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Calculus 1 : How to find rate of change

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Example Questions

Example Question #3051 : Calculus

The rate of change of the radius of a sphere is given by the equation $8e^{-2t}$ . If the initial length of the radius is , what is the rate of change of the sphere's volume at time t=ln(2) ?

Possible Answers:

$49\pi$

$98\pi$

$392\pi$

$56\pi$

$8\pi$

Correct answer:

$392\pi$

Explanation:

The rate of change of the radius of a sphere is given by the equation $8e^{-2t}$ . Written in mathematical terms

$\frac{dr(t)}{dt}=8e^{-2t}$

The formula for the length of the radius can be found by taking the integral of this function with respect to time

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-4e^{-2t}+C$

This constant of integration can be found by utilizing the initial condition

$r(0)=-4e^{0}+C=4$

-4+C=4

C=8

$r(t)=8-4e^{-2t}$

The volume of a sphere is given by the equation

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

The rate of change of the volume can be found by taking the derivative of this equation with respect to time

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

$\frac{dV(t)}{dt}=4 \pi (8-4e^{-2t})^2 (8e^{-2t})$

$\frac{dV(ln(2))}{dt}=4 \pi (8-4e^{-2ln(2)})^2 (8e^{-2ln(2)})$

$\frac{dV(ln(2))}{dt}=392\pi$

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Example Question #142 : Rate Of Change

The rate of change of the radius of a sphere is $9e^{-t}$ . If the sphere has an initial radius of , what is the rate of change of the sphere's surface area at time t=ln(3) ?

Possible Answers:

$288\pi$

$18\pi$

$36\pi$

$144\pi$

$72\pi$

Correct answer:

$288\pi$

Explanation:

To say that the rate of change of the radius of a sphere is $9e^{-t}$ means

$\frac{dr(t)}{dt}=9e^{-t}$

The equation for the length of the radius can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$r(t)=-9e^{-t}+C$

The constant of integration can be found by utilizing the initial condition:

$r(0)=-9e^{0}+C=6$

-9+C=6

C=15

$r(t)=15-9e^{-t}$

The surface area of a sphere is given by the equation

$A=4\pi r^2$

The rate of change of this area can be found by taking the derivative of the equation with respect to time:

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

$\frac{dA(t)}{dt}=8\pi (15-9e^{-t}) (9e^{-t})$

$\frac{dA(ln(3))}{dt}=8\pi (15-9e^{-ln(3)}) (9e^{-ln(3)})$

$\frac{dA(ln(3))}{dt}=288\pi$

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Example Question #143 : Rate Of Change

The rate of growth of the sides of a cube is $16e^{-4t}$ . If the cube has an initial volume of , what is the rate of growth of the surface area at time $t=ln(\sqrt{2})$ ?

Possible Answers:

336

672

112

Correct answer:

336

Explanation:

Saying that the rate of growth of the sides of a cube is $16e^{-4t}$ , means in mathematical terms

$\frac{ds(t)}{dt}=16e^{-4t}$

The length of the sides at any time can be found by integrating this equation with respect to time:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$s(t)=-4e^{-4t}+C$

To find this constant of integration, we'll need the value for the length of the sides at some known point of time. We're told the initial volume; we can use this to find the initial lengths of the cube's sides:

V_0=s_0^3=64

s_0=4

$s(0)=-4e^{-0}+C=4$

-4+C=4

C=8

$s(t)=8-4e^{-4t}$

The surface area of a cube is given by the equation

A=6s^2

The rate of change of the area can be found by taking the derivative of this equation with respect to time:

$\frac{dA}{dt}=12s\frac{ds}{dt}$

$\frac{dA(t)}{dt}=12(8-4e^{-4t})(16e^{-4t})$

$\frac{dA(ln(\sqrt{2}))}{dt}=12(8-4e^{-4ln(\sqrt{2})})(16e^{-4ln(\sqrt{2})})$

$\frac{dA(ln(\sqrt{2}))}{dt}=336$

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Example Question #144 : Rate Of Change

The rate of change of the sides of a square is $6e^{-2t}$ . If the sides have initial lengths of , what is the rate of growth of the square's area at time $t=ln(\sqrt{2})$ ?

Possible Answers:

$\frac{9}{4}$

$\frac{25}{4}$

$\frac{25}{8}$

$\frac{75}{4}$

$\frac{3}{4}$

Correct answer:

$\frac{75}{4}$

Explanation:

We're told that the rate of growth of the sides of the square is $6e^{-2t}$

Which is to say $\frac{ds(t)}{dt}=6e^{-2t}$

The function for the sides can thus be found by integrating this function:

$\int e^{nt}dt=\frac{e^{nt}}{n}+C$

$s(t)=-3e^{-2t}+C$

To find this constant of integration, use the initial condition

s(0)=-3^0+C=4

-3+C=4

C=7

$s(t)=-3e^{-2t}+7$

Now, consider the area of a square

A=s^2

The rate of change of the area can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dA}{dt}=2s\frac{ds}{dt}$

$\frac{dA(t)}{dt}=2(-3e^{-2t}+7)(6e^{-2t})$

$\frac{dA(ln(2))}{dt}=2(-3e^{-2ln(2)}+7)(6e^{-2ln(2)})$

$\frac{dA(ln(2))}{dt}=\frac{75}{4}$

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Example Question #3051 : Calculus

Find the derivative of the following function.

f(x)=3x^4+2x^2-17x+15

Possible Answers:

12x^3+4x^2-17

3x^4+2x^2-17x+15

12x^3+2x^2-17x+15

12x^3+4x-17

12x^3+4x

Correct answer:

12x^3+4x-17

Explanation:

Because this function can be written in four seperate terms of addition (3x^4, 2x^2, -17x, 15) we are allowed to take the derivative of each component individually.

Using the power rule $(\frac{\mathrm{d} }{\mathrm{d} x}(x^n)=nx^{n-1})$ , we can derive each component individually to give us our answer,

12x^3+4x-17 (the 15 disappears because it is just a constant)

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Example Question #3052 : Calculus

Find the average rate of change of the function given by

f(x)=3x^2+sin(x) from x=0 to $x=\pi$

Possible Answers:

$-3\pi$

$3\pi^2$

$3\pi$

Correct answer:

$3\pi$

Explanation:

The average rate of change $\Delta f$ is given by

$\Delta f=\frac{f(b)-f(a)}{b-a}$ , where and are the boundaries between which we're trying to determine the average.

For this function, $b=\pi$ and a=0

f(x)=3x^2+sin(x)

Plugging into the equation, the average rate of change $\Delta f$ can be given by:

$\Delta f=\frac{f(\pi)-f(0)}{\pi-0}$

$\Delta f=\frac{(3(\pi )^2+sin(\pi))-(3(0)^2+sin(0))}{\pi-0}=\frac{3\pi^2}{\pi}=3\pi$

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Example Question #3053 : Calculus

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the radius?

Possible Answers:

$3\pi$

$\frac{9}{\pi}$

Correct answer:

Explanation:

Let's begin by writing the equations for the volumeof a sphere with respect to the sphere's radius:

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

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, let's solve for a radius that satisfies it the problem statement: What is the surface area of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the radius?

$4\pi r^2 \frac{dr}{dt}=(3) \frac{dr}{dt}$

$r^2=\frac{3}{4\pi}$

$r=\sqrt{\frac{3}{4\pi}}$

Now to solve for the surface area:

$A=4\pi r^2$

$A=4\pi (\sqrt{\frac{3}{4\pi}})^2$

A=3

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Example Question #148 : Rate Of Change

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is four times the rate of growth of the surface area?

Possible Answers:

Correct answer:

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

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

$A=4\pi r^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, let's solve for a radius that satisfies the question: What is the radius of the sphere at the instance the rate of growth of the volume is four times the rate of growth of the surface area?

$4\pi r^2 \frac{dr}{dt}=(4)8\pi r \frac{dr}{dt}$

r =(4)2

r=8

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Example Question #141 : Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the surface area?

Possible Answers:

$144\pi$

$288\pi$

$312\pi$

$72\pi$

$6\pi$

Correct answer:

$288\pi$

Explanation:

Let's begin by writing the equations for the volume and surface area of a sphere with respect to the sphere's radius:

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

$A=4\pi r^2$

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, let's solve for a radius that satisfies the question given: What is the volume of the sphere at the instance the rate of growth of the volume is three times the rate of growth of the surface area?

$4\pi r^2 \frac{dr}{dt}=(3)8\pi r \frac{dr}{dt}$

r =(3)2

r=6

Now to solve for the volume:

$V=\frac{4}{3}\pi (6)^3=288\pi$

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Example Question #141 : Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to four times the rate of growth of one of its sides?

Possible Answers:

$2\sqrt{3}$

$4\sqrt{3}$

Correct answer:

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to four times the rate of growth of one of its sides:

$3s^2\frac{ds}{dt}=4\frac{ds}{dt}$

$s^2=\frac{4}{3}$

$s=\frac{2}{\sqrt{3}}$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$

$d=\sqrt{3\frac{4}{3}}$

$d=\sqrt{4}$

d=2

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3051 : Calculus

Example Question #142 : Rate Of Change

Example Question #143 : Rate Of Change

Example Question #144 : Rate Of Change

Example Question #3051 : Calculus

Example Question #3052 : Calculus

Example Question #3053 : Calculus

Example Question #148 : Rate Of Change

Example Question #141 : Rate Of Change

Example Question #141 : Rate Of Change

All Calculus 1 Resources

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