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Calculus 1 : Calculus

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Example Questions

Example Question #161 : How To Find Rate Of Change

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

Possible Answers:

$8\pi$

$\frac{1 }{2}$

$2\pi$

Correct answer:

$\frac{1 }{2}$

Explanation:

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

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

D=2r

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

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

$\frac{dD}{dt}=2\frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of shrinkage of the diameter is four times times the rate of shrinkage of the volume, let's solve for a radius that satisfies it.

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

$r^2 =\frac{2}{16\pi}$

$r=\sqrt{\frac{1}{8\pi}}$

The surface area of a sphere is

$A=4\pi r^2$

$A=4\pi (\sqrt{\frac{1}{8\pi}})^2$

$A=\frac{4\pi }{8\pi}$

$A=\frac{1 }{2}$

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Example Question #162 : How To Find Rate Of Change

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

Possible Answers:

$\frac{\pi}{1296}$

$\frac{1}{10368\pi}$

$\frac{1}{10368\pi^2}$

$\frac{\pi}{10368}$

$\frac{1}{1296\pi^2}$

Correct answer:

$\frac{1}{1296\pi^2}$

Explanation:

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

D=2r

$A=4\pi r^2$

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

$\frac{dD}{dt}=2 \frac{dr}{dt}$

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

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 diameter is three times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

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

$r =\frac{2}{24\pi}$

$r=\frac{1}{12\pi}$

The volume of a sphere is given by the equation:

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

$V=\frac{4}{3}\pi (\frac{1}{12\pi})^3$

$V=\frac{1}{1296\pi^2}$

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Example Question #162 : 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 half the rate of shrinkage of the surface area?

Possible Answers:

0.5

$\pi$

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 half the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{2})8\pi r \frac{dr}{dt}$

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

r^2=r

r=1

The diameter is simply twice the radius:

D=2

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Example Question #171 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the length of the tetrahedron's sides at the time the rate of growth of its volume is a third the rate of growth of its surface area?

Possible Answers:

$\frac{4\sqrt{6}}{3}$

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

$3\sqrt{6}$

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

$\frac{3\sqrt{6}}{4}$

Correct answer:

$\frac{4\sqrt{6}}{3}$

Explanation:

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

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

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

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

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

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 volume is a third the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

$s^2=(\frac{1}{3})4\sqrt{6}s$

$s=(\frac{1}{3})4\sqrt{6}$

$s=\frac{4\sqrt{6}}{3}$

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Example Question #172 : Rate Of Change

A regular tetrahedron is growing in size. What is the volume of the tetrahedron at the time the rate of growth of its volume is a $\sqrt{6}$ times the rate of growth of its surface area?

Possible Answers:

1152

$\frac{576}{\sqrt{2}}$

$576\sqrt{2}$

$\frac{1152}{\sqrt{2}}$

$1152\sqrt{2}$

Correct answer:

$1152\sqrt{2}$

Explanation:

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

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

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

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

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

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 volume is a $\sqrt{6}$ times the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\sqrt{6})2\sqrt{3}s\frac{ds}{dt}$

$s^2=(\sqrt{6})4\sqrt{6}s$

$s=(\sqrt{6})4\sqrt{6}$

s=24

To find the volume then:

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

$V=\frac{24^3}{6\sqrt{2}}$

$V=1152\sqrt{2}$

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Example Question #173 : 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 voume is a fourth the rate of growth of its sides?

Possible Answers:

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

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

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

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

$4\sqrt{6}$

Correct answer:

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

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 voume is a fourth the rate of growth of its sides, solve for the corresponding length of the tetrahedron's sides:

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

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

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

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

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

The height of a tetrahedron is given by the equation:

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

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

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

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Example Question #172 : How To Find Rate Of Change

A regular tetrahedron is growing in size. What is the length of the tetrahedron's sides at the time the rate of growth of its height is twice the rate of growth of its surface area?

Possible Answers:

$4\sqrt{2}$

$4\sqrt{3}$

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

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

$\frac{\sqrt{2}}{48}$

Correct answer:

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

Explanation:

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

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

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

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

$\frac{dh}{dt}=\frac{\sqrt{6}}{3}\frac{ds}{dt}$

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

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 height is twice the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{\sqrt{6}}{3}\frac{ds}{dt}=(2)2\sqrt{3}s\frac{ds}{dt}$

$\frac{\sqrt{6}}{3}=4\sqrt{3}s$

$s=\frac{\sqrt{2}}{12}$

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Example Question #173 : How To Find Rate Of Change

The radius of a sphere is equal to the length of a cube's side. If the radius of the sphere begins to grow at twice the rate of the cube's side, by what factor is the rate of change of the sphere's surface area different than the rate of change of the cube's surface area?

Possible Answers:

$\frac{2\pi}{3}$

$\frac{\pi}{3}$

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

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

$\frac{3}{2\pi}$

Correct answer:

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

Explanation:

Since this problem involves the comparison of surface areas, begin by writing the surface area equations of the two shapes:

$A_s=4\pi r^2$

A_c=6s^2

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

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

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

Now we're given a relationship in the problem statement, namely that $\frac{dr}{dt}=2\frac{ds}{dt}$ and r=s . Rewrite accordingly:

$\frac{dA_s}{dt}=16\pi s \frac{ds}{dt}$

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

The factor by which the rate of change of the sphere is different can be found by taking the ratio of these two values:

$\phi = \frac{16\pi s\frac{ds}{dt}}{12s\frac{ds}{dt}}=\frac{4\pi}{3}$

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Example Question #174 : How To Find Rate Of Change

The radius of a sphere is equal to half the length of a cube's side. If the radius of the sphere begins to grow at twice the rate of the cube's side, by what factor is the rate of change of the sphere's volume different than the rate of change of the cube's volume?

Possible Answers:

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

$\frac{2\pi}{3}$

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

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

$\frac{3}{2\pi}$

Correct answer:

$\frac{2\pi}{3}$

Explanation:

Since this problem involves the comparison of volumes, begin by writing the volume equations of the two shapes:

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

V_c=s^3

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

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

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

Now we're given a relationship in the problem statement, namely that $\frac{dr}{dt}=2\frac{ds}{dt}$ and $r=\frac{1}{2}s$ . Rewrite accordingly:

$\frac{dV_s}{dt}=4\pi(\frac{1}{2}s)^2 (2\frac{ds}{dt})=2\pi s^2\frac{ds}{dt}$

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

The factor by which the rate of change of the sphere is different can be found by taking the ratio of these two values:

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

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Example Question #3081 : Calculus

The width of a rectangle increases three times as fast as its length. How does the rate of change of the rectangle's area compare to that of the rate of change of the width when the length of the rectangle is half of its width?

Possible Answers:

$\frac{1}{3}w$

$\frac{5}{6}w$

$\frac{3}{2}w$

$\frac{2}{3}w$

Correct answer:

$\frac{5}{6}w$

Explanation:

Begin by writing the expression for the area of a rectangle:

A=wl

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

$\frac{dA}{dt}=w\frac{dl}{dt}+l\frac{dw}{dt}$

Now, we're told two things:

The width of a rectangle increases three times as fast as its length: $\frac{dw}{dt}=3\frac{dl}{dt}$

The length of the rectangle is half of its width: $l=\frac{1}{2}w$

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

$\frac{dA}{dt}=w(\frac{1}{3}\frac{dw}{dt})+(\frac{1}{2}w)\frac{dw}{dt}$

$\frac{dA}{dt}=\frac{5}{6}w\frac{dw}{dt}$

The rate of change of the area is $\frac{5}{6}w$ times the rate of change of the rate of change of the width.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #161 : How To Find Rate Of Change

Example Question #162 : How To Find Rate Of Change

Example Question #162 : Rate Of Change

Example Question #171 : How To Find Rate Of Change

Example Question #172 : Rate Of Change

Example Question #173 : Rate Of Change

Example Question #172 : How To Find Rate Of Change

Example Question #173 : How To Find Rate Of Change

Example Question #174 : How To Find Rate Of Change

Example Question #3081 : Calculus

All Calculus 1 Resources

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