// Numbas version: exam_results_page_options {"name": "MID-TRIMESTER EXAM 2020 T3", "metadata": {"description": "", "licence": "None specified"}, "duration": 7800, "percentPass": "0", "showQuestionGroupNames": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-shuffled", "pickQuestions": "20", "questionNames": ["", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""], "questions": [{"name": "Finding the formula for a shifted function", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Fig13-exp1.png", "/srv/numbas/media/question-resources/Fig13-exp1.png"], ["question-resources/Fig15-exp3.png", "/srv/numbas/media/question-resources/Fig15-exp3.png"], ["question-resources/Fig20-recip2.png", "/srv/numbas/media/question-resources/Fig20-recip2.png"], ["question-resources/Fig16-sin1.png", "/srv/numbas/media/question-resources/Fig16-sin1.png"], ["question-resources/Fig17-sin2.png", "/srv/numbas/media/question-resources/Fig17-sin2.png"], ["question-resources/Fig18-sin3.png", "/srv/numbas/media/question-resources/Fig18-sin3.png"], ["question-resources/Fig19-recip1.png", "/srv/numbas/media/question-resources/Fig19-recip1.png"], ["question-resources/Fig23-quad2.png", "/srv/numbas/media/question-resources/Fig23-quad2.png"], ["question-resources/Fig21-recip3.png", "/srv/numbas/media/question-resources/Fig21-recip3.png"], ["question-resources/Fig14-exp2.png", "/srv/numbas/media/question-resources/Fig14-exp2.png"], ["question-resources/Fig22-quad1.png", "/srv/numbas/media/question-resources/Fig22-quad1.png"], ["question-resources/Fig24-quad3.png", "/srv/numbas/media/question-resources/Fig24-quad3.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Owen Jepps", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1195/"}], "variablesTest": {"maxRuns": 100, "condition": ""}, "ungrouped_variables": ["originalF", "shiftedF", "Fs", "figList"], "variable_groups": [], "parts": [{"displayType": "radiogroup", "distractors": ["", "", "", ""], "marks": 0, "prompt": "

{image(figlist[Fs[0]])}

If the blue curve represents is {originalF[Fs[0]]}, then the black dotted function is

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{shiftedF[Fs[0]][0]}

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{shiftedF[Fs[0]][1]}

", "

{shiftedF[Fs[0]][2]}

", "

{shiftedF[Fs[0]][3]}

{image(figlist[Fs[1]])}

If the blue curve represents is {originalF[Fs[1]]}, then the black dotted function is

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{shiftedF[Fs[1]][0]}

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{shiftedF[Fs[1]][1]}

", "

{shiftedF[Fs[1]][2]}

", "

{shiftedF[Fs[1]][3]}

{image(figlist[Fs[2]])}

If the blue curve represents is {originalF[Fs[2]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[2]][0]}

", "

{shiftedF[Fs[2]][1]}

", "

{shiftedF[Fs[2]][2]}

", "

{shiftedF[Fs[2]][3]}

{image(figlist[Fs[3]])}

If the blue curve represents is {originalF[Fs[3]]}, then the black dotted function is

", "minMarks": 0, "displayColumns": 0, "scripts": {}, "type": "1_n_2", "showCorrectAnswer": true, "choices": ["

{shiftedF[Fs[3]][0]}

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{shiftedF[Fs[3]][1]}

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{shiftedF[Fs[3]][2]}

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{shiftedF[Fs[3]][3]}

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harder functions

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[
'$e^x$','$e^x$','$e^x$',
'$\\\\sin(x)$','$\\\\sin(x)$','$\\\\sin(x)$',
\"$1/x$\", \"$1/x$\", \"$1/x$\",
\"$x^2$\", \"$x^2$\", \"$x^2$\"
]

", "name": "shiftedF"}}, "functions": {}, "metadata": {"licence": "None specified", "description": ""}, "rulesets": {}, "tags": [], "statement": "

For each of the following blue functions $f(x)$, identify the formula for the black dotted funtion.

", "advice": "

Look over the summary and examples in the lecture slides. For a function $f(x)$, and positive constant $c$

--- $f(x\\pm c)$ shifts $f(x)$ to the left or right by $c$

--- $f(x) \\pm c$ shifts $f(x)$ up or down by $c$

--- $f(cx)$ rescales $f(x)$ in the $x$-direction [shrinks if $c>1$, or expands if $c<1$]

--- $c f(x)$ rescales $f(x)$ by $c$ in the $y$-direction

", "preamble": {"js": "", "css": ""}, "type": "question"}, {"name": "General function inversion [harder]", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Owen Jepps", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1195/"}], "functions": {}, "variables": {"hn": {"templateType": "anything", "name": "hn", "description": "", "definition": "shuffle(0..len(harderF)-1)", "group": "Ungrouped variables"}, "harderF": {"templateType": "anything", "name": "harderF", "description": "

harder functions

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inverses of harder functions

", "definition": "[\n 'sqrt((x+1)/2)',\n '((x-2)/3)^3',\n '3(e^x-1)',\n '(1-ln(x))/2',\n '(x^2+5)/2',\n 'sqrt(2(x^2-5))',\n '(arccos(x)-pi)/2',\n '4(tan(x)-2)'\n]", "group": "Ungrouped variables"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "rulesets": {}, "variable_groups": [], "tags": [], "ungrouped_variables": ["harderF", "harderI", "hn"], "advice": "

To find the inverse of $f(x)$, write $f(y)=x$, and then keep re-arranging this equation until you reach $y=\\ldots$. During the re-arranging, remember that anything you do to the left-hand side must be done to the right-hand side as well. This process \"undoes\" the operations on $y$, in the reverse order in which they are done when calculating $f(y)$.

So, if $f(x)=\\cos(x/2+1)$, we write

$f(y) = \\cos(y/2+1) = x$

and then re-arrange until we only have $y$ on the left-hand side. To calculate the left-hand side, we halve $y$, add 1, then find the cosine, so we undo this in the reverse order:

--- first, we take the inverse-cosine -- the arccos -- of both side, to get

$ y/2+1 = \\arccos(x)$

--- second, we subtract 1 from both side to get

$ y/2 = \\arccos(x) - 1$

--- finally, we double both sides to get

$ y = 2(\\arccos(x) - 1) = 2\\arccos(x) - 2$

So if $f(x)=\\cos(x/2+1)$, then the inverse function $f^{-1}(x)=2\\arccos(x)-2$

", "metadata": {"licence": "None specified", "description": ""}, "statement": "

For each of the following functions $f(x)$, give the inverse function $f^{-1}(x)$.

To write the function, you can write, e.g., $\\sqrt{x-1}$ as sqrt(x-1), $x^n$ as x^n, $\\cos(x/2)$ as cos(x/2)

", "preamble": {"js": "", "css": ""}, "parts": [{"prompt": "

If {harderF[hn[0]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[1]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[2]]}, then the inverse function $f^{-1}(x)=$ [[0]]

If {harderF[hn[3]]}, then the inverse function $f^{-1}(x)=$ [[0]]

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Testing your knowledge of the domain and range of some common important functions

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Give the domain of the function $y=\\var{funcNames[d0]}$

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$\\var{xIntervals[correctDomain[d0]]}$

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$\\var{xIntervals[dN0[0]]}$

", "

$\\var{xIntervals[dN0[1]]}$

", "

$\\var{xIntervals[dN0[2]]}$

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Give the range of the function $y=\\var{funcNames[r0]}$

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$\\var{yIntervals[correctRange[r0]]}$

", "

$\\var{yIntervals[rN0[0]]}$

", "

$\\var{yIntervals[rN0[1]]}$

", "

$\\var{yIntervals[rN0[2]]}$

"], "matrix": ["1/2", 0, 0, 0], "distractors": ["", "", "", ""]}]}, {"name": "Identifying the formula for piecewise-defined functions [QUIZ]", "extensions": [], "custom_part_types": [], "resources": [["question-resources/Fig02-piece2.png", "/srv/numbas/media/question-resources/Fig02-piece2.png"], ["question-resources/Fig01a-piece1a.png", "/srv/numbas/media/question-resources/Fig01a-piece1a.png"], ["question-resources/Fig03a-piece3a.png", "/srv/numbas/media/question-resources/Fig03a-piece3a.png"]], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Owen Jepps", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1195/"}], "statement": "

Testing your ability to identify the formulas defining piecewise-defined functions

", "rulesets": {}, "parts": [{"scripts": {}, "choices": ["

{options[whichFig][0]}

", "

{options[whichFig][1]}

", "

{options[whichFig][2]}

", "

{options[whichFig][3]}

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{image(figList[whichFig])}

The graph above is a plot of the formula

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{image(figList[whichFunc[0]])}

", "choices": ["

Odd

", "

Even

", "

Neither

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{image(figList[whichFunc[1]])}

", "choices": ["

Odd

", "

Even

", "

Neither

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For each graph, indicate whether the function is odd, even or neither. Note that the grids have the same units in the $x$ and $y$ directions

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Even functions are symmetric about the $y$-axis ($f(-x)=f(x)$), while odd functions are antisymmetric about the $y$-axis ($f(-x)=-f(x)$). However, functions can be neither symmetric nor antisymmetric.

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For each graph, identify the plotted function. Note that the grids have the same units in the $x$ and $y$ directions

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0: '$x$',

1: '$\\\\frac{x}{2}+3$',

2: '$4-x$',

3: '$x^2$',

4: '$x^3$',

5: '$\\\\sqrt{x]$',

6: '$\\\\sin x$',

7: '$\\\\cos x$',

8: '$\\\\tan x$',

9: '$e^x$',

10: '$\\\\ln x$',

11: '$\\\\frac{1}{x}$']

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Input as a fraction or an integer.

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$\\displaystyle \\simplify{Limit((x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}),x,{a}) }=\\;$[[0]] (input as a fraction or as an integer).

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Find the limit of the following function.

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Note that on putting $x=\\var{a}$ into $\\displaystyle \\simplify{(x + { -a}) / (x ^ 2 + { -a -b} * x + {a * b}) }$ we get a $0/0$ case and so we have to do more work.

You can factorise $\\simplify{x ^ 2 + { -a -b} * x + {a * b}}$ and then see what happens.

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For the function below

$\\simplify{f}(\\simplify{x})=\\left\\{\\begin{align}&\\simplify{1},&& \\text{ for } \\simplify{x}< \\var{b0}, \\\\&\\simplify{{b}-{a}x},&& \\text{ for } \\var{b0}\\leq \\simplify{x}< \\var{b1},\\\\ &\\simplify{{c}x^2-{d}},&& \\text{ for } x\\geq \\var{b1},\\end{align}\\right.$

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Find the value of $f(\\var{Input1})$ [[0]]

Find the value of $f(\\var{Input2})$ [[1]]

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Solve an exponential equation

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Determine the value for \$x\$ that satisfies the equation: \$\\var{a}e^{\\var{m}x+\\var{c}}=\\var{k}\$

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\$\\var{k}=\\var{a}e^{\\var{m}x+{\\var{c}}}\$

\$\\frac{\\var{k}}{\\var{a}}=e^{\\var{m}x+\\var{c}}\$

\$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)=\\var{m}x+\\var{c}\$

\$ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}=\\var{m}x\$

\$\\frac{ln\\left(\\frac{\\var{k}}{\\var{a}}\\right)-\\var{c}}{\\var{m}}=x\$

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Input your answer correct to three decimal places.

\$x = \$ [[0]]

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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

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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 $\\dfrac{dy}{dx}$.

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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)}\\]

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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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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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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.

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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"}, {"name": "Differentiation 18 - Applications", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Harry Flynn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/976/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

Another practical application of differentiation

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

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.

", "advice": "

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

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

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

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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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A basic introduction to differentiation

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

Differentiate the following polynomials.

Note: some questions may not include all the possible terms.

", "advice": "

If $y=ax^n$,

$\\frac{dy}{dx}=anx^{n-1}$ for all rational $n$.

We'll take one of the terms from Part a as an example:

$\\var{cc[0]}x^\\var{cp}$

All we have to do to terms where $x$ is to a power of anything is times the coefficient of $x$ by the original power, and then take one away from the original power.

If you are not familiar with this kind of work, these instructions may sound confusing, but it is much easier once you have seen it in practice.

We take

$\\var{cc[0]}x^\\var{cp}$

and times $\\var{cc[0]}$ by $\\var{cp}$, to get

$(\\var{cc[0]}*\\var{cp})x^\\var{cp}=\\simplify{{cc[0]}*{cp}x^{cp}}$.

We then subtract one from the original power, $\\var{cp}$.

This gives us the final answer of

$\\simplify{{cc[0]}*{cp}x^{cp-1}}$.

Remember, don't be confused if there is no coefficient. The fact the term is there means the coefficient must be $1$, but we don't tend to write it out as, for example $1x$, we just say $x$.

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$\\simplify[basic,zeroterm,zerofactor,unitfactor]{{ac[1]}x^{ap}+{bc[1]}x^{bp}+{cc[1]}x^{cp}+{dc[1]}x^{dp}+{ec[1]}x+{fc[1]}}$

$f'(x)=$ [[0]]

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

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$f(x) = \\var{a}$

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$f(x) = \\var{b}x^2 -\\var{c}x+\\var{d}$

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$f(x) = x^2(1-\\var{f}x)$

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$f(x) = -\\frac{\\var{g}}{x^\\var{h}}$

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$f(x) = \\sqrt{x}$

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$f(x) = x^2 + e^x$

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Some basic derivatives.

rebelmaths

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Differentiate the following with respect to $x$.

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Let the function $f$ be given by $\\displaystyle f(t)=\\simplify{({a3} * t + {b3}) / ({a2} * t ^ 2 + {2 * b2 * a2} * t + {c2 + a2 * b2 ^ 2}) }$

", "advice": "

Graph of $f$.

{plotf(a2,b2,c2,a3,b3,stat1,stat2)}

Note that $\\simplify{{a2} * t ^ 2 + {2 * b2 * a2} * t + {c2 + a2 * b2 ^ 2} ={a2}*(t+{b2})^2+{c2}} \\gt 0$.

Hence the denominator of $f(t) \\neq 0,\\;\\forall t \\in \\mathbb{R}$ and so $f$ is continuous at all points in $\\mathbb{R}$.

This means that in part a) we can take the limit by simply subsituting $t=\\var{d2}$ into the expression for $f(t)$ and we get:

\\[\\lim_{x \\to \\var{d2}}f(t)=\\simplify[all,!otherNumbers,fractionNumbers,!collectNumbers]{({a3} * {d2} + {b3}) / ({a2} * {d2}^ 2 + {2 * b2 * a2} * {d2}+ {c2 + a2 * b2 ^ 2})={v}/{w} }\\]

Similarly in part b) we have :

\\[\\lim_{x \\to a}f(t)=\\simplify[all,!collectNumbers,!otherNumbers,fractionNumbers]{({a3} * a + {b3}) / ({a2} * a^ 2 + {2 * b2 * a2} * a+ {c2 + a2 * b2 ^ 2})}\\]

As noted above we can find this limit by simply putting $t=a$ into the formula for the function as $f$ is continuous at all points in $\\mathbb{R}$.

", "rulesets": {}, "variables": {"c2": {"name": "c2", "group": "Ungrouped variables", "definition": "random(1..5)", "description": "", "templateType": "anything"}, "a3": {"name": "a3", "group": "Ungrouped variables", "definition": "random(-9..9 except 0)", "description": "", "templateType": "anything"}, "v": {"name": "v", "group": "Ungrouped variables", "definition": "a3*d2+b3", "description": "", "templateType": "anything"}, "w": {"name": "w", "group": "Ungrouped variables", "definition": "a2*d2^2+2*a2*b2*d2+c2+a2*b2^2", "description": "", "templateType": "anything"}, "b2": {"name": "b2", "group": "Ungrouped variables", "definition": "-random(1..6)", "description": "", "templateType": "anything"}, "d2": {"name": "d2", "group": "Ungrouped variables", "definition": "random(0..5)", "description": "", "templateType": "anything"}, "stat1": {"name": "stat1", "group": "Ungrouped variables", "definition": "(2*a2*b3+sqrt(4*a2^2*b3^2+4*a3*a2*(a3*c2-2*b3*b2*a2+a2*b2^2*a3)))/(-2*a3*a2)", "description": "", "templateType": "anything"}, "b3": {"name": "b3", "group": "Ungrouped variables", "definition": "random(-9..9)", "description": "", "templateType": "anything"}, "a2": {"name": "a2", "group": "Ungrouped variables", "definition": "random(1..3)", "description": "", "templateType": "anything"}, "stat2": {"name": "stat2", "group": "Ungrouped variables", "definition": "(2*a2*b3-sqrt(4*a2^2*b3^2+4*a3*a2*(a3*c2-2*b3*b2*a2+a2*b2^2*a3)))/(-2*a3*a2)", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["w", "stat1", "b3", "a3", "a2", "b2", "v", "c2", "d2", "stat2"], "variable_groups": [], "functions": {"plotf": {"parameters": [["a2", "number"], ["b2", "number"], ["c2", "number"], ["a3", "number"], ["b3", "number"], ["stat1", "number"], ["stat2", "number"]], "type": "html", "language": "javascript", "definition": "var f = function(t){ return (a3*t+b3)/(a2*t*t+2*a2*b2*t+c2+a2*b2*b2); };\nvar m1=Math.min(stat1,stat2);\nvar m2=Math.max(stat1,stat2);\nvar f1=f(stat1);\nvar f2=f(stat2);\nvar a=Math.abs(f1);\nvar b=Math.abs(f2);\nvar M=Math.max(a,b);\nvar div = Numbas.extensions.jsxgraph.makeBoard('300px','300px', {axis:true,showNavigation:false,boundingbox:[m1-10,M+2,m2+10,-M-2]});\n\nvar brd=div.board;\n\nvar plot = brd.create('functiongraph',[f,m1-10,m2+10]);\n\n//brd.create('text',[c,-2,c]);\n//var i1 = brd.create('integral', [[0, c], plot]);\n\nreturn div;"}}, "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 following limit:

$\\displaystyle \\simplify{Limit(f(t),t,{d2}) }= \\;$[[0]].

Enter your answer as a fraction or an integer and not as a decimal.

Enter as a fraction or an integer and not as a decimal.

"}, "valuegenerators": []}], "sortAnswers": false}, {"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": "

Also find:

$\\displaystyle \\simplify{Limit(f(t),t,a) }= \\;$[[0]], where $a \\in \\mathbb{R}$ is any point.

", "gaps": [{"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, "answer": "({a3}*a+{b3})/({a2}*a^2+{2*a2*b2}*a+{c2+a2*b2^2})", "answerSimplification": "all", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 0.001, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "singleLetterVariables": false, "allowUnknownFunctions": true, "implicitFunctionComposition": false, "valuegenerators": [{"name": "a", "value": ""}]}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Maria's copy of Implicit differentiation 1 (Basic)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Clodagh Carroll", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/384/"}, {"name": "JPO AddMath", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/2346/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "

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

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

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

", "advice": "

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

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

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

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

$\\ $

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

", "rulesets": {}, "variables": {"a": {"name": "a", "group": "Ungrouped variables", "definition": "Random(-6..6 except 0)", "description": "", "templateType": "anything"}, "c": {"name": "c", "group": "Ungrouped variables", "definition": "Random(-3..3)", "description": "", "templateType": "anything"}, "constant": {"name": "constant", "group": "Ungrouped variables", "definition": "{a}*{c}-{b}*{d}", "description": "", "templateType": "anything"}, "d": {"name": "d", "group": "Ungrouped variables", "definition": "if(c=0, random(-3..3),0)", "description": "", "templateType": "anything"}, "b": {"name": "b", "group": "Ungrouped variables", "definition": "Random(-6..6 except 0 except a except -a)", "description": "", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["a", "b", "c", "d", "constant"], "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": "

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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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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Given $f(x)=(x+b)^n$. Find the gradient and equation of the chord between $(a,f(a))$ and $(a+h,f(a+h))$ for randomised values of $a$, $b$ and $h$.

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

Let $f(x)=\\simplify[std]{(x+{b})^{n}}$. What are the gradient and equation of the chord/secant between $(\\var{a},f(\\var{a}))$ and $(\\simplify[std]{{a}+{h}},f(\\simplify[std]{{a}+{h}}))$?

", "advice": "

Given two points $(a,f(a))$ and $(a+h,f(a+h))$ on the graph of the function $y=f(x)$.
Then the chord is the straight line between these two points and has the equation \\[y-f(a)=m(x-a)\\] where $m$ is the gradient of the chord.
The gradient is given by dividing the change in $y$ by the change in $x$.
Hence for this example \\[m = \\frac{f(a+h)-f(a)}{h} = \\frac{f(\\var{a+h})-f(\\var{a})}{\\var{h}} = \\var{d1} = \\var{val}\\] to 3 decimal places.
Hence the equation of the chord is of the form $y=\\var{d1}x+b$ for some constant $b$.
But we know that when $x=\\var{a}$ then $y=f(\\var{a}) = \\var{a+b}^\\var{n}=\\var{(a+b)^n}$
So \\[b=\\var{(a+b)^n}-\\var{d1}\\times\\var{a} = \\var{d}=\\var{val1}\\] to 3 decimal places

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The gradient $m =$ [[0]] (input your answer to 3 decimal places).

The equation of the chord/secant is $y=ax+b$ where:

$a= \\;$[[1]] and $b=\\; $[[2]]

Enter both values $a$ and $b$ correct to 3 decimal places.

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Identifying some of the basic properties (intercepts, asymptotes, quadrants) of a right hyperbola.

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You are given the equation of a function \\[\\simplify[all]{y={k}/(x-{a})+{b}}.\\]

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The vertical asymptote corresponds to the value of $x$ that results in attempting to divide by $0$. For the equation

\\[\\simplify[all]{y={k}/(x-{a})+{b}}\\]

This means the equation of the vertical asymptote is $x=\\var{a}$.

The horizontal asymptote corresponds to the value of $y$ that results from $x$ approaching infinity. As $x$ gets really really large, the fraction $\\simplify{{k}/(x-{a})}$ gets really really close to zero. The bigger $x$ gets, the closer $\\simplify{{k}/(x-{a})}$ gets to zero, and the closer $\\simplify[all]{y={k}/(x-{a})+{b}}$ gets to $y=\\var{b}$.

This means the equation of the horizontal asymptote is $y=\\var{b}$.

Notice that in our equation, $\\var{k}$ is multiplying the fraction $\\simplify[all]{1/(x-{a})}$. It is a general fact that because $\\var{k}$ is positive the graph will be in the top right and bottom left parts of the plane. negative the graph will be in the top left and bottom right parts of the plane. We can see this as follows:

By substituting into the equation an $x$ value that is to the right of the vertical asymptote $x=\\var{a}$, we will find the resulting $y$ value is above below the horizontal asymptote $y=\\var{b}$.
By substituting into the equation an $x$ value that is to the left of the vertical asymptote $x=\\var{a}$, we will find the resulting $y$ value is above below the horizontal asymptote $y=\\var{b}$.

To find the $y$-intercept, let $x=0$ and solve for $y$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$y$	$=$	$\\simplify{{k}/(0-{a})+{b}}$
	$=$	$\\simplify[fractionNumbers]{{-k/a+b}}$

To find the $x$-intercept, let $y=0$ and solve for $x$:

\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n

$0$	$=$	$\\simplify{{k}/(x-{a})+{b}}$
$\\var{-b}$	$=$	$\\simplify{{k}/(x-{a})}$
$\\simplify{{-b}(x-{a})}$	$=$	$\\var{k}$
$\\simplify{x-{a}}$	$=$	$\\simplify{{k}/({-b})}$
$x$	$=$	$\\simplify[fractionNumbers]{{-k/b+a}}$

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The graph of this function has a vertical asymptote at $x=$ [[0]].

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The graph of this function has a horizontal asymptote at $y=$ [[0]].

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The $y$-intercept is at $y=$[[0]].

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The $x$-intercept is at $x=$ [[0]].

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