{t1})} \\\\ &=\\var{ansa} ~~\\text{[2.dp]}\\end{split}\\]

Therefore, the rate of change of the temperature when $t=\\var{t1}$ is $\\frac{dT}{dt}=\\var{ansa} ^\\circ C/min$.

b) When the temperature reaches maximum, the rate of change of the temperature will be $0$. So, to find the time when the temperature reaches its maximum we must solve $\\frac{dT}{dt}=0$. Using our answer from a):

\\[ \\begin{split} \\var{a}e^{-t}(1-t)&=0 \\end{split}\\]

Since, $\\var{a}e^{-t} > 0$ for every value of $t$, this equation can only be equal to zero when

\\[ \\begin{split} 1-t &=0 \\\\ t&=1 \\end{split}\\]

Thus, the maximum temperature will be reached when $t=1$.

Finally, to calculate the temperature we need to substitute $t=1$ in the original function $T$:

\\[ \\begin{split} T&=\\var{a}te^{-t} \\\\\\\\ T_{max}&=\\var{a}\\times 1\\times e^{-1} \\\\&=\\frac{\\var{a}}{e} \\\\ &=\\var{ansb} ~~\\text{[2.d.p.]} \\end{split}\\]

Thus, the maximum temperature of the mixture is $\\var{ansb} ^\\circ C$ above room temperature.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "random(50..80)", "description": "", "templateType": "anything", "can_override": false}, "t1": {"name": "t1", "group": "Ungrouped variables", "definition": "random(3..10)", "description": "", "templateType": "anything", "can_override": false}, "Ansa": {"name": "Ansa", "group": "Ungrouped variables", "definition": "precround({a}e^(-t1)(1-{t1}),2)", "description": "", "templateType": "anything", "can_override": false}, "Ansb": {"name": "Ansb", "group": "Ungrouped variables", "definition": "precround({a}/e,2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "t1", "Ansa", "Ansb"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the rate of change of the temperature $\\var{t1}$ minutes after the metal is added.

When $t=\\var{t1}$ minutes then $\\frac{dT}{dt}=$ [[0]] $^\\circ C/min$.

Give your answer to two decimal places, if needed.

Calculate the maximum temperature of the mixture.

The maximum temperature of the mixture is [[0]] $^\\circ C$ above room temperature.

Give your answer to two decimal places, if needed.

The temperature, $T$°C of the air at height metres inside a blast furnace is modelled by the equation:

$ T=\\var{a}z^2+\\var{b}z+\\var{c} $

", "advice": "

a) We first need to differentiate the formula of the curve in respect of $z$.

\\[ \\frac{dT}{dz} = 2 \\times (\\var{a}) z+ \\var{b}+0 \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}}z+\\var{b} \\]

Now to find the gradient at the point where $z=\\var{z_1}$ we need to substitute the value in the equation we found in the previous step.

\\[ \\frac{dT}{dz}=\\simplify{2*{a}}\\times \\var{z_1} +\\var{b} \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}*{z_1}}+\\var{b} \\\\ \\implies \\frac{dT}{dz}= \\simplify{2*{a}*{z_1}+{b}} \\]

b) The height at which the temperature reaches the maximum can be calculated by solving the equation:

\\[ \\frac{dT}{dz}=0\\]

Therefore:

\\[ \\simplify{2*{a}}z+\\var{b}=0 \\\\ \\implies \\simplify{2*{a}}z= - \\var{b} \\\\ \\implies z = \\frac{-\\var{b}}{\\simplify{2*{a}}} \\\\ \\implies z= \\simplify{{-{b}}/{2*{a}}} \\\\ z=\\var{ans_b_rounded}~ \\text{(2.d.p.)} \\]

c) Now that we know that $z=\\var{ans_b_rounded}$ is the height that results in the maximum temperature, we can substitute this value in the original equation of the curve and calculate the maximum temperature.

\\[ T_{max}= \\var{a} \\times \\var{ans_b_rounded}^2 + \\var{b} \\times \\var{ans_b_rounded} + \\var{c} \\\\ \\implies T_{max}=\\var{a} \\times \\var{ans_b_rounded^2}+ \\var{b} \\times \\var{ans_b_rounded}+\\var{c} \\\\ \\implies T_{max}=\\simplify{{a*ans_b_rounded^2}}+ \\simplify{{b*ans_b_rounded}}+\\var{c} \\\\ \\implies T_{max}=\\simplify{{a*ans_b_rounded^2+b*ans_b_rounded+c}}\\]

So, if rounded, $T_{max}=\\var{ans_c_rounded}$.

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Find the rate of decrease of the temperature with height at the point where $z=\\var{z_1}$.

When $z=\\var{z_1}$ then $\\frac{dT}{dz}=$ [[0]].

Find the height at which the temperature is highest.

The temperature is highest when $z=$ [[0]].

Give your answer rounded to 2 decimal places, if needed.

Find the maximum temperature in the blast furnace.

The maximum temperarure is $T_{max}=$[[0]]$^\\circ C$.

Give your answer rounded to 2 decimal places, if needed.

The discharge of water into a stream, $D$ $dm^3/s$, affects the growth of the fish in the stream.

The average length of the fish, $x$ $cm$, is modelled by the curve:

$ x=\\var{a}D^2+\\var{b}D+\\var{c} $

", "advice": "

a) We first need to find differentiate the formula of the curve in respect of D.

\\[ \\frac{dx}{dD} = 2 \\times (\\var{a}) D+ \\var{b}+0 \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}}D+\\var{b} \\]

Now to find the gradient at the point where $D=\\var{d_1}$ we need to substitute the value in the equation we found in the previous step.

\\[ \\frac{dx}{dD}=\\simplify{2*{a}}\\times \\var{d_1} +\\var{b} \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}*{d_1}}+\\var{b} \\\\ \\implies \\frac{dx}{dD}= \\simplify{2*{a}*{d_1}+{b}} \\]

b) The discharge that results in the maximum average length of the fish can be calculated by solving the equation:

\\[ \\frac{dx}{dD}=0\\]

Therefore:

\\[ \\simplify{2*{a}}D+\\var{b}=0 \\\\ \\implies \\simplify{2*{a}}D= - \\var{b} \\\\ \\implies D = \\frac{-\\var{b}}{\\simplify{2*{a}}} \\\\ \\implies D= \\simplify{{-{b}}/{2*{a}}} \\]

Or, if rounded,

\\[ D=\\var{ans_b_rounded}\\]

c) Now that we know that $D=\\var{ans_b_rounded}$ is the discharge that results in the maximum average length we can substitute this value in the original equation of the curve and calculate the maximum length.

So,

\\[ x_{max}= \\var{a} \\times \\var{ans_b_rounded}^2 + \\var{b} \\times \\var{ans_b_rounded} + \\var{c} \\\\ \\implies x_{max}=\\var{a} \\times \\var{ans_b_rounded^2}+ \\var{b} \\times \\var{ans_b_rounded}+\\var{c} \\\\ \\implies x_{max}=\\simplify{{a*ans_b_rounded^2}}+ \\simplify{{b*ans_b_rounded}}+\\var{c} \\\\ \\implies x_{max}=\\simplify{{a*ans_b_rounded^2+b*ans_b_rounded+c}}\\]

So, $x_{max}=\\var{ans_c_rounded}$.

", "rulesets": {}, "builtin_constants": {"e": true, "pi,\u03c0": true, "i": true}, "constants": [], "variables": {"a": {"name": "a", "group": "Problem posing", "definition": "-random ( 50 .. 200)", "description": "", "templateType": "anything", "can_override": false}, "b": {"name": "b", "group": "Problem posing", "definition": "random(50..200)", "description": "", "templateType": "anything", "can_override": false}, "c": {"name": "c", "group": "Problem posing", "definition": "Random(10..80)", "description": "", "templateType": "anything", "can_override": false}, "d_1": {"name": "d_1", "group": "Problem posing", "definition": "random (0.2 .. 0.75 #0.05 )", "description": "", "templateType": "anything", "can_override": false}, "ans_a": {"name": "ans_a", "group": "Answers", "definition": "2*{a}*{d_1}+{b}", "description": "", "templateType": "anything", "can_override": false}, "ans_b": {"name": "ans_b", "group": "Answers", "definition": "-b/(2a)", "description": "", "templateType": "anything", "can_override": false}, "ans_c": {"name": "ans_c", "group": "Answers", "definition": "a*ans_b_rounded^2+b*ans_b_rounded+c", "description": "", "templateType": "anything", "can_override": true}, "ans_b_rounded": {"name": "ans_b_rounded", "group": "Answers", "definition": "precround(ans_b,2)", "description": "", "templateType": "anything", "can_override": false}, "ans_c_rounded": {"name": "ans_c_rounded", "group": "Answers", "definition": "precround({ans_c},2)", "description": "", "templateType": "anything", "can_override": false}}, "variablesTest": {"condition": "(ans_c_rounded<100) and (ans_a>0)", "maxRuns": "100"}, "ungrouped_variables": [], "variable_groups": [{"name": "Problem posing", "variables": ["a", "b", "c", "d_1"]}, {"name": "Answers", "variables": ["ans_a", "ans_b", "ans_c", "ans_b_rounded", "ans_c_rounded"]}], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the gradient of this curve at the point where $D=\\var{d_1}$.

", "answer": "{ans_a}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the discharge which results in the largest average length of the fish.

Give your answer rounded to 2 decimal places.

", "answer": "{ans_b_rounded}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}, {"type": "jme", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Calculate the maximum average length of the fish.

Give your answer rounded to 2 decimal places.

", "answer": "{ans_c_rounded}", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "caseSensitive": false, "valuegenerators": []}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Max and Min 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Calculus", "calculus", "classifying stationary points", "finding global maxima and minima", "finding local mamima and minima", "finding the stationary points", "optimisation", "optimising functions", "third derivative test for maximum or minimum"], "advice": "\n

First derivative

Differentiating we have:

\\[\\begin{eqnarray*} g'(x)&=&\\simplify{(x-{b})^3+3*(x-{a})*(x-{b})^2}\\\\ &=&\\simplify{(x-{b})^2(3*(x-{a})+x-{b})}\\\\ &=&\\simplify{4*(x-{k})*(x-{b})^2} \\end{eqnarray*} \\] and we have factorised the expression.

Stationary points

These are given by solving $g'(x)=0 \\Rightarrow x=\\var{k},\\;\\;\\mbox{or }x=\\var{b}$

Therefore the least stationary point is $x=\\var{k}$ and the greatest is $x=\\var{b}$ and we see that both stationary points are in $I$.

Second derivative.

The second derivative is given by:
\\[\\begin{eqnarray*} g''(x)&=&\\simplify{4*(x-{b})^2+8*(x-{k})(x-{b})}\\\\ &=&\\simplify{4*(x-{b})(3*x-{2*k+b})} \\end{eqnarray*} \\]

Local Minimum

At the stationary point $x=\\var{k}$ we have $g''(\\var{k})=\\var{4*(k-b)^2} \\gt 0$.

Hence $x=\\var{k}$ is a local minimum.

The other stationary point

The value at $x=\\var{b}$ is $g(\\var{b})= 0$.

Hence this test fails at this point and we proceed to use the third derivative to see in more information can be gained.

Third derivative

We see that $g'''(x)=\\simplify{8*(3*x-{k+2*b})}$.

Testing the stationary point using the third derivative gives:

$g'''(\\var{b})=\\var{8*(b-k)} \\neq 0$.

Therefore there cannot be an extremum point at $x=\\var{b}$.

Finding the global maximum and minimum on $I$

First we find the values at the endpoints of the interval $I=[\\var{l},\\var{m}]$ are:

$g(\\var{l})=\\var{valbegin}$.

$g(\\var{m})=\\var{valend}$.

Global Maximum

To find the global maximum note that we are only concerned with the values of $g$ on the interval $I$ and since $g$ does not have a local maximum on $I$ it must take its maximum value at one of the end points of $I$.

We see from the values at the end points obtained above that the global maximum value on $I$ is at $x=\\var{xma}$.

We have $g(\\var{xma})=\\var{gma}$.

Global Minimum.

$g$ has only one local minimum on $I$ at $x=\\var{k}$ and so this must be the global minimum on $I$.

We have $g(\\var{k})=\\var{(k-a)*(k-b)^3}$.

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n

Input the first derivative of $g$ here, factorised into a product of two factors in the form $g'(x)=c(x-a)(x-b)^2$for suitable integers $a$, $b$ and $c$:

\n \n

$g'(x)=\\;\\;$[[0]]

\n \n ", "gaps": [{"notallowed": {"message": "

Factorise the expression

", "showstrings": false, "strings": ["x^2", "x^3"], "partialcredit": 0.0}, "checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "4*(x + {-k}) * (x + {-b})^2", "type": "jme", "musthave": {"message": "

Factorise the expression

", "showstrings": false, "strings": ["(", ")"], "partialcredit": 0.0}}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

Find the stationary points of $g$, here thought of as $g:\\mathbb{R} \\rightarrow \\mathbb{R}$.

\n \n

Least stationary point: [[0]]

\n \n

Greatest stationary point: [[1]]

\n \n

Do both these stationary points lie in the interval $I$ ? [[2]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{k}", "type": "jme"}, {"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "{b}", "type": "jme"}, {"maxanswers": 0.0, "matrix": [1.0, 0.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

Yes

", "

"], "marks": 0.0, "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", ""], "displaycolumns": 0.0, "type": "1_n_2", "minmarks": 0.0}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

Input the second derivative of $g$:

\n \n

$g''(x)=\\;\\;$ [[0]]

\n \n

Using $g''(x)$, determine more information about the stationary points:

\n \n

Least stationary point is: (Choose one of the following)
[[1]]

\n \n

Greatest stationary point is: (Choose one of the following)
[[2]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "4*(x-{b})*(3*x-{2*k+b})", "type": "jme"}, {"maxanswers": 0.0, "matrix": [1.0, 0.0, 0.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

A local minimum.

", "

A local maximum.

", "

Uncertain as the second derivative test fails.

"], "marks": 0.0, "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", ""], "displaycolumns": 0.0, "type": "1_n_2", "minmarks": 0.0}, {"maxanswers": 0.0, "matrix": [0.0, 0.0, 1.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

A local minimum.

", "

A local maximum.

", "

Uncertain as the second derivative test fails.

"], "marks": 0.0, "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", ""], "displaycolumns": 0.0, "type": "1_n_2", "minmarks": 0.0}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

Using the third derivative answer the following questions:
$g'''(x) = \\;\\;$[[0]]

\n \n

If $a$ is the least stationary point then $g'''(a) =\\;\\;$[[1]]

\n \n

If $b$ is the other stationary point then $g'''(b) =\\;\\;$[[2]]

\n \n

This information tells us that: (Choose one of the following).
[[3]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "8*(3*x-{k+2*b})", "type": "jme"}, {"minvalue": "{third1}", "type": "numberentry", "maxvalue": "{third1}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{third2}", "type": "numberentry", "maxvalue": "{third2}", "marks": 0.5, "showPrecisionHint": false}, {"maxanswers": 0.0, "matrix": [0.0, 1.0, 0.0, 0.0, 0.0], "shufflechoices": true, "minanswers": 0.0, "choices": ["

{k} is not a local minimum or a local maximum.

", "

{b} is not a local minimum or a local maximum.

", "

{b} is not a local minimum or a local maximum and neither is {k}.

", "

{k} is a local maximum and {b} is a local minimum.

", "

{k} is a local minimum and {b} is a local maximum.

"], "marks": 0.0, "displaytype": "radiogroup", "maxmarks": 0.0, "distractors": ["", "", "", "", ""], "displaycolumns": 0.0, "type": "1_n_2", "minmarks": 0.0}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

What are the following values at the end points of the interval $I$ ?

\n \n

$g(\\var{l})=\\;\\;$ [[0]]

\n \n

$g(\\var{m})=\\;\\;$ [[1]]

\n \n

Input both to 2 decimal places.

\n \n ", "gaps": [{"minvalue": "{valbegin}", "type": "numberentry", "maxvalue": "{valbegin}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{valend}", "type": "numberentry", "maxvalue": "{valend}", "marks": 0.5, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

Global Maximum

\n \n

At what value of $x \\in I$ does $g$ have a global maximum ?

\n \n

$x=\\;\\;$ [[0]]

\n \n

Value of $g$ at this global maximum = [[1]].

\n \n

Global Minimum

\n \n

At what value of $x \\in I$ does $g$ have a global minimum ?

\n \n

$x=\\;\\;$ [[2]]

\n \n

Value of $g$ at this global minimum = [[3]].

\n \n ", "gaps": [{"minvalue": "{xma}", "type": "numberentry", "maxvalue": "{xma}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{gma}", "type": "numberentry", "maxvalue": "{gma}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{k}", "type": "numberentry", "maxvalue": "{k}", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "{gmi}", "type": "numberentry", "maxvalue": "{gmi}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

Let $I=[\\var{l},\\var{m}]$ be an interval and let $g: I \\rightarrow \\mathbb{R}$ be a function defined on this interval
given by :\\[g(x) = \\simplify{(x-{a})*(x-{b})^3}\\]

\n \n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "s1*random(1..9)", "name": "a"}, "c": {"definition": "random(3,6)", "name": "c"}, "b": {"definition": "4*k-3*a", "name": "b"}, "d": {"definition": "if(c=3,random(-9..9),random(-9..9#2))", "name": "d"}, "gma": {"definition": "(xma-a)*(xma-b)^3", "name": "gma"}, "k": {"definition": "a+random(1..6)", "name": "k"}, "m": {"definition": "b+random(1..3)", "name": "m"}, "xma": {"definition": "if(valend > valbegin,m,l)", "name": "xma"}, "l": {"definition": "k-random(1..3)", "name": "l"}, "valend": {"definition": "(m-a)*(m-b)^3", "name": "valend"}, "gmi": {"definition": "(k-a)*(k-b)^3", "name": "gmi"}, "valbegin": {"definition": "(l-a)*(l-b)^3", "name": "valbegin"}, "third2": {"definition": "8*(b-k)", "name": "third2"}, "third1": {"definition": "16*(k-b)", "name": "third1"}, "s1": {"definition": "random(1,-1)", "name": "s1"}}, "metadata": {"notes": "\n \t\t

9/07/2012:

\n \t\t

Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t", "description": "

$I$ compact interval, $g:I\\rightarrow I$, $g(x)=(x-a)(x-b)^2$. Stationary points in interval. Find local and global maxima and minima of $g$ on $I$.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": " Optimisation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "TEAME UCC", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/351/"}], "functions": {}, "ungrouped_variables": ["c", "b", "d"], "tags": ["rebelmaths"], "preamble": {"css": "", "js": ""}, "advice": "", "rulesets": {}, "parts": [{"prompt": "

Find two numbers whose difference is {b} and whose product is minimum.

Smaller number : [[0]]. Larger number: [[1]].

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "variableReplacements": [], "maxValue": "-{b}/2", "minValue": "-{b}/2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{b}/2", "minValue": "{b}/2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "

Find two positive numbers whose product is $\\simplify{{c}^2}$ and whose sum is minimum.

Smaller number: [[0]]. Larger number: [[1]].

(If the numbers are the same size still be sure to fill in both boxes.)

", "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "gaps": [{"allowFractions": false, "variableReplacements": [], "maxValue": "{c}", "minValue": "{c}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}, {"allowFractions": false, "variableReplacements": [], "maxValue": "{c}", "minValue": "{c}", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "showCorrectAnswer": true, "scripts": {}, "marks": 0, "type": "gapfill"}, {"prompt": "

The sum of two positive numbers is {d}. What is the smallest possible value for the sum of their squares?

", "allowFractions": false, "variableReplacements": [], "maxValue": "{d}^2/2", "minValue": "{d}^2/2", "variableReplacementStrategy": "originalfirst", "correctAnswerFraction": false, "showCorrectAnswer": true, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "", "variable_groups": [], "variablesTest": {"maxRuns": 100, "condition": ""}, "variables": {"c": {"definition": "random(5..12)", "templateType": "anything", "group": "Ungrouped variables", "name": "c", "description": ""}, "b": {"definition": "2*random(10..30)", "templateType": "anything", "group": "Ungrouped variables", "name": "b", "description": ""}, "d": {"definition": "2*random(3..6)", "templateType": "anything", "group": "Ungrouped variables", "name": "d", "description": ""}}, "metadata": {"description": "

Optimisation using calculus

rebel

rebelmaths

Rebel

", "licence": "None specified"}, "type": "question", "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Maria's copy of Maclaurin series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

Find the first 3 terms in the MacLaurin series for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

You are asked to find the first 3 terms in the MacLaurin series for $f(x)=(\\simplify[all]{{a}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$ and $x$ is near $0$.

", "advice": "

The first three terms in the MacLaurin series are given by $a+bx+cx^2$ where $\\displaystyle a=f(0),\\;\\;b=f'(0),\\;\\;c=\\frac{f''(0)}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(0)=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(0)=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(0)}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the MacLaurin series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2} \\]

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First 3 terms = ?

Input the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.

Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.

", "answer": "{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2", "answerSimplification": "all,fractionNumbers,!collectNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 1e-06, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "valuegenerators": [{"name": "x", "value": ""}]}], "type": "question"}, {"name": "Maria's copy of Taylor series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

Find the first 3 terms in the Taylor series at $x=c$ for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

You are asked to find the first 3 terms in the Taylor series at $x=\\var{c}$ for $f(x)=(\\simplify[all]{{a-b*c}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$.

", "advice": "

The first three terms in the Taylor series are given by $\\simplify[all]{a+b(x-{c})+c(x-{c})^2}$ where $\\displaystyle a=f(\\var{c}),\\;\\;b=f'(\\var{c}),\\;\\;c=\\frac{f''(\\var{c})}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a-b*c}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a-b*c}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(\\var{c})=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(\\var{c})=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(\\var{c})}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the Taylor series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2} \\]

", "rulesets": {}, "variables": {"s1": {"name": "s1", "group": "Ungrouped variables", "definition": "random(1,-1)", "description": "", "templateType": "anything"}, "a": {"name": "a", "group": "Ungrouped variables", "definition": "random(1,4,8,9,16,27,32,25,36,49)", "description": "", "templateType": "anything"}, "tm1": {"name": "tm1", "group": "Ungrouped variables", "definition": "tm0*b", "description": "", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "s1*switch(a=1,random(2..9),a=4,random(3,5,7,9),a=8,random(1,3,5,7,9),a=9,random(1,2,4,5,7,8),a=16,random(1,3,5,7,9),a=32,random(1,3,5,7,9),a=25,random(1,2,4,6,7,9),a=27,random(1,2,4,5,7,8),a=36,random(1,5,7,9),random(1,2,3,4,5,8,9))", "description": "", "templateType": "anything"}, "tm0": {"name": "tm0", "group": "Ungrouped variables", "definition": "a^(1/n)", "description": "", "templateType": "anything"}, "tm2": {"name": "tm2", "group": "Ungrouped variables", "definition": "-(n-1)*tm1*b", "description": "", "templateType": "anything"}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "templateType": "anything"}, "n": {"name": "n", "group": "Ungrouped variables", "definition": "if(a=4 or a=9 or a=25 or a=36 or a=49,2,if(a=8 or a=27,3,if(a=32,5,if(a=16,random(2,4),random(2..5)))))", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "tm0", "tm2", "b", "tm1", "s1", "c", "n"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "jme", "useCustomName": false, "customName": "", "marks": 4, "showCorrectAnswer": true, "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "adaptiveMarkingPenalty": 0, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "prompt": "

Input the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.

Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.

", "answer": "{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2", "answerSimplification": "all,fractionNumbers,!collectNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 1e-06, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "notallowed": {"strings": ["!", "."], "showStrings": false, "partialCredit": 0, "message": "

Do not input factorials or decimals in the Taylor series.

"}, "valuegenerators": [{"name": "x", "value": ""}]}], "type": "question"}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form 0/0", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "functions": {}, "parts": [{"scripts": {}, "type": "gapfill", "gaps": [{"checkingtype": "absdiff", "type": "jme", "showCorrectAnswer": false, "vsetrangepoints": 5, "showFeedbackIcon": true, "expectedvariablenames": [], "variableReplacements": [], "answer": "{{choice2}[-2]}", "variableReplacementStrategy": "originalfirst", "scripts": {}, "marks": 1, "showpreview": true, "checkvariablenames": false, "checkingaccuracy": "0.0000000001", "vsetrange": [0, 1]}], "showCorrectAnswer": true, "prompt": "

As {choice2[0]}, {choice2[2]} approaches [[0]]

Note: infinity is simply entered by typing infinity

", "showFeedbackIcon": true, "marks": 0, "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}], "tags": [], "advice": "

As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)=0$, $\\lim_{x\\rightarrow a}g(x)=0$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:

\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]

(If this is still an indeterminate form we can (of course) repeat the procedure)

So by applying L'Hospital's rule we now need to determine what {choice2[3]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[4]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is {choice2[-1]}.

", "statement": "

This question is about limits of indeterminate forms.

", "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Just what the title says, I guess.

"}, "variables": {"pc": {"name": "pc", "definition": "pl[2]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "na": {"name": "na", "definition": "nl[0]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "nb": {"name": "nb", "definition": "nl[1]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "pe": {"name": "pe", "definition": "pl[4]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "poly_sin": {"name": "poly_sin", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{((x-{na})*(x-{nb}))/({nc}sin(x-{na}))}\\$','\\$\\\\simplify{(2x-{na+nb})/({nc}cos(x-{na}))}\\$',(na-nb)/nc,'\\$\\\\simplify[fractionNumbers]{{(na-nb)/nc}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{na+nb}x+{na*nb})/({nc}sin(x-{na}))}\\$','\\$\\\\simplify{(2x-{na+nb})/({nc}cos(x-{na}))}\\$',(na-nb)/nc,'\\$\\\\simplify[fractionNumbers]{{(na-nb)/nc}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{na})^2/({nc}sin(x-{na}))}\\$','\\$\\\\simplify{(2(x-{na}))/({nc}cos(x-{na}))}\\$',0,'\\$0\\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{nb}x)/({nc}sin(x))}\\$','\\$\\\\simplify{(2x-{nb})/({nc}cos(x))}\\$',-nb/nc,'\\$\\\\simplify[fractionNumbers]{{-nb/nc}}\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{x^2/({nc}sin(x))}\\$','\\$\\\\simplify{(2x)/({nc}cos(x))}\\$',0,'\\$ 0\\$'],\n \n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{na})^{pa}/({nc}sin(x-{na}))}\\$','\\$\\\\simplify{({pa}(x-{na})^{pa-1})/({nc}cos(x-{na}))}\\$',0,'\\$ 0\\$']\n\n]", "templateType": "anything", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

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{{-2*nb/gcd(-2*nb,pa)}/({pa/gcd(-2*nb,pa)}*pi)}

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

", "group": "Ungrouped variables"}, "pa": {"name": "pa", "definition": "pl[0]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "poly_cos": {"name": "poly_cos", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{((x-{na})*(x-{nb}))/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{(2x-{na+nb})/({-nc}sin(x-{na}))}\\$',(na-nb)/(-nc)*infinity,'\\$\\\\simplify{{(na-nb)/(-nc)*infinity}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{((x-{na})*(x-{nb}))/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{(2x-{na+nb})/({-nc}sin(x-{na}))}\\$',(na-nb)/(nc)*infinity,'\\$\\\\simplify{{(na-nb)/(nc)*infinity}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{na+nb}x+{na*nb})/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{(2x-{na+nb})/({-nc}sin(x-{na}))}\\$',(na-nb)/(-nc)*infinity,'\\$\\\\simplify{{(na-nb)/(-nc)*infinity}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{na+nb}x+{na*nb})/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{(2x-{na+nb})/({-nc}sin(x-{na}))}\\$',(na-nb)/(nc)*infinity,'\\$\\\\simplify{{(na-nb)/(nc)*infinity}}\\$'], \n \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{na})^2/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{(2(x-{na}))/({-nc}sin(x-{na}))}\\$','\\$\\\\simplify{(2)/({-nc}cos(x-{na}))}\\$',-2/nc,'\\$\\\\simplify[fractionNumbers]{{-2/nc}}\\$'],\n\n['\\$x\\\\rightarrow 0^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{nb}x)/({nc}cos(x)-{nc})}\\$','\\$\\\\simplify{(2x-{nb})/({-nc}sin(x))}\\$',(nb/nc)*infinity,'\\$\\\\simplify{{(nb/nc)*infinity}}\\$'], \n['\\$x\\\\rightarrow 0^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2-{nb}x)/({nc}cos(x)-{nc})}\\$','\\$\\\\simplify{(2x-{nb})/({-nc}sin(x))}\\$',(-nb/nc)*infinity,'\\$\\\\simplify{{(-nb/nc)*infinity}}\\$'], \n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x^2)/({nc}cos(x)-{nc})}\\$','\\$\\\\simplify{(2x)/({-nc}sin(x))}\\$','\\$\\\\simplify{(2)/({-nc}cos(x))}\\$',(-2/nc),'\\$\\\\simplify[fractionNumbers]{{-2/nc}}\\$'], \n \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{(x-{na})^{pa}/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{({pa}(x-{na})^{pa-1})/({-nc}sin(x-{na}))}\\$','\\$\\\\simplify[unitPower,zeroPower]{({pa*(pa-1)}(x-{na})^{pa-2})/({-nc}cos(x-{na}))}\\$',if(pa=2,-2/nc,0),if(pa=2,'\\$\\\\simplify[fractionNumbers]{{-2/nc}}\\$','\\$ 0\\$')]\n\n]", "templateType": "anything", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

", "group": "Ungrouped variables"}, "choice1": {"name": "choice1", "definition": "random(sin_poly,cos_poly,poly_sin,poly_cos,trig_trig,log_sin,log_cos,log_tan,sin_log,cos_log,tan_log)", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "pL": {"name": "pL", "definition": "shuffle(2..12)[0..5]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "ne": {"name": "ne", "definition": "nl[4]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "sin_poly": {"name": "sin_poly", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/((x-{na})*(x-{nb}))}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2x-{na+nb})}\\$',nc/(na-nb),'\\$\\\\simplify[fractionNumbers]{{nc/(na-nb)}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x^2-{na+nb}x+{na*nb})}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2x-{na+nb})}\\$',nc/(na-nb),'\\$\\\\simplify[fractionNumbers]{{nc/(na-nb)}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^2}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2(x-{na}))}\\$',0,'\\$0\\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/(x^2-{nb}x)}\\$','\\$\\\\simplify{({nc}cos(x))/(2x-{nb})}\\$',-nc/nb,'\\$\\\\simplify[fractionNumbers]{{-nc/nb}}\\$'], \n['\\$x\\\\rightarrow 0^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/x^2}\\$','\\$\\\\simplify{({nc}cos(x))/(2x)}\\$',nc*infinity,'\\$\\\\simplify{{nc*infinity}}\\$'],\n['\\$x\\\\rightarrow 0^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/x^2}\\$','\\$\\\\simplify{({nc}cos(x))/(2x)}\\$',-nc*infinity,'\\$\\\\simplify{-{nc*infinity}}\\$'],\n \n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^{pa}}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/({pa}(x-{na})^{pa-1})}\\$',nc*infinity,'\\$\\\\simplify{{nc*infinity}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^{pa}}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/({pa}(x-{na})^{pa-1})}\\$',if(mod(pa,2)=0,-nc*infinity,nc*infinity),if(mod(pa,2)=0,'\\$\\\\simplify{{-nc*infinity}}\\$','\\$\\\\simplify{{nc*infinity}}\\$')] \n\n]", "templateType": "anything", "description": "

using a list to keep track of important things

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using a list to keep track of important things

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using a list to keep track of important things

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using a list to keep track of important things

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using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

", "group": "Ungrouped variables"}, "nc": {"name": "nc", "definition": "nl[2]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "nL": {"name": "nL", "definition": "shuffle(-12..12 except 0)[0..5]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "alt1", "sin_poly", "cos_poly", "poly_sin", "poly_cos", "trig_trig", "log_sin", "log_cos", "log_tan", "sin_log", "cos_log", "tan_log", "choice1", "choice2"], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question"}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form infinity/infinity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "advice": "

As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)=\\infty$, $\\lim_{x\\rightarrow a}g(x)=\\infty$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:

\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]

(If this is still an indeterminate form we can (of course) repeat the procedure)

", "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "log_power", "power_log", "exp_power", "power_exp", "exp_log", "log_exp", "log_root", "root_log", "choice1", "choice2"], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "type": "numberentry", "minValue": "choice2[-1]", "allowFractions": false, "mustBeReducedPC": 0, "scripts": {}, "mustBeReduced": false, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "choice2[-1]", "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "marks": 1, "correctAnswerStyle": "plain"}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "marks": 0, "prompt": "

As {choice2[0]}, {choice2[2]} approaches [[0]]

Note: infinity is simply entered by typing infinity

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using a list to keep track of important things

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using a list to keep track of important things

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"}, "log_exp": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}ln({pb}x+{nb}))/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*pb}/({pb}x+{nb}))/({pc}*{pa}*e^({pa}x+{nb}))={pd*pb}/({pc}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))}\\$',0], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}ln({pb}x+{nb}))/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({-pd*pb}/({pb}x+{nb}))/({pc}*{pa}*e^({pa}x+{nb}))={-pd*pb}/({pc}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))}\\$',0], \n\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({pd}ln({-pb}x+{nb}))/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*-pb}/({-pb}x+{nb}))/({pc}*{-pa}*e^({-pa}x+{nb}))={pd*-pb}/({pc}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))}\\$',0],\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({-pd}ln({-pb}x+{nb}))/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*pb}/({-pb}x+{nb}))/({pc}*{-pa}*e^({-pa}x+{nb}))={pd*pb}/({pc}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))}\\$',0]\n]", "name": "log_exp", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

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"}, "pc": {"templateType": "anything", "definition": "pl[2]", "name": "pc", "group": "Ungrouped variables", "description": ""}, "choice2": {"templateType": "anything", "definition": "switch(choice1=0,random(log_power),choice1=1,random(power_log),choice1=2,random(exp_power),choice1=3,random(power_exp),choice1=4,random(exp_log),choice1=5,random(log_exp),choice1=6,random(log_root),choice1=7,random(root_log),'')", "name": "choice2", "group": "Ungrouped variables", "description": ""}, "nc": {"templateType": "anything", "definition": "nl[2]", "name": "nc", "group": "Ungrouped variables", "description": ""}, "pb": {"templateType": "anything", "definition": "pl[1]", "name": "pb", "group": "Ungrouped variables", "description": ""}, "power_log": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({pd*pb}x^{pb-1}))/({pc*pa})}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({-pd*pb}x^{pb-1}))/({pc*pa})}\\$',-infinity], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({-pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,-infinity,infinity)],\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,infinity,-infinity)]\n\n\n]\n", "name": "power_log", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}, "nb": {"templateType": "anything", "definition": "nl[1]", "name": "nb", "group": "Ungrouped variables", "description": ""}, "pL": {"templateType": "anything", "definition": "shuffle(1..12)[0..5]", "name": "pL", "group": "Ungrouped variables", "description": ""}, "na": {"templateType": "anything", "definition": "nl[0]", "name": "na", "group": "Ungrouped variables", "description": ""}, "exp_log": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}e^({pa}x+{nb})+{nc})/({pc}ln({pb}x+{nb}))}\\$','\\$\\\\simplify{({pd}*{pa}*e^({pa}x+{nb}))/({pc*pb}/({pb}x+{nb}))=({pd}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))/{pc*pb}}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}e^({pa}x+{nb})+{nc})/({pc}ln({pb}x+{nb}))}\\$','\\$\\\\simplify{({-pd}*{pa}*e^({pa}x+{nb}))/({pc*pb}/({pb}x+{nb}))=({-pd}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))/{pc*pb}}\\$',-infinity], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({pd}e^({-pa}x+{nb})+{nc})/({pc}ln({-pb}x+{nb}))}\\$','\\$\\\\simplify{({pd}*{-pa}*e^({-pa}x+{nb}))/({pc*-pb}/({-pb}x+{nb}))=({pd}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))/{pc*-pb}}\\$',infinity],\n ['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({-pd}e^({-pa}x+{nb})+{nc})/({pc}ln({-pb}x+{nb}))}\\$','\\$\\\\simplify{({-pd}*{-pa}*e^({-pa}x+{nb}))/({pc*-pb}/({-pb}x+{nb}))=({pd}*{pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))/{pc*-pb}}\\$',-infinity]\n]", "name": "exp_log", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}, "pe": {"templateType": "anything", "definition": "pl[4]", "name": "pe", "group": "Ungrouped variables", "description": ""}, "nL": {"templateType": "anything", "definition": "shuffle(-12..12 except 0)[0..5]", "name": "nL", "group": "Ungrouped variables", "description": ""}, "choice1": {"templateType": "anything", "definition": "random(0..7)", "name": "choice1", "group": "Ungrouped variables", "description": ""}, "root_log": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({pa}x+{nb},{pb}))/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})^(1/{pb}))/({pc*pb})}\\$',infinity], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({-pa}x+{nb},{pb}))/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})^(1/{pb}))/({pc*pb})}\\$',infinity], \n \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({pa}x+{nb},{pb}))/({-pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))/({-pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})^(1/{pb}))/({-pc*pb})}\\$',-infinity], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({-pa}x+{nb},{pb}))/({-pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))/({pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})^(1/{pb}))/({-pc*pb})}\\$',-infinity]\n\n]", "name": "root_log", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}, "nd": {"templateType": "anything", "definition": "nl[3]", "name": "nd", "group": "Ungrouped variables", "description": ""}, "log_root": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}ln({pa}x+{nb}))/(root({pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({pc*pa}/({pa}x+{nb}))/({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))=({pc*pb})/(({pa}x+{nb})^(1/{pb}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}ln({-pa}x+{nb}))/(root({-pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({-pc*pa}/({-pa}x+{nb}))/({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))=({pc*pb})/(({-pa}x+{nb})^(1/{pb}))}\\$',0], \n \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}ln({pa}x+{nb}))/(root({pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({-pc*pa}/({pa}x+{nb}))/({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))=({-pc*pb})/(({pa}x+{nb})^(1/{pb}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}ln({-pa}x+{nb}))/(root({-pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({pc*pa}/({-pa}x+{nb}))/({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))=({-pc*pb})/(({-pa}x+{nb})^(1/{pb}))}\\$',0]\n\n]", "name": "log_root", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}, "pd": {"templateType": "anything", "definition": "pl[3]", "name": "pd", "group": "Ungrouped variables", "description": ""}, "ne": {"templateType": "anything", "definition": "nl[4]", "name": "ne", "group": "Ungrouped variables", "description": ""}, "pa": {"templateType": "anything", "definition": "pl[0]", "name": "pa", "group": "Ungrouped variables", "description": ""}, "log_power": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}ln({pa}x+{nb}))/({pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({pc*pa}/({pa}x+{nb}))/({pd*pb}x^{pb-1})=({pc*pa})/(({pa}x+{nb})*({pd*pb}x^{pb-1}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{-\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pc}ln({-pa}x+{nb}))/({-pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({-pc*pa}/({-pa}x+{nb}))/({-pd*pb}x^{pb-1})=({-pc*pa})/(({-pa}x+{nb})*({-pd*pb}x^{pb-1}))}\\$',0]\n, \n ['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}ln({pa}x+{nb}))/({pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({-pc*pa}/({pa}x+{nb}))/({pd*pb}x^{pb-1})=({-pc*pa})/(({pa}x+{nb})*({pd*pb}x^{pb-1}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({-pc}ln({-pa}x+{nb}))/({-pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({pc*pa}/({-pa}x+{nb}))/({-pd*pb}x^{pb-1})=({pc*pa})/(({-pa}x+{nb})*({-pd*pb}x^{pb-1}))}\\$',0]\n]", "name": "log_power", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Just what the title says, I guess.

"}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "statement": "

This question is about limits of indeterminate forms.

", "type": "question"}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form 0*infinity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "advice": "

As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. We rewrite this product as a quotient {choice2[4]} so that it is an indeterminate form of type {choice2[3]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)$ and $\\lim_{x\\rightarrow a}g(x)$ both equal $\\infty$ or both equal $0$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:

\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]

(If this is still an indeterminate form we can (of course) repeat the procedure)

So by applying L'Hospital's rule we now need to determine what {choice2[5]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[6]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is {choice2[-1]}.

", "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "alt1", "power_exp", "log_power", "exp_power", "log_sin", "log_tan", "tan_linear", "choice1", "choice2"], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"expectedvariablenames": [], "checkingtype": "absdiff", "vsetrangepoints": 5, "vsetrange": [0, 1], "scripts": {}, "answer": "{{choice2}[-2]}", "variableReplacementStrategy": "originalfirst", "showpreview": true, "type": "jme", "showCorrectAnswer": false, "checkingaccuracy": "0.0000000001", "showFeedbackIcon": true, "checkvariablenames": false, "variableReplacements": [], "marks": 1}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "marks": 0, "prompt": "

As {choice2[0]}, {choice2[2]} approaches [[0]]

Note: infinity is simply entered by typing infinity

"}], "variables": {"na": {"templateType": "anything", "definition": "nl[0]", "name": "na", "group": "Ungrouped variables", "description": ""}, "log_sin": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-{na}))(sin(x-{na}))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-{na}))/(1/(sin(x-{na})))}\\$','\\$\\\\simplify{(({pa})/((x-{na}))/(-cos(x-{na})/sin(x-{na})^2))=(-{pa}sin(x-{na})^2)/((x-{na})cos(x-{na}))}\\$','\\$\\\\simplify{(-{2*pa}sin(x-{na})cos(x-{na}))/(-(x-{na})sin(x-{na})+cos(x-{na}))}\\$',0,'\\$0\\$'], \n['\\$x\\\\rightarrow 0^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-0))(sin(x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-0))/(1/(sin(x-0)))}\\$','\\$\\\\simplify{(({pa})/((x-0))/(-cos(x-0)/sin(x-0)^2))=(-{pa}sin(x-0)^2)/((x-0)cos(x-0))}\\$','\\$\\\\simplify{(-{2*pa}sin(x-0)cos(x-0))/(-(x-0)sin(x-0)+cos(x-0))}\\$',0,'\\$0\\$'],\n['\\$x\\\\rightarrow 0^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-0))(sin(pi*x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-0))/(1/(sin(pi*x-0)))}\\$','\\$\\\\simplify{(({pa})/((x-0))/(-pi*cos(pi*x-0)/sin(pi*x-0)^2))=(-{pa}sin(pi*x-0)^2)/(pi*(x-0)cos(pi*x-0))}\\$','\\$\\\\simplify{(-{2*pa}*pi*sin(pi*x-0)cos(pi*x-0))/(-pi^2*(x-0)sin(pi*x-0)+pi*cos(pi*x-0))}\\$',0,'\\$0\\$'],\n['\\$x\\\\rightarrow 1^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-1))(sin(pi*x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-1))/(1/(sin(pi*x-0)))}\\$','\\$\\\\simplify{(({pa})/((x-1))/(-pi*cos(pi*x-0)/sin(pi*x-0)^2))=(-{pa}sin(pi*x-0)^2)/(pi*(x-1)cos(pi*x-0))}\\$','\\$\\\\simplify{(-{2*pa}*pi*sin(pi*x-0)cos(pi*x-0))/(-pi^2*(x-1)sin(pi*x-0)+pi*cos(pi*x-0))}\\$',0,'\\$0\\$']\n \n]", "name": "log_sin", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

"}, "log_tan": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-{na}))(tan(x-{na}))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-{na}))/(cot(x-{na}))}\\$','\\$\\\\simplify{(({pa})/((x-{na})))/(-cosec(x-{na})^2)=(-{pa}sin(x-{na})^2)/((x-{na}))}\\$','\\$\\\\simplify{(-{2*pa}sin(x-{na})cos(x-{na}))}\\$',0,'\\$0\\$'], \n['\\$x\\\\rightarrow 0^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-0))(tan(x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-0))/(cot(x-0))}\\$','\\$\\\\simplify{(({pa})/((x-0))/(-cosec(x-0)^2))=(-{pa}sin(x-0)^2)/(x-0)}\\$','\\$\\\\simplify{(-{2*pa}sin(x-0)cos(x-0))/1}\\$',0,'\\$0\\$'],\n['\\$x\\\\rightarrow 0^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-0))(tan(pi*x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-0))/(cot(pi*x-0))}\\$','\\$\\\\simplify{(({pa})/((x-0))/(-pi*cosec(pi*x-0)^2))=(-{pa}sin(pi*x-0)^2)/(pi*(x-0))}\\$','\\$\\\\simplify{(-{2*pa}*pi*sin(pi*x-0)cos(pi*x-0))/(pi)}\\$',0,'\\$0\\$'],\n['\\$x\\\\rightarrow 1^+\\$','\\$-\\\\infty\\\\cdot 0\\$','\\$\\\\simplify{({pa}ln(x-1))(tan(pi*x-0))}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pa}ln(x-1))/(cot(pi*x-0))}\\$','\\$\\\\simplify{(({pa})/((x-1))/(-pi*cosec(pi*x-0)^2))=(-{pa}sin(pi*x-0)^2)/(pi*(x-1))}\\$','\\$\\\\simplify{(-{2*pa}*pi*sin(pi*x-0)cos(pi*x-0))/(pi)}\\$',0,'\\$0\\$']\n \n]", "name": "log_tan", "group": "Ungrouped variables", "description": "

using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.

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using a list to keep track of important things

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using a list to keep track of important things

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using a list to keep track of important things

what x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches, latex display of what y approaches.

"}, "pL": {"templateType": "anything", "definition": "shuffle(2..12)[0..5]", "name": "pL", "group": "Ungrouped variables", "description": ""}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "

Just what the title says, I guess.

"}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "statement": "

This question is about limits of indeterminate forms.

", "type": "question"}, {"name": "Maria's copy of W1 Copy of Displacement-time to velocity time graphs", "extensions": ["jsxgraph"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Stuart Turner", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/997/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "parts": [{"showCorrectAnswer": true, "prompt": "

The graph below is the displacement-time graph of a body.

{plot(q[0],1,{a[0]},{b[0]},{c[0]},{d[0]})}

Select the graph that shows the corresponding velocity-time graph.

", "displayType": "checkbox", "maxAnswers": "1", "shuffleChoices": true, "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "matrix": ["1", 0, 0, 0], "minAnswers": "1", "distractors": ["", "", "", ""], "displayColumns": 0, "variableReplacements": [], "maxMarks": "1", "type": "m_n_2", "choices": ["

{plot(q[0],2,{a[0]},{b[0]},{c[0]},{d[0]})}

", "

{plot(q[0],3,{a[0]},{b[0]},{c[0]},{d[0]})}

", "

{plot(q[0],4,{a[0]},{b[0]},{c[0]},{d[0]})}

", "

{plot(q[0],5,{a[0]},{b[0]},{c[0]},{d[0]})}

"], "warningType": "none", "marks": 0, "minMarks": "0", "scripts": {}}, {"showCorrectAnswer": true, "prompt": "

The graph below is the displacement-time graph of a body.

{plot(q[1],1,{a[1]},{b[1]},{c[1]},{d[1]})}

Select the graph that shows the corresponding velocity-time graph.

{plot(q[1],2,{a[1]},{b[1]},{c[1]},{d[1]})}

", "

{plot(q[1],3,{a[1]},{b[1]},{c[1]},{d[1]})}

", "

{plot(q[1],4,{a[1]},{b[1]},{c[1]},{d[1]})}

", "

{plot(q[1],5,{a[1]},{b[1]},{c[1]},{d[1]})}

"], "warningType": "none", "marks": 0, "minMarks": 0, "scripts": {}}, {"showCorrectAnswer": true, "prompt": "

The graph below is the displacement-time graph of a body.

{plot(q[2],1,{a[2]},{b[2]},{c[2]},{d[2]})}

Select the graph that shows the corresponding velocity-time graph.

{plot(q[2],2,{a[2]},{b[2]},{c[2]},{d[2]})}

", "

{plot(q[2],3,{a[2]},{b[2]},{c[2]},{d[2]})}Choice 2

", "

{plot(q[2],4,{a[2]},{b[2]},{c[2]},{d[2]})}

", "

{plot(q[2],5,{a[2]},{b[2]},{c[2]},{d[2]})}

"], "warningType": "none", "marks": 0, "minMarks": 0, "scripts": {}}, {"showCorrectAnswer": true, "prompt": "

The graph below is the displacement-time graph of a body.

{plot(q[3],1,{a[3]},{b[3]},{c[3]},{d[3]})}

Select the graph that shows the corresponding velocity-time graph.

{plot(q[3],2,{a[3]},{b[3]},{c[3]},{d[3]})}

", "

{plot(q[3],3,{a[3]},{b[3]},{c[3]},{d[3]})}

", "

{plot(q[3],4,{a[3]},{b[3]},{c[3]},{d[3]})}

", "

{plot(q[3],5,{a[3]},{b[3]},{c[3]},{d[3]})}

"], "warningType": "none", "marks": 0, "minMarks": 0, "scripts": {}}], "advice": "", "ungrouped_variables": ["a", "b", "c", "d", "q"], "variables": {"d": {"description": "", "group": "Ungrouped variables", "name": "d", "definition": "[b[0]+random(-2..2 except 0),\n b[1]+random(-2..2 except 0),\n b[2]+random(-2..2 except 0),\n b[3]+random(-2..2 except 0)]", "templateType": "anything"}, "c": {"description": "", "group": "Ungrouped variables", "name": "c", "definition": "shuffle([-2,-1,1,2])", "templateType": "anything"}, "q": {"description": "", "group": "Ungrouped variables", "name": "q", "definition": "[1,3,random(2..3),2]", "templateType": "anything"}, "a": {"description": "", "group": "Ungrouped variables", "name": "a", "definition": "shuffle([-2,-1,1,2])", "templateType": "anything"}, "b": {"description": "", "group": "Ungrouped variables", "name": "b", "definition": "shuffle([-2,-1,1,2])", "templateType": "anything"}}, "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Displacement-time graphs are given and the student should select the correct velocity-time graphs from a list. Includes linear, piecewise linear and quadratic displacement-time functions.

"}, "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "", "functions": {"plot": {"parameters": [["q", "number"], ["n", "number"], ["a", "number"], ["b", "number"], ["c", "number"], ["d", "number"]], "language": "javascript", "definition": "//The variable q randomises which style of graph you get: linear, quadratic or piecewise linear\n//The variable n controls whether you are plotting the question, the correct answer or one of three incorrect answers.\n//a,b,c and d are variables that vary parameters in the graphs, e.g. gradient, etc.\n\n\n\n//Maximum and minimum of graph goes here\n\nvar x_min = -5;\nvar x_max = 5;\nvar y_min = -6;\nvar y_max = 6;\n\n\n// Make the JSXGraph board.\nvar div = Numbas.extensions.jsxgraph.makeBoard('400px','400px',\n {boundingBox: [x_min,y_max,x_max,y_min], \n axis: false,\n showNavigation: false,\n grid: true\n});\n\n\n// div.board is the object created by JSXGraph, which you use to \n// manipulate elements\nvar board = div.board; \n\n// create the x-axis and y-axis\nvar xaxis = board.create('axis',[[0,0],[1,0]],\n {name:'t', \n\t\t\t withLabel: true, \n\t\t\t label: {position: 'rt', \n\t\t\t\t\t offset: [-15, 10] // (in pixels)\n\t\t\t\t\t }\n });\n\n// create the y-axis\nvar yaxis = board.create('axis',[[0,0],[0,1]]);\n\n\n\n \n// FUNCTIONS DEFINED HERE\n\n\nswitch(q) {\n case 1:\n var f1 = function(x) {return b*(x-a)+c;}\n var f2 = function(x) {return b;}\n var f3 = function(x) {return -b;}\n var f4 = function(x) {return b+1;}\n var f5 = function(x) {return b*(x-a)+c;}\n break;\n \n case 2:\n var f1 = function(x) {return a*(x+b)*(x+d);}\n var f2 = function(x) {return a*(2*x+b+d);}\n var f3 = function(x) {return -a*(2*x+b+d);}\n var f4 = function(x) {return a*(2*(x-1)+b+d);}\n var f5 = function(x) {return -a*(2*(x-1)+b+d);}\n break;\n \n case 3:\n var f1 = function(x) {\n if ((x-a)<0)\n return b*(x-a)+c;\n else\n return d*(x-a)+c;\n }\n var f2 = function(x) {\n if ((x-a)<0)\n return b;\n else\n return d;\n }\n var f3 = function(x) {\n if ((x-a-1)<0)\n return b;\n else\n return d;\n }\n var f4 = function(x) {\n if ((x-a)<0)\n return d;\n else\n return b;\n }\n var f5 = function(x) {\n if ((x-a)<0)\n return b-1;\n else\n return d-1;\n }\n break;\n \n} \n \n \n \n\n\n// PLOT THE FUNCTION\n\nswitch(n) {\n case 1:\n board.create('functiongraph', [f1], {strokeWidth:3});\n break;\n \n case 2:\n board.create('functiongraph', [f2], {strokeWidth:3});\n break;\n \n case 3:\n board.create('functiongraph', [f3], {strokeWidth:3});\n break;\n \n case 4:\n board.create('functiongraph', [f4], {strokeWidth:3});\n break;\n \n case 5:\n board.create('functiongraph', [f5], {strokeWidth:3});\n break;\n}\n\n\nreturn div;", "type": "html"}}, "preamble": {"js": "", "css": ""}, "tags": [], "variable_groups": [], "rulesets": {}, "type": "question"}, {"name": "Maria's copy of Volume of revolution 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": ["Calculus", "calculus", "definite integration", "diagram", "integral", "integration", "rotation about an axis", "rotation about y-axis", "volume integral", "volume of revolution"], "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "variables": {"m": {"name": "m", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..8)"}, "d1": {"name": "d1", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "if(td1<=c1,c1+1,td1)"}, "vol": {"name": "vol", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "precround(tvol,2)"}, "tol": {"name": "tol", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "0"}, "tvol": {"name": "tvol", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "((a1)/(2*m^2))*(exp(2*d1/a1)-exp(2*c1/a1))"}, "td1": {"name": "td1", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..5)"}, "c1": {"name": "c1", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..5)"}, "a1": {"name": "a1", "description": "", "templateType": "anything", "group": "Ungrouped variables", "definition": "random(2..8)"}}, "variablesTest": {"maxRuns": 100, "condition": ""}, "functions": {}, "ungrouped_variables": ["vol", "m", "a1", "tol", "td1", "tvol", "c1", "d1"], "type": "question", "metadata": {"description": "

Rotate the graph of $y=a\\ln(bx)$ by $2\\pi$ radians about the $y$-axis between $y=c$ and $y=d$. Find the volume of revolution.

", "licence": "Creative Commons Attribution 4.0 International", "notes": "\n \t\t

3/07/2012:

\n \t\t

Added tags.

\n \t\t

Improved display in statement and prompts, and Advice.

\n \t\t

No tolerance allowed in second part answer. Set new tolerance variable tol=0.

\n \t\t

Checked calculation.

\n \t\t

20/07/2012:

\n \t\t

Added description.

\n \t\t

Changed question format to hopefully make it clear that it is the multiple of $\\pi$ wanted.

\n \t\t

Checked calculation again.

\n \t\t

Should have a diagram of the volume - or a schematic version of revolving a function about the $y$-axis.

\n \t\t

25/07/2012:

\n \t\t

Added tags.

\n \t\t

In the Advice section moved \\Rightarrow so that it is at the beginning of the line instead of the end of the previous line.

\n \t\t

Question appears to be working correctly.

\n \t\t

\n \t\t"}, "showQuestionGroupNames": false, "statement": "

Consider the function \\[y=f(x)=\\var{a1}\\ln(\\var{m}x)\\]

", "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "questions": [], "pickQuestions": 0}], "advice": "\n

a)
Given $y=\\var{a1}\\ln(\\var{m}x)$ we rearrange so that $x$ is the subject of the equation:
\\[\\begin{eqnarray*} y&=&\\var{a1}\\ln(\\var{m}x)\\\\ \\Rightarrow \\ln(\\var{m}x) &=&\\frac{y}{\\var{a1}}\\\\ \\Rightarrow \\var{m}x&=& e^{\\frac{y}{\\var{a1}}}\\\\ \\Rightarrow x&=&\\frac{1}{\\var{m}}e^{\\frac{y}{\\var{a1}}} \\end{eqnarray*} \\]
Hence \\[g(y)=\\frac{1}{\\var{m}}e^{\\frac{y}{\\var{a1}}}\\]

b)
The volume of revolution is given by:
\\[V=\\pi\\int_{\\var{c1}}^{\\var{d1}}g(y)^2\\;dy\\]
Using the expression for $g(y)$ from the first part we have:
\\[\\begin{eqnarray*} V&=&\\pi\\int_{\\var{c1}}^{\\var{d1}}\\frac{1}{\\var{m^2}}\\simplify[std]{e^(2y/{a1})}\\;dy\\\\ &=&\\simplify[std]{({a1}/{2*m^2})*pi}\\left[\\simplify[std]{e^(2y/{a1})}\\right]_{\\var{c1}}^{\\var{d1}}\\\\ \\\\&=&\\var{tvol}\\pi = \\var{vol}\\pi \\end{eqnarray*} \\]
Hence the multiple of $\\pi$ to two decimal places is $\\var{vol}$.

\n ", "variable_groups": [], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "marks": 0, "gaps": [{"showpreview": true, "answersimplification": "std", "expectedvariablenames": [], "checkingaccuracy": 0.001, "showCorrectAnswer": true, "answer": "(1 / {m}) * Exp(y / {a1})", "type": "jme", "scripts": {}, "vsetrangepoints": 5, "marks": 1, "checkingtype": "absdiff", "vsetrange": [0, 1], "checkvariablenames": false}], "scripts": {}, "prompt": "\n \n \n

Rewrite this function in the form $x=g(y)$ , where $g(y)$ is a function of $y$ only.

\n \n \n \n

$x=g(y)=\\;\\;$[[0]]

\n \n \n "}, {"type": "gapfill", "showCorrectAnswer": true, "marks": 0, "gaps": [{"minValue": "vol-tol", "type": "numberentry", "showCorrectAnswer": true, "showPrecisionHint": false, "maxValue": "vol+tol", "marks": 1, "correctAnswerFraction": false, "scripts": {}, "allowFractions": false}], "scripts": {}, "prompt": "\n

Hence, find the volume of revolution, $V$ obtained as follows:

Rotate $y=f(x)$ by $2\\pi$ radians about the $y$-axis, between the limits of $y=\\var{c1}$ and $y=\\var{d1}$.

The volume $V$ can be written as a multiple of $\\pi$, $V=m\\pi$, where:

$m=\\;\\;$[[0]]. Input $m$ to two decimal places.

\n "}], "preamble": {"css": "", "js": ""}}, {"name": "Maria's copy of Volume of revolution 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "ungrouped_variables": ["a", "c", "b", "tv", "v", "sb", "sa"], "variablesTest": {"maxRuns": 100, "condition": ""}, "preamble": {"css": "", "js": ""}, "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepsPenalty": 0, "steps": [{"marks": 0, "prompt": "

Recall that if $V$ is the volume generated between the limits $x=a$ and $x=b$ then $\\displaystyle V=\\pi\\int_a^by^2\\;dx$.

", "variableReplacements": [], "scripts": {}, "type": "information", "variableReplacementStrategy": "originalfirst", "showCorrectAnswer": true}], "prompt": "\n

Find the volume of this object.

$V=\\;\\;$[[0]]

Enter your answer to 3 decimal places.

Click on Show steps for information on volumes of revolution. You will not lose any marks.

\n ", "marks": 0, "scripts": {}, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "minValue": "V", "showPrecisionHint": false, "marks": 1, "scripts": {}, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "maxValue": "V", "showCorrectAnswer": true, "allowFractions": false}], "showCorrectAnswer": true}], "advice": "

Recall that if $V$ is the volume generated between the limits $x=a$ and $x=b$ by rotating the function about the $x$-axis then $\\displaystyle V=\\pi\\int_a^by^2\\;dx$.

So we have:
\\[\\begin{eqnarray*} V&=&\\pi\\int_{\\var{c}\\pi}^{\\var{c+2}\\pi}\\simplify[std]{{a^2}(cos(x)+{b})^2}\\;dx\\\\ &=&\\var{a^2}\\pi\\int_{\\var{c}\\pi}^{\\var{c+2}\\pi}\\simplify[std]{cos(x)^2+{2*b}*cos(x)+{b^2}}\\;dx\\\\ &=&\\var{a^2}\\pi\\left[\\simplify[std]{((1 / 4) Sin(2*x) + (1 / 2) * x + {2 * b} * Sin(x) + {b ^ 2} * x)}\\right]_{\\var{c}\\pi}^{\\var{c+2}\\pi}\\\\ \\end{eqnarray*}\\]
Here we have used the identity $\\cos(x)^2=\\frac{1}{2}(1+\\cos(2x))$ in order to integrate $\\cos(x)^2$.

Since $\\sin(n\\pi)=0$ for all integers $n$ we see that:
\\[\\begin{eqnarray*} V&=&\\var{a^2}\\pi\\frac{\\var{1+2b^2}}{\\var{2}}\\left(\\var{c+2}\\pi-\\var{c}\\pi\\right)\\\\ &=&\\var{a^2*(1+2b^2)}\\pi^2\\\\ &=&\\var{V}\\mbox{ to 3 decimal places} \\end{eqnarray*} \\]

", "variable_groups": [], "question_groups": [{"name": "", "pickQuestions": 0, "pickingStrategy": "all-ordered", "questions": []}], "type": "question", "functions": {}, "showQuestionGroupNames": false, "metadata": {"description": "

Rotate $y=a(\\cos(x)+b)$ by $2\\pi$ radians about the $x$-axis between $x=c\\pi$ and $x=(c+2)\\pi$. Find the volume of revolution.

", "licence": "Creative Commons Attribution 4.0 International", "notes": "\n \t\t

3/07/2012:

\n \t\t

Added tags.

\n \t\t

Checked calculations.

\n \t\t

Improved display of statement, prompt and Advice.

\n \t\t

Wanted to put the Hint into Show steps - but cannot create Steps at present.

\n \t\t

No tolerance allowed. Must be exact to three decimal places.

\n \t\t

Note the use of $\\cos(x)^2$ instead of the standard $\\cos^2(x)$ as best to be consistent as we cannot use $\\cos^2(x)$ if any jme calculation is involved.

\n \t\t

20/07/2012:

\n \t\t

Added description.

\n \t\t

Added Show steps hint.

\n \t\t

Checked description.

\n \t\t

Perhaps the tolerance should be 1, not 0.001 given the magnitude of the answer.

\n \t\t

25/07/2012:

\n \t\t

Added tags.

\n \t\t

Question appears to be working correctly.

\n \t\t

\n \t\t"}, "variables": {"a": {"description": "", "name": "a", "templateType": "anything", "definition": "sa*random(2..9)", "group": "Ungrouped variables"}, "sa": {"description": "", "name": "sa", "templateType": "anything", "definition": "random(1,-1)", "group": "Ungrouped variables"}, "c": {"description": "", "name": "c", "templateType": "anything", "definition": "random(2..9)", "group": "Ungrouped variables"}, "sb": {"description": "", "name": "sb", "templateType": "anything", "definition": "random(1,-1)", "group": "Ungrouped variables"}, "v": {"description": "", "name": "v", "templateType": "anything", "definition": "precround(tV,3)", "group": "Ungrouped variables"}, "tv": {"description": "", "name": "tv", "templateType": "anything", "definition": "pi^2*a^2*(1+2*b^2)", "group": "Ungrouped variables"}, "b": {"description": "", "name": "b", "templateType": "anything", "definition": "sb*random(1..9)", "group": "Ungrouped variables"}}, "tags": ["Calculus", "calculus", "definite integration", "diagram", "integral", "integration", "rotation about an axis", "rotation about x axis", "volume integral", "volume of revolution"], "statement": "

Consider the solid object that is obtained when the function: \\[y=\\simplify[std]{{a}(cos(x)+{b})}\\] is rotated by $2\\pi$ radians about the $x$-axis between the limits $x=\\var{c}\\pi$ and $x=\\var{c+2}\\pi$

"}, {"name": "Maria's copy of Maximum/minimum Cylinder", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Untitled2.jpg", "/srv/numbas/media/question-resources/Untitled2.jpg"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Frank Doheny", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/789/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

Problem on a closed cylindrical tank having minimum surface area

", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

A closed cylindrical tank is to be built having a volume of \$\\var{v}\$ cm³.

Determine the required height, \$h\$, and radius, \$r\$, if the total surface area is to be a minimum.

", "advice": "

\$\\pi r^2h=\\var{v}\$

\$h=\\frac{\\var{v}}{\\pi r^2}\$

The total surface area is to be a minimum.

Lid + curved surface area + base

\$A=\\pi r^2+2\\pi rh+\\pi r^2\$

\$A=2\\pi r^2+2\\pi r\\left(\\frac{\\var{v}}{\\pi r^2}\\right)\$

\$A=2\\pi r^2+\\simplify{2*{v}}r^{-1}\$

\$\\frac{dA}{dr}=4\\pi r-\\simplify{2{v}}r^{-2}=0\$

\$4\\pi r=\\simplify{2*{v}}/{r^2}\$

\$r^3=\\frac{\\var{v}}{2\\pi}\$

\$r=\\simplify{({v}/(2*pi))^(1/3)}\$

From the second line we have the relation \$h=\\frac{\\var{v}}{\\pi r^2}\$ to get

\$h=2*\\simplify{({v}/(2*pi))^(1/3)}\$

Input the cyinder height, correct to two decimal places.

\$h = \$ [[0]]

Input the required cylinder radius, correct to two decimal places.

\$r = \$ [[1]]

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Input the value for \$x\$ correct to one decimal place.

\$x = \$ [[0]]

", "scripts": {}, "variableReplacements": [], "type": "gapfill", "variableReplacementStrategy": "originalfirst", "gaps": [{"correctAnswerFraction": false, "precisionPartialCredit": 0, "scripts": {}, "precisionMessage": "You have not given your answer to the correct precision.", "variableReplacements": [], "allowFractions": false, "showCorrectAnswer": true, "precisionType": "dp", "minValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6", "showPrecisionHint": false, "precision": "1", "marks": 1, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "strictPrecision": false, "maxValue": "({w}+{l}-sqrt({w}^2+{l}^2-{w}*{l}))/6"}], "showCorrectAnswer": true}], "advice": "

The length of the box is \$\\var{l}-2x\$, the width is \$\\var{w}-2x\$ and the height is \$x\$.

The volume is then given by

\$V=(\\var{l}-2x).(\\var{w}-2x).x\$

\$V=\\simplify{4x^3-2*({w}+{l})x^2+{l}*{w}x}\$

\$\\frac{dV}{dx}=\\simplify{12x^2-4*({w}+{l})x+{l}*{w}}\$

This is a quadratic equation.

\$x=\\frac{\\simplify{4*({w}+{l})}\\pm\\sqrt(\\simplify{16*({w}+{l})^2-48*{w}*{l}})}{24}\$

\$x=\\frac{\\simplify{{w}+{l}}\\pm\\sqrt(\\simplify{{w}^2-{w}*{l}+{l}^2})}{6}\$

\$\\frac{d^2V}{dx^2}=\\simplify{24x-4*({w}+{l})}\$

when \$x=\\simplify{({w}+{l}-sqrt({w}^2-{w}*{l}+{l}^2))/6}\$ \$\\frac{d^2V}{dx^2}<0\$ and therefore is the value that gives a maximum.

", "variable_groups": [], "tags": [], "functions": {}, "metadata": {"description": "

Maximising the volume of a rectangular box

", "licence": "Creative Commons Attribution 4.0 International"}, "variables": {"w": {"description": "", "name": "w", "templateType": "randrange", "definition": "random(10..28#2)", "group": "Ungrouped variables"}, "l": {"description": "", "name": "l", "templateType": "randrange", "definition": "random(30..100#5)", "group": "Ungrouped variables"}}, "statement": "

Determine the value of \$x\$ that will maximise the volume of this box.

", "type": "question"}, {"name": "Applications of differentiation", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "Vicky Hall", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/659/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "", "advice": "

Parts A and B

Part C

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Is the stationary point a maximum?

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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]

Firstly, differentiate.

$\\displaystyle \\frac{dy}{dx}=$ [[1]]

Gradient at $x=\\var{d}\\;$ is [[0]]

You have not given your answer to the correct precision.

Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

$y=\\simplify {{f}x^2+{g}x+{h}}$

Firstly, find the first and second derivatives $y$.

$\\displaystyle \\frac{dy}{dx}=$ [[2]]

$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

$x$-coordinate of the turning point $=$ [[0]]

$y$-coordinate of the turning point $=$ [[1]]

The turning point is a [[4]]

You have not given your answer to the correct precision.

", "strictPrecision": false, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}, {"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "g^2/(4*f)-g^2/(2*f)+h", "maxValue": "g^2/(4*f)-g^2/(2*f)+h", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "precisionType": "dp", "precision": "2", "precisionPartialCredit": 0, "precisionMessage": "

You have not given your answer to the correct precision.

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maximum

", "

minimum

"], "matrix": ["if(maximum, 1, 0)", "if(maximum, 0, 1)"], "distractors": ["", ""]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always", "type": "question"}, {"name": "Implicit Differentiation product", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Gareth Woods", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/978/"}, {"name": "Aoife O'Brien", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1693/"}, {"name": "Antonia Wilmot-Smith", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2410/"}, {"name": "Leticija Dubickaite", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2461/"}, {"name": "Kevin Bohan", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3363/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "\n \t\t

Implicit differentiation.

\n \t\t

Given $x^2+y^2+dxy +ax+by=c$ find $\\displaystyle \\frac{dy}{dx}$ in terms of $x$ and $y$.

\n \t\t

Also find two points on the curve where $x=0$ and find the equation of the tangent at those points.

\n \t\t

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "

You are given the following relation between $x$ and $y$
\\[\\simplify{x^2+y^2+{d}x y+{a}x+{b}y}=\\var{c}\\]
where $y=f(x)$. Find $\\dfrac{dy}{dx}$.

", "advice": "

Hint:

Note that we regard $y$ as a function of $x$. Hence we have (using the Chain Rule): $\\displaystyle \\frac{d(y^2)}{dx} = 2y\\frac{dy}{dx}$. And, using the Product Rule: $\\displaystyle \\frac{d(xy)}{dx} = y+x\\frac{dy}{dx}$.

Now differentiate both sides of the relation with respect to $x$. Below is a worked solution to the problem.

a) By differentiating both sides of the equation implicitly we get
\\[2x + \\simplify[all,!collectNumbers]{2y*Diff(y,x,1) +{d}(y+x*Diff(y,x,1))+ {a} + {b} *Diff(y,x,1)} = 0\\]
Collecting terms in $\\displaystyle\\frac{dy}{dx}$ and rearranging the equation we get
\\[( \\simplify[all,!collectNumbers]{({b} + 2y+{d}x)} )\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{{ -a} -2x-{d}y}\\] and hence on further rearranging:

\\[\\frac{dy}{dx} = \\simplify[all,!collectNumbers]{({ - a} - 2 * x-{d}y) / ({b} + (2 * y)+{d}x)}\\]

", "rulesets": {"std": ["all", "fractionNumbers"]}, "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "-random(2..9)", "description": "", "templateType": "anything"}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "random(3..9 except -a+1)", "description": "", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "c-1", "description": "", "templateType": "anything"}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "random(-3..3 except[0,1])", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "c", "b", "d"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

Input your answer here:

$\\displaystyle \\frac{dy}{dx}= $ [[0]]

Input all numbers as integers not as decimals.

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Input all numbers as integers or as fractions, not as decimals.

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Implicit differentiation question with customised feedback to catch some common errors.

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Find the gradient of the curve $\\simplify{{a}x}+x^2y^2=\\simplify{{constant}+{b}y}$ at the point $(\\var{c},\\var{d})$.

", "advice": "

$\\frac{d}{dx}(\\var{a}x+x^2y^2)=\\frac{d}{dx}(\\simplify{{constant}+{b}y})$

$\\var{a}+2x.y^2+x^2.2y.\\frac{dy}{dx}=\\var{b}.\\frac{dy}{dx}$

$(\\simplify{2x^2y-{b}}) \\frac{dy}{dx}=\\simplify{-{a}-2x*y^2}$

$\\frac{dy}{dx}=\\frac{\\simplify{-{a}-2x*y^2}}{\\simplify{2x^2*y-{b}}} =\\simplify{-(2x*y^2+{a})/(2x^2*y-{b})}$

$\\ $

$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=\\simplify{{a}/{b}}$

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Hint: Type $2x^2y$ as 2x^2*y

$\\frac{dy}{dx}=$ [[0]]

", "gaps": [{"type": "jme", "useCustomName": false, "customName": "", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "malrules:\n [\n [\"3/y+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1+3/y)\", \"There are two main errors here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"],\n [\"-y/(x+1/y^3)\", \"Be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/y^3+y+x\", \"This is an implicit differentiation question. Therefore you need to consider $y$ as having something to do with $x$. Therefore, whenever you differentiate part of this expression involving $y$ you need the chain rule and so need to include $\\\\frac{dy}{dx}$. For example, when using the product rule to differentiate $xy$ (which you spotted - well done!), you get $x \\\\cdot \\\\frac{dy}{dx} + y \\\\cdot 1 = x \\\\frac{dy}{dx}+y$. Similarly, when differentiating $\\\\ln \\\\left( y^3 \\\\right)$, recall that $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$\"],\n [\"1/(1/y^3+1)\", \"There are three things to watch here. Firstly, since this is implicit differentiation, you are thinking of $y$ as having something to do with $x$. This means you need the product rule to differentiate $xy$, since $x \\\\cdot y$ is really $x \\\\times $ (something to do with $x$). Secondly, be careful when differentiating $\\\\ln \\\\left( y^3 \\\\right)$. Remember, $\\\\frac{d}{dx} \\\\left( \\\\ln x \\\\right)=\\\\frac{1}{x}$ but $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( f(x) \\\\right) \\\\right)=\\\\frac{1}{f(x)} \\\\cdot f'(x)$. Therefore $\\\\frac{d}{dx} \\\\left( \\\\ln \\\\left( y^3 \\\\right) \\\\right) = \\\\frac{1}{y^3} \\\\cdot \\\\frac{d}{dx} \\\\left(y^3 \\\\right) = \\\\frac{1}{y^3} \\\\cdot 3y^2 \\\\cdot \\\\frac{dy}{dx}$. Finally, what is the derivative of the right hand side? Don't forget to differentiate $\\\\textbf{both}$ sides.\"]\n ]\n\n\nparsed_malrules: \n map(\n [\"expr\":parse(x[0]),\"feedback\":x[1]],\n x,\n malrules\n )\n\nagree_malrules (Do the student's answer and the expected answer agree on each of the sets of variable values?):\n map(\n len(filter(not x ,x,map(\n try(\n resultsEqual(unset(question_definitions,eval(studentexpr,vars)),unset(question_definitions,eval(malrule[\"expr\"],vars)),settings[\"checkingType\"],settings[\"checkingAccuracy\"]),\n message,\n false\n ),\n vars,\n vset\n )))$\\frac{dy}{dx} \\bigg|_{(\\var{c},\\var{d})}=$[[0]]

(Answer in fraction form if necessary)

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