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Rate of change problem involving velocity & acceleration

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\n

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\n

Calculate the average rate of change over an interval in a graph between points $(x0,y0) = (\\var{x0},\\var{y0})$ and $(x1,y1)= (\\var{x1},\\var{y1})$

\n

Average rate = [[0]]

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If it takes {t2} hours to drive a distance of {km2} km on a motorway, what would be your average speed in km/h?

\n

Average speed = [[0]] km/h

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Simple derivatives

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"", "templateType": "anything"}, "tem41": {"name": "tem41", "group": "Ungrouped variables", "definition": "random(10..38)", "description": "", "templateType": "anything"}, "t4": {"name": "t4", "group": "Ungrouped variables", "definition": "random(3..8)", "description": "", "templateType": "anything"}, "aa": {"name": "aa", "group": "Ungrouped variables", "definition": "random(-9..9 except [-1,0,1])", "description": "", "templateType": "anything"}, "bb": {"name": "bb", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "templateType": "anything"}, "bb2": {"name": "bb2", "group": "Ungrouped variables", "definition": "random(1..9)", "description": "", "templateType": "anything"}, "aa2": {"name": "aa2", "group": "Ungrouped variables", "definition": "random(-9..9 except [-1,0,1])", "description": "", "templateType": "anything"}, "nn": {"name": "nn", "group": "Ungrouped variables", "definition": "random(2..9)", "description": "", "templateType": "anything"}, 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Find derivative of $y=\\var{ann4}x^\\var{nn4}+\\var{bnn4}x^\\var{nn5} +\\var{cnn4}$

\n

 [[0]]

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More work on differentiation with trigonometric functions

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Differentiate the following trigonometric functions using the chain rule.

\n

Do not write out $dy/dx$; only input the differentiated right hand side of each equation.

", "advice": "

If you don't know how to differentiate trigonometric functions, please see 'Differentiation 4 - Trigonometric Functions'.

\n

\n

These questions use the chain rule.

\n

The earlier questions are easy to do by inspection, e.g using Part a:

\n

$y=sin(\\var{c[0]}x)$.

\n

We differentiate the term(s) inside the function, here the term is $\\var{c[0]}x$.

\n

Then we derive $sin$ of any function, giving us $cos$.

\n

Putting our results together, we get

\n

$\\var{c[0]}cos(\\var{c[0]}x)$.

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coefficients

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$y=-5\\cos(\\var{c[3]}x)+\\sin(\\var{c[4]}x)$

\n

$\\frac{dy}{dx}=$ [[0]]

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Differentiating exponentials and Logs

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Differentiate the following.

\n

Do not write out $dy/dx$; only input the differentiated right hand side of each equation.

\n

Remember to enclose all single powers inside a bracket, for example, $e^{2x}$ is inputted as $e$^$(2x)$, or use $\\ln(2)$ instead of $\\ln2.$

", "advice": "

The key fact to understand here is that the differentiate of $e^x$ is $e^x$.

\n

This can be proven by looking at evaluating limits etc. but it is not necessary to do so at this stage.

\n

The basic steps to differentiate an exponential function are:

\n

Differentiate the power of $e$, for example in Part b, $y=\\var{c[1]}e^{\\var{p[1]}x}$, you would differentiate $\\var{p[1]}x$.

\n

In this example, it is $\\var{p[1]}$.

\n

Then multiply the coefficient of $e$ by this result.

\n

Here, you would find $\\simplify{{c[1]}{p[1]}e^({p[1]}x)}$.

\n

This is your final answer for the derivative.

\n

\n

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 $x$.

\n

\n

Basic formulas:

\n

$\\frac{d}{dx} e^x = e^x$

\n

$\\frac{d}{dx} e^{u(x)} = e^{u(x)}\\frac{d}{dx} u$

\n

$\\frac{d}{dx} a^x = a^x \\ln(a)$

\n

$\\frac{d}{dx} a^{u(x)} = a^{u(x)} \\ln(a) \\frac{d}{dx} u$

\n

$\\frac{d}{dx} \\ln(x) = \\frac{1}{x} ~~ (x>0)$

\n

$\\frac{d}{dx} \\ln|x| = \\frac{1}{x} ~~ (x\\neq 0)$

\n

$\\frac{d}{dx} \\log_a(x) = \\frac{1}{x \\ln a} ~~ (a>0, a \\neq 1)$

\n

$\\frac{d}{dx} x^x = x^x (1+\\ln x)$

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$y=\\var{c[1]}e^{\\var{p[1]}x}$

\n

$\\frac{dy}{dx}=$ [[0]]

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Calculating gradients - polynomials

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Given $y=\\var{a[1]}x^2+\\var{b[1]}x+\\var{c[1]},$ first calculate the differential $y^\\prime (x) = \\frac{dy}{dx}$ and then gradient $y^\\prime (\\var{x[1]}) $ at point $x = \\var{x[1]}$

\n

$y^\\prime (x) =$ [[0]]

\n

$y^\\prime (\\var{x[1]}) =$ [[1]]

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A graph (of a cubic) is given. The question is to determine the number of roots and number of stationary points the graph has. Non-calculator. Advice is given.

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Finding the number of roots and turning points based on a graph.

", "advice": "

(i) Definition: A 'root' of a function $f(x)$ is a value of $x$ which makes $f(x)=0$.  Visually a root can be found be seeing when the $y$-coordinate of the graph is $0$, i.e., when the graph crosses the $x$-axis. Therefore, to count the roots, you need to count how many times the graph crosses the $x$-axis.  In this question, the graph crosses the $x$-axis $\\var{num_roots}$ time(s), so there are $\\var{num_roots}$ roots.

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(ii) Definition: A 'stationary point' of a function is a point on the graph where $f'(x)=0$.  Remember that $f'$ tells us the gradient of $f$, so visually a stationary point is where the gradient of the curve is 0.  In this question, there is/are $\\var{num_stat}$ place(s) where the gradient of the graph is $0$, so the answer is $\\var{num_stat}$.

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(iii) There are 3 types of stationary points: maximum points, minimum points and points of inflection.

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Consider what happens to the gradient at a maximum point. It is positive just before the maximum point, zero at the maximum point, then negative just after the maximum point.

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Just before a minimum point the gradient is negative, at the minimum the gradient is zero and just after the minimum point it is positive.

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The number of roots.

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Random amount of vertifical shift for sake of variability.

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Number of stationary points

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Random amount of horizontal shift to create variability.

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Coefficient of x^3

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{plotgraph(num_stat,num_roots, a, hshift, vshift)}

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Above is the graph of some function $f$.

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How many roots does $~~f(x)=0~~$ have? [[0]]

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How many stationary points does $f(x)$ have? [[1]]

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Finding the coordinates and determining the nature of the stationary points on a polynomial function

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\\[ \\simplify{y = 2x^3-3{(x12+x22)}x^2+6{x12*x22}x+{c02}} \\]

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Determine the coordinates and the nature of the stationary points.

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First Derivative; $y^{\\prime}(x) =$ [[4]]

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Give values of x where stationary points occur: smallest-$x_1$ =[[6]],    largest-$x_2$ = [[7]]

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Second Derivative is $~ y^{\\prime\\prime}(x) =$ [[5]]

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Values of Second derivative at stationary points: 

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$y^{\\prime\\prime}(x_1) = $[[8]]

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$y^{\\prime\\prime}(x_2) = $[[9]]

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Minimum point: $\\big($ [[0]] $ , $ [[1]] $\\big)$ and maximum point: $\\big($ [[2]] $ , $ [[3]] $\\big)$

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Enter fractions in their simplest form.

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