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Calculus 1 : Rate of Change

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Example Questions

Example Question #191 : How To Find Rate Of Change

If the position of an object at time is represented by the function p(t) = 3t^2-12t+9 , when does the object stop moving (i.e. The velocity is zero)?

Possible Answers:

t=3

t=1 or t=3

The velocity is never .

t=1

t=2

Correct answer:

t=2

Explanation:

When the velocity is , that means p'(t)=0 . That is, 6t-12=0 . So t=2 .

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Example Question #192 : How To Find Rate Of Change

The width of a rectangular prism increases half as fast as its length and a third as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the length, height, and width are equal?

Possible Answers:

$\frac{1}{4}w^2$

5w^2

6w^2

$\frac{1}{6}w^2$

3w^2

Correct answer:

6w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

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

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases half as fast as its length and a third as fast as its height: $\frac{dw}{dt}=\frac{1}{2}\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{3}\frac{dh}{dt}$

The width when the length, height, and width are equal: w=l=h

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w^2(2\frac{dw}{dt})+w^2(3\frac{dh}{dt})+w^2\frac{dw}{dt}$

$\frac{dV}{dt}=6w^2\frac{dw}{dt}$

The rate of change of the volume is 6w^2 times the rate of change of the rate of change of the width.

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Example Question #193 : How To Find Rate Of Change

The width of a rectangular prism increases twice as fast as its length and half as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the width is half the length, which is half the height?

Possible Answers:

20w^2

22w^2

14w^2

9w^2

7w^2

Correct answer:

14w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

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

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases twice as fast as its length and half as fast as its height: $\frac{dw}{dt}=2\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{2}\frac{dh}{dt}$

The width is half the length, which is half the height: $w=\frac{1}{2}l,l=\frac{1}{2}h\rightarrow w=\frac{1}{4}h$

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w(4w)(\frac{1}{2}\frac{dw}{dt})+w(2w)(2\frac{dw}{dt})+(4w)(2w)\frac{dw}{dt}$

$\frac{dV}{dt}=2w^2\frac{dw}{dt}+4w^2\frac{dw}{dt}+8w^2\frac{dw}{dt}$

$\frac{dV}{dt}=14w^2\frac{dw}{dt}$

The rate of change of the volume is 14w^2 times the rate of change of the rate of change of the width.

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Example Question #194 : How To Find Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameter of the sphere at the instance the rate of shrinkage of the volume is equal to the rate of shrinkage of the surface area?

Possible Answers:

$8\pi$

$4\pi$

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 can be found by taking the derivative of each side of the equation 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, the rate of shrinkage of the volume is equal to the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

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

r =2

Since the diameter is twice this, d=4

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Example Question #195 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its sides is equal to the rate of growth of its volume?

Possible Answers:

$\frac{\sqrt{3}}{\sqrt[4]{2}}$

$\frac{2\sqrt[4]{2}}{{3}}$

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

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

$\frac{6\sqrt[4]{2}}{\sqrt{3}}$

Correct answer:

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

Rates of change can then be found by taking the derivative of each property with respect to time:

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

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering, so given our problem condition, the rate of growth of its sides is equal to the rate of growth of its volume, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=\frac{ds}{dt}$

$s^2=2\sqrt{2}=\sqrt{8}$

$s=\sqrt[4]{8}$

The height of a tetrahedron is given by the equation:

$h=\frac{\sqrt{6}}{3}s$

$h=\frac{\sqrt{6}}{3}\sqrt[4]{8}=\frac{2\sqrt[4]{2}}{\sqrt{3}}$

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Example Question #196 : How To Find Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's surface is equal to 1.5 times the rate of growth of its volume?

Possible Answers:

$\frac{3}{4}$

$\frac{6}{5}$

$\frac{4}{3}$

Correct answer:

$\frac{4}{3}$

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 surface is equal to 1.5 times the rate of growth of its volume:

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

$s=\frac{12}{9}=\frac{4}{3}$

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Example Question #197 : How To Find Rate Of Change

A cube is growing in size. What is the length of the sides 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 diagonal?

Possible Answers:

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

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

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

$\frac{\sqrt[4]{3}}{4}$

$\frac{1}{\sqrt[4]{3}}$

Correct answer:

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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 diagonal:

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

$s^2=\frac{(2)\sqrt{3}}{3}=\frac{2}{\sqrt{3}}$

$s=\frac{\sqrt{2}}{\sqrt[4]{3}}$

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Example Question #198 : How To Find Rate Of Change

A cube is growing in size. What is the area of one of the cube's faces at the time that the rate of growth of the cube's volume is equal to twice the rate of growth of its sides?

Possible Answers:

$6\sqrt{\frac{2}{3}}$

$\frac{2}{3}$

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

Correct answer:

$\frac{2}{3}$

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 twice the rate of growth of its sides

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

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

The area of one of those cube's faces is the length of its sides squared:

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

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Example Question #197 : How To Find Rate Of Change

A square is growing in area. What is the length of the square's sides at the moment that the rate of growth of the area is the same as the rate of growth of the square's diagonal?

Possible Answers:

$\frac{1}{2}$

$\sqrt{2}$

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

Correct answer:

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

Explanation:

Start by writing the equations for a square's dimensions.

A=s^2

$d=s\sqrt{2}$

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

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

$\frac{dd}{dt}=\sqrt{2}\frac{ds}{dt}$

Now, solve for the length of the side of the square to satisfy the problem condition, the rate of growth of the area is the same as the rate of growth of the square's diagonal:

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

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

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Example Question #198 : How To Find Rate Of Change

Use normal rules of differentiation to find the derrivative f'(0) of the function

$f(x)=2^{5x}+(2x^2+3)^3$

Possible Answers:

$5\ln 2$

$2 \ln 5$

$-5 \ln 2$

$-2 \ln 5$

Correct answer:

$5\ln 2$

Explanation:

Taking the derrivative,

$f'(x)=\frac{d}{dx}[2^{5x}+(2x^2+3)^3]$

We separate the summed terms:

$=\frac{d}{dx}[2^{5x}]+\frac{d}{dx}[(2x^2+3)^3]$

Using the chain rule:

$=2^{5x}\ln 2\frac{d}{dx}[5x]+2(2x^2+3)^2\frac{d}{dx}[(2x^2+3)]$

$=2^{5x}\ln 2*5+3(2x^2+3)^2* 4x$

Simplifying

$=2^{5x}5\ln 2+12x(2x^2+3)^2$

Now, evaluating at f'(0) :

$=2^{5(0)}5\ln 2+12(0)(2(0)^2+3)^2$

$=(1)5 \ln 2+(0)$

$=5 \ln 2$

which is our answer.

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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 #191 : How To Find Rate Of Change

Example Question #192 : How To Find Rate Of Change

Example Question #193 : How To Find Rate Of Change

Example Question #194 : How To Find Rate Of Change

Example Question #195 : How To Find Rate Of Change

Example Question #196 : How To Find Rate Of Change

Example Question #197 : How To Find Rate Of Change

Example Question #198 : How To Find Rate Of Change

Example Question #197 : How To Find Rate Of Change

Example Question #198 : How To Find Rate Of Change

All Calculus 1 Resources

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