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Calculus 1 : How to find rate of change

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Example Questions

Example Question #3131 : Calculus

The rate of change of a cylinder's radius is equal to the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is five times the height?

Possible Answers:

$\frac{19}{32}r$

$\frac{13}{32}r$

$\frac{7}{32}r$

$\frac{17}{32}r$

$\frac{23}{32}r$

Correct answer:

$\frac{7}{32}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to the rate of change of its height: $\frac{dr}{dt}=\frac{dh}{dt}$

The radius is five times the height r=5h

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(\frac{1}{5}r)\frac{dr}{dt}+\pi r^2 \frac{dr}{dt}=\frac{7\pi r^2}{5}\frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (\frac{1}{5}r) )\frac{dr}{dt}+2\pi r \frac{dr}{dt}=\frac{32\pi r}{5}\frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{\frac{7\pi r^2}{5}\frac{dr}{dt}}{\frac{32\pi r}{5}\frac{dr}{dt}}=\frac{7}{32}r$

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Example Question #222 : Rate Of Change

The rate of change of a cylinder's radius is equal to half the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is a third the height?

Possible Answers:

$\frac{1}{7}r$

$\frac{4}{7}r$

$\frac{2}{7}r$

$\frac{7}{2}r$

Correct answer:

$\frac{4}{7}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to half the rate of change of its height: $\frac{dr}{dt}=\frac{1}{2}\frac{dh}{dt}$

The radius is a third the height: $r=\frac{1}{3}h$

Using these properties, rewrite the rate equations:

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

$\frac{dA}{dt}=(4\pi r+2\pi (3r) )\frac{dr}{dt}+2\pi r (2\frac{dr}{dt})=14\pi r \frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{8\pi r^2\frac{dr}{dt}}{14\pi r \frac{dr}{dt}}=\frac{4}{7}r$

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Example Question #223 : Rate Of Change

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height. How does the rate of change of the cylinder's volume compare to the rate of change of its surface area when the radius is a fifth of the height?

Possible Answers:

$\frac{5}{8}r$

$\frac{3}{8}r$

$\frac{9}{8}r$

$\frac{1}{8}r$

$\frac{17}{8}r$

Correct answer:

$\frac{5}{8}r$

Explanation:

To approach this problem, begin by defining the cylinder's volume and surface area in terms of its height and radius:

$V=\pi r^2h$

$A=2\pi r^2 +2\pi rh$

Cylinderdimensions

Rates of change can be found by deriving, then, with respect to time:

$\frac{dV}{dt}=2\pi rh\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt}+2\pi r \frac{dh}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi h )\frac{dr}{dt}+2\pi r \frac{dh}{dt}$

We're told two things:

The rate of change of a cylinder's radius is equal to a fifth of the rate of change of its height: $\frac{dr}{dt}=\frac{1}{5}\frac{dh}{dt}$
The radius is a fifth of the height: $r=\frac{1}{5}h$

Using these properties, rewrite the rate equations:

$\frac{dV}{dt}=2\pi r(5r)\frac{dr}{dt}+\pi r^2 (5\frac{dr}{dt})=15\pi r^2 \frac{dr}{dt}$

$\frac{dA}{dt}=(4\pi r+2\pi (5r) )\frac{dr}{dt}+2\pi r (5\frac{dr}{dt})=24\pi r \frac{dr}{dt}$

The comparison between the volume and surface area can be found by taking the ratio of the two:

$\phi = \frac{15\pi r^2 \frac{dr}{dt}}{24\pi r \frac{dr}{dt}}=\frac{5}{8}r$

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Example Question #311 : Rate

A regular tetrahedron is growing in size. What is the length of the sides of the tetrahedron at the time the rate of growth of its volume is one thirteenth the rate of growth of its surface area?

Possible Answers:

$\frac{4\sqrt{6}}{13}$

$\frac{2\sqrt{6}}{13}$

$\frac{12}{13}$

$\frac{24}{13}$

$\frac{4}{13}$

Correct answer:

$\frac{4\sqrt{6}}{13}$

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 one thirteenth the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(\frac{1}{13})2\sqrt{3}s\frac{ds}{dt}$

$s^2=(\frac{1}{13})4\sqrt{6}s$

$s=\frac{4\sqrt{6}}{13}$

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Example Question #312 : Rate

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its volume is ten times the rate of growth of its surface area?

Possible Answers:

$80\sqrt{3}$

$120\sqrt{3}$

$40\sqrt{3}$

Correct answer:

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 ten times the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

$\frac{s^2}{2\sqrt{2}}\frac{ds}{dt}=(10)2\sqrt{3}s\frac{ds}{dt}$

$s^2=(10)4\sqrt{6}s$

$s=40\sqrt{6}$

The height of a tetrahedron is given by the equation:

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

$h=\frac{\sqrt{6}}{3}(40\sqrt{6}s)$

h=80

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Example Question #226 : 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 $\frac{3\sqrt{3}}{2}$ times rate of growth of its surface area?

Possible Answers:

$972\sqrt{2}$

$812\sqrt{2}$

1944

812

$456\sqrt{2}$

Correct answer:

1944

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 $\frac{3\sqrt{3}}{2}$ times rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

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

$s=18\sqrt{2}$

Now find the volume:

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

$V=\frac{(18\sqrt{2})^3}{6\sqrt{2}}$

V=1944

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Example Question #313 : Rate

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 thirteen times the rate of growth of its surface area?

Possible Answers:

$\sqrt{13}$

$2\sqrt{13}$

Correct answer:

Explanation:

Begin by writing s=52 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 thirteen times the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(13)12s\frac{ds}{dt}$

s=(13)4

s=52

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Example Question #3141 : Calculus

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 twice the rate of growth of the area of one of its sides??

Possible Answers:

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

$\frac{8\sqrt{3}}{3}$

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

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

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

Correct answer:

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

Explanation:

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

V=s^3

a=s^2

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

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

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

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 the area of one of its sides:

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

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

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

The diagonal of a cube is given by the equation:

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

$d=s\sqrt{3}$

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

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Example Question #229 : Rate Of Change

A cube is diminishing in size. What is the perimeter of one of the faces of the cube at the time that the rate of shrinkage of the cube's volume is equal to a sixith the rate of shrinkage of its surface area?

Possible Answers:

$\frac{4}{5}$

$\frac{8}{5}$

$\frac{2}{3}$

$\frac{8}{3}$

$\frac{4}{9}$

Correct answer:

$\frac{8}{3}$

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 shrinkage of the cube's volume is equal to a sixith the rate of shrinkage of its surface area

$3s^2\frac{ds}{dt}=(\frac{1}{6})12s\frac{ds}{dt}$

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

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

The perimter of the face of cube is given as:

P=4s

$P=\frac{8}{3}$

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Example Question #3142 : Calculus

A cube is diminishing in size. What is the length of the diagonal of the cube at the time that the rate of shrinkage of the cube's volume is equal to $5\sqrt{3}$ times the rate of shrinkage of the diagonal?

Possible Answers:

$5\sqrt{5}$

$\sqrt{15}$

$5\sqrt{3}$

$3\sqrt{15}$

$3\sqrt{5}$

Correct answer:

$\sqrt{15}$

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 shrinkage of the cube's volume is equal to $5\sqrt{3}$ times the rate of shrinkage of the diagonal:

$3s^2\frac{ds}{dt}=(5\sqrt{3})\sqrt{3}\frac{ds}{dt}$

s^2=5

$s=\sqrt{5}$

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

$d=\sqrt{15}$

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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 #3131 : Calculus

Example Question #222 : Rate Of Change

Example Question #223 : Rate Of Change

Example Question #311 : Rate

Example Question #312 : Rate

Example Question #226 : Rate Of Change

Example Question #313 : Rate

Example Question #3141 : Calculus

Example Question #229 : Rate Of Change

Example Question #3142 : Calculus

All Calculus 1 Resources

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