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Calculus 1 : Rate of Change

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Example Questions

Example Question #1902 : Functions

Find the rate of change of a function f(x)=7x-8 from x_1=5 to x_2=6 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(5)=7(5)-8=35-8=27

f(x_2)=f(6)=7(6)-8=42-8=34

Plugging our values into the average rate of change formula, we get:

$\frac{(34-27)}{(6-5)}=\frac{7}{1}=7$ .

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Example Question #1903 : Functions

Find the rate of change of a function f(x)=10x+12 from x_1=7 to x_2=9 .

Possible Answers:

Correct answer:

Explanation:

We can solve by utilizing the formula for the average rate of change: $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ .

Solving for f(x) at our given points:

f(x_1)=f(7)=10(7)+12=70+12=82

f(x_2)=f(9)=10(9)+12=90+12=102

Plugging our values into the average rate of change formula, we get:

$\frac{(102-82)}{(9-7)}=\frac{20}{2}=10$ .

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Example Question #2932 : Calculus

A spherical balloon is inflating at rate of 0.1 cubic meters per second. Find the rate of change of the surface area of the balloon (in square meters per second) when the balloon has a radius of 1 meter.

Possible Answers:

$-\frac{1}{5}$

$\frac{1}{5}$

$\frac{1}{40\pi }$

$8\pi$

Correct answer:

$\frac{1}{5}$

Explanation:

The volume of a sphere is given by

$V=\frac{4\pi }{3}r^3$ , where r is the radius of the sphere.

The surface area of a sphere is given by $A=4\pi r^2$ (r is the radius of the sphere). The rate of change of the volume of the sphere is found by taking the first derivative of the function for volume:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$ , where $\frac{dr}{dt}$ is the change of the radius with respect to time.

This derivative was found by using the power rule

$\frac{d}{dt}x^n=nx^{n-1}\frac{dx}{dt}$ .

The rate of change of the surface area of the sphere is found the same way, instead using the function for surface area:

$\frac{dA}{dt}=8\pi r \frac{dr}{dt}$ .

The balloon's volume is increasing at a constant rate of 0.1 cubic meters per second when the radius of the balloon is 1 meter, thus plugging in these values into the $\frac{dV}{dt}$ equation gives us the following:

$0.1 \frac{m^3}{sec}=4\pi (1^2)\frac{dr}{dt}$ . Solving for $\frac{dr}{dt}$ , we get $\frac{dr}{dt}=\frac{1}{40\pi }$ meters per second. Since we are solving for the rate of change of surface area of the balloon, we plug this value into its equation along with the radius we want (1 m). Solving for the rate of change of the surface area we get:

$\frac{dA}{dt}=8\pi (1)\left(\frac{1}{40\pi }\right)=\frac{1}{5}$ square meters per second.

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Example Question #2933 : Calculus

The spherical bottom of a snowman is melting such that its volume is decreasing at a rate of $3\frac{cm^3}{hr}$ .

What is the rate of change of the radius, to the nearest whole number, when the radius is 24cm ?

Possible Answers:

$18.8\frac{cm}{hr}$

$20\frac{cm}{hr}$

$19\frac{cm}{hr}$

$18\frac{cm}{hr}$

Correct answer:

$19\frac{cm}{hr}$

Explanation:

First, find the rate of increase in the radius $\left(\frac{dr}{dt}\right)$ :

$\frac{3\frac{cm^3}{hr}}{24cm}=\frac{1}{8}\frac{cm^2}{hr}$

Next, take the derivative of the area of a circle ( $A=\pi\cdot r^2$ ) by using the Power Rule ( $x^n=nx^{n-1}$ ):

$\frac{dA}{dt}=(2)(\pi)(r)\left(\frac{dr}{dt}\right)$

Lastly, plug in the givens and simplify:

$\frac{dA}{dt}=2\pi(24)\left(\frac{1}{8}\right)=6\pi\approx 18.85 \approx19\frac{cm}{hr}$

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Example Question #25 : How To Find Rate Of Change

For which of the following functions is the average rate of change along the interval [-2, 2] equal to the instentaneous rate of change at x=0 ?

f(x)=x

$g(x)=x^{2}$

$h(x)=x^{3}$

Possible Answers:

$f(x)\textup{ and }h(x)$

$f(x)\textup{\textup{ only}}$

$f(x),g(x),\textup{and }h(x)$

$f(x) \textup{\textup{ and }} g(x)$

Correct answer:

$f(x) \textup{\textup{ and }} g(x)$

Explanation:

The average rate of change of a function f(x) on an interval [a,b] is given by

$\frac{\Delta f(x)}{\Delta x}=$ $\frac{f(b)-f(a)}{b-a}$

while the instantaneous rate of change at a is given by

$\frac{\mathrm{d} }{\mathrm{d} x}f(a)$ .

Finding each of these on the interval [-2, 2] for the equations given:

$f'(0)=1, \frac{\Delta f(x)}{\Delta x}=1$

$g'(0)=0, \frac{\Delta g(x)}{\Delta x}=0$

$h'(0)=0, \frac{\Delta h(x)}{\Delta x}=4$

We can see that the two rates are equal for f(x) and g(x), but not h(x).

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Example Question #1906 : Functions

Find $\frac{dy}{dx}$ for the equation 8y^2 - 4xy + 3 = 12

Possible Answers:

x + 1

y + x(1 + 4y)

$\frac{x+y}{4y}$

$\frac{y}{4y - x}$

$\frac{x-y}{x + 2y}$

Correct answer:

$\frac{y}{4y - x}$

Explanation:

To evaluate, first take the derivatives of both sides of the equation. This gives

$16y\frac{dy}{dx} - 4x\frac{dy}{dx} - 4y = 0$

By factoring and simple algebraic simplification, the equation becomes

$\frac{dy}{dx} = \frac{y}{x - 4y}$

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Example Question #2934 : Calculus

A spherical balloon with radius r = 4in is being inflated with air. Its radius is expanding at a rate of $2 \frac{in}{min}$ . What is the change in volume of the sphere (in $\frac{in^3}{min}$ )?

Possible Answers:

$25\pi \frac{in^3}{min}$

$256\pi \frac{in^3}{min}$

$128\pi \frac{in^3}{min}$

$64\pi \frac{in^3}{min}$

$80\pi \frac{in}{min}$

Correct answer:

$128\pi \frac{in^3}{min}$

Explanation:

The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$

To find the change in volume, differentiate the volume with respect to time.

$\frac{dV}{dt} =4\pi r^2 \frac{dr}{dt}$

Simply plug in the values given in the problem to find the rate of change of the volume.

$\frac{dV}{dt} = 4\pi(4)^2(2) = 128\pi$ $\frac{in^3}{min}$

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Example Question #21 : Rate Of Change

For the relation $\small x^2+y^4=2$ , compute $\small y'$ using implicit differentiation.

Possible Answers:

$\small y'=-\frac{x}{2y^3}$

$\small \small y'=-\frac{2x^4}{y}$

$\small \small y'=\frac{x}{2y^3}$

$\small \small y'=-\frac{x}{4y^3}$

Correct answer:

$\small y'=-\frac{x}{2y^3}$

Explanation:

To calculate $\small y'$ using implicit differentiation, we differentiate everything in the relation $\small x^2+y^4=2$ and solve for $\small y'$ :

$\small \frac{d}{dx}(x^2+y^4)=\frac{d}{dx}(2)$

$\small \small \implies 2x+4y^3\cdot y'=0$

$\small \small \iff 4y^3 y'=-2x$

$\small y'=-\frac{2x}{4y^3}=-\frac{x}{2y^3}$

So we have $\small y'=-\frac{x}{2y^3}$ .

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Example Question #22 : Rate Of Change

If b(t) models the number of books a library has a function of time (in days), at what rate was the library gaining or losing books when $t=\frac{3\pi}{4}$ ?

b(t)=6cos(x)+200x

Possible Answers:

The library was gaining 19.576 books per day.

The library was losing 19.576 books per day.

The library was gaining 195.76 books per day.

The library was losing 195.76 books per day.

Correct answer:

The library was gaining 195.76 books per day.

Explanation:

If b(t) models the rate at which a library buys and sells books, at what rate was the library gaining or losing books when $t=\frac{3\pi}{4}$ ?

b(t)=6cos(x)+200x

We need to find an instantaneous rate of change. Rates of change can be found with derivatives. Start by finding b'(t), and then using our given t value to find our final answer:

b(t)=6cos(x)+200x

b'(t)=-6sin(x)+200

$b'\left(\frac{3\pi}{4}\right)=-6sin\left(\frac{3\pi}{4}\right)+200=195.76$

So the library was gaining 195.76 books per day.

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Example Question #30 : How To Find Rate Of Change

Abel wants to fill a spherical ball with air to bring with him to the park. It has an initial volume of 450in^3 , and when Abel attaches the pumps, he sets it to inflate at a rate of $0.2\frac{in^3}{s}$ . How quickly does the point where the nozzle connects move from the center of the basketball?

Possible Answers:

$\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{3\pi}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{1}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{\pi}{180\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

$\frac{3}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Correct answer:

$\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Explanation:

To solve this problem, first understand what it's asking. When it talks about the distance between the center of the basketball and the point of connection, it's talking about a radius, and when it asks about the movement of this point, it's talking about a change in the radius.

First, write an equation for the radius with respect to the volume:

$\frac{4}{3}\pi r^3=V$

$r=\sqrt[3]{\frac{3}{4\pi}V}$

Afterwards, derive each side with respect to time, and plug in known values.

$\frac{dr}{dt}=\frac{1}{3}(\frac{3}{4\pi})(\frac{3}{4\pi}V)^{-\frac{2}{3}}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{4\pi \sqrt[\frac{2}{3}]{\frac{3}{4\pi}V}}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{180\pi\sqrt[\frac{2}{3}]{\frac{25}{2\pi}}}\frac{in}{s}$

Another approach is to derive the original volume equation:

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$

However, since $r=\sqrt[3]{\frac{3}{4\pi}V}$ , this also reduces to:

$\frac{dr}{dt}=\frac{1}{4\pi \sqrt[\frac{2}{3}]{\frac{3}{4\pi}V}}\frac{dV}{dt}$

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1902 : Functions

Example Question #1903 : Functions

Example Question #2932 : Calculus

Example Question #2933 : Calculus

Example Question #25 : How To Find Rate Of Change

Example Question #1906 : Functions

Example Question #2934 : Calculus

Example Question #21 : Rate Of Change

Example Question #22 : Rate Of Change

Example Question #30 : How To Find Rate Of Change

All Calculus 1 Resources

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