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Calculus 1 : How to find rate of change

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Example Questions

Example Question #151 : How To Find 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 half the rate of growth of one of its sides?

Possible Answers:

$\frac{1}{2}$

$\frac{1}{6}$

$\sqrt{\frac{1}{2}}$

$\frac{\sqrt{3}}{6}$

Correct answer:

$\sqrt{\frac{1}{2}}$

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 half the rate of growth of one of its sides:

$3s^2\frac{ds}{dt}=\frac{1}{2}\frac{ds}{dt}$

$s^2=\frac{1}{6}$

$s=\sqrt{\frac{1}{6}}$

The diagonal of a cube is given by the equation:

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

$d=\sqrt{\frac{3}{6}}$

$d=\sqrt{\frac{1}{2}}$

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Example Question #152 : How To Find 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 surface area is equal to five times the rate of growth of the length of a side?

Possible Answers:

$13\sqrt{3}$

$\frac{12}{5}\sqrt{3}$

$\frac{5}{12}\sqrt{3}$

$\frac{5}{12}\sqrt{2}$

$13\sqrt{2}$

Correct answer:

$\frac{5}{12}\sqrt{3}$

Explanation:

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

A=6s^2

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

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

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's surface area is equal to five times the rate of growth of the length of a side:

$12s\frac{ds}{dt}=5\frac{ds}{dt}$

$s=\frac{5}{12}$

The diagonal of a cube is given by the equation:

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

$d=s\sqrt{3}$

$d=\frac{5}{12}\sqrt{3}$

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Example Question #151 : Rate Of Change

A cube is growing in size. What is the volume 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 its surface area?

Possible Answers:

512

2048

1024

4096

256

Correct answer:

4096

Explanation:

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

V=s^3

A=6s^2

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

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

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

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 its surface area

$3s^2\frac{ds}{dt}=(4)12s\frac{ds}{dt}$

s=(4)4

s=16

Finding the volume:

V=16^3

V=4096

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Example Question #3061 : Calculus

A cube is growing in size. What is the length of a side of the cube at the time that the rate of growth of the cube's volume is equal to thrice the rate of growth of its surface area?

Possible Answers:

Correct answer:

Explanation:

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

V=s^3

A=6s^2

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

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

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

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 thrice the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(3)12s\frac{ds}{dt}$

s=(3)4

s=12

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Example Question #151 : How To Find Rate Of Change

A cube is inexplicably growing in size, worrying several scientists. 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 twice the rate of growth of its surface area?

Possible Answers:

$8\sqrt{3}$

$4\sqrt{3}$

Correct answer:

$8\sqrt{3}$

Explanation:

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

V=s^3

A=6s^2

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

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

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

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 twice the rate of growth of its surface area

$3s^2\frac{ds}{dt}=(2)12s\frac{ds}{dt}$

s=(2)4

s=8

The diagonal of a cube is given by the equation:

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

$d=s\sqrt{3}$

$d=8\sqrt{3}$

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Example Question #151 : Rate Of Change

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 surface area is six times the rate of growth of the radius?

Possible Answers:

$12\pi$

$4\pi$

$\frac{9}{4\pi}$

$9\pi$

$\frac{9\pi}{4}$

Correct answer:

$\frac{9}{4\pi}$

Explanation:

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

$A=4\pi r^2$

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

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

The rate of change of the radius is going to be the same for the sphere, whether talking about the volume, surface area, or radius itself. So given our problem conditions, namely that the surface area of the sphere at the instance the rate of growth of the surface area is six times the rate of growth of the radius, let's solve for a radius that satisfies it.

$8\pi r \frac{dr}{dt}=6 \frac{dr}{dt}$

$r =\frac{6}{8\pi}$

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

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

$A=\frac{9}{4\pi}$

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Example Question #151 : How To Find Rate Of Change

Find the derivative of the function $f(x)=\frac{-7}{2} x^4+7x^3+6x^{-2}$ using normal rules of differentiation:

Possible Answers:

$f'(x)=7 x^2+\frac{7}{2}x^2+6x^{-3}$

$f'(x)=-14x^3+21x^2-12x^{-1}$

$f'(x)=\frac{7}{2} x^3-7x^2-12x^{-2}$

$f'(x)=21 x^3+14x^2-12x^{-3}$

$f'(x)=-14x^3+21x^2-12x^{-3}$

Correct answer:

$f'(x)=-14x^3+21x^2-12x^{-3}$

Explanation:

We'll try to find the derivative

$f'(x)=\frac{d}{dx}[\frac{-7}{2} x^4+7x^3+6x^{-2}]$

The opperation separating the terms is addition. Hence we can apply the derrivative on all of the terms separately.

$f'(x)=\frac{d}{dx}[\frac{-7}{2}x^4]+\frac{d}{dx}[7x^3]+\frac{d}{dx}[6x^{-2}]$

These terms are all polynomials, so we apply the rule $\frac{d}{dx}[x^a]=ax^{a-1}$ :

$=(\frac{-7}{2}*4 x^3a)+(7*3x^2)+(6*(-2)x^{-3})$

Simplifying, we have:

$=-14x^3+21x^2-12x^{-3}$ Which is our answer.

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Example Question #3071 : Calculus

Use normal rules of differentiation to find the derrivative of the function $f(x)=8x^2\ln x$ .

Possible Answers:

$16x\ln x -8x$

$4x(2 \ln x+1)$

$24\ln x(\frac{1}{x} +1)$

$8x(2\ln x +1)$

Correct answer:

$8x(2\ln x +1)$

Explanation:

We notice that this is the product of two functions, so we'll use the product rule.

$f'(x)=\frac{d}{dx}[8x^2\ln x]$

$=\frac{d}{dx}[8x^2] \ln x+8x^2 \frac{d}{dx}[\ln x]$

simplifying:

$=16x \ln x+8x^2 (\frac{1}{x})$

simplying:

$=16x \ln x+8x$

factoring out , we have

$=8x(2 \ln x+1)$

which is our answer.

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Example Question #152 : How To Find Rate Of Change

Use normal rules of differentiation to find the derrivative of the function $f(x)=\frac{\arctan x}{\tan x}$ .

Possible Answers:

$\frac{\cos^2x}{1+x^2}}$

This function is not differentiable at any point on its domain.

$\frac{\cos^2x- \arctan x}{\tan x (1+x^2)}$

$\frac{\sec x(\frac{1}{1+x^2})-sec^2x}{\tan^2x}$

$\cot x \csc x (\frac{\sin x}{1+x^2}-\arctan x \sec x)$

Correct answer:

$\cot x \csc x (\frac{\sin x}{1+x^2}-\arctan x \sec x)$

Explanation:

We take the derrivative of the function:

$f'(x)=\frac{d}{dx}[\frac{\arctan x}{\tan x}]$

We notice that this is the quotient of two functions, so we apply the quotient rule:

$=\frac{\frac{d}{dx}[\arctan x]\tan x -\arctan x\frac{d}{dx}{[\tan x}]}{\tan^2 x}$

$=\cot^2 x(\frac{1}{1+x^2}\tan x -\arctan x \sec^2 x)$

Factoring out a $\sec x$

$=\cot^2 x \sec x(\frac{\sin x}{1+x^2}-\arctan x \sec x)$ .

Simplifying, we have

$=\cot x \csc x(\frac{\sin x}{1+x^2}-\arctan x \sec x)$

which is our answer.

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Example Question #2041 : Functions

Use normal rules of differentiation to find the derrivative of the function

$f(x)=\frac{\arcsin x}{e^x}$ .

Possible Answers:

$e^{x}(\arcsin x-\frac{1}{\sqrt{1-x^2}})$

$-e^x \arcsin x$

$e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x)$

$\frac{1}{e^x}(\frac{1}{\sqrt{1-x^2}})$

This function is not differentiable at any point in its domain.

Correct answer:

$e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x)$

Explanation:

We take the derrivative

$f'(x)=\frac{d}{dx}[\frac{\arcsin x}{e^x}]$

This function is the quotient of two functions, so we use the quotient rule:

$= \frac{\frac{d}{dx}[\arcsin x] e^x-\arcsin x\frac{d}{dx}[e^x]}{(e^x)^2}$

$= \frac{\frac{1}{\sqrt{1-x^2}} e^x-\arcsin x e^x}{(e^{2x})}$

Simplifying, we have

$=e^{-x}(\frac{1}{\sqrt{1-x^2}}-\arcsin x )$

which is our answer.

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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 #151 : How To Find Rate Of Change

Example Question #152 : How To Find Rate Of Change

Example Question #151 : Rate Of Change

Example Question #3061 : Calculus

Example Question #151 : How To Find Rate Of Change

Example Question #151 : Rate Of Change

Example Question #151 : How To Find Rate Of Change

Example Question #3071 : Calculus

Example Question #152 : How To Find Rate Of Change

Example Question #2041 : Functions

All Calculus 1 Resources

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