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Calculus 1 : Rate of Change

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Example Questions

Example Question #521 : Rate Of Change

A regular tetrahedron is decreasing in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length 30?

Possible Answers:

$15\sqrt{2}$

$90\sqrt{2}$

$30\sqrt{2}$

$120\sqrt{2}$

$60\sqrt{2}$

Correct answer:

$90\sqrt{2}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its surface area and height, in terms of the length of its sides:

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

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

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

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

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

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

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

$\phi=3(30)\sqrt{2}=90\sqrt{2}$

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Example Question #3435 : Calculus

A regular tetrahedron is decreasing in size. What is the ratio of the rate of change of the surface area of the tetrahedron to the rate of change of its height when its sides have length $\frac{\sqrt{2}}{9}$ ?

Possible Answers:

$\frac{2}{9}$

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

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

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

$\frac{2}{3}$

Correct answer:

$\frac{2}{3}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its surface area and height, in terms of the length of its sides:

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

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

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

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

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

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

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

$\phi=3(\frac{\sqrt{2}}{9})\sqrt{2}=\frac{2}{3}$

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Example Question #521 : How To Find Rate Of Change

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its surface area and height, in terms of the length of its sides:

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

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

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

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

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

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

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

$\phi=3(\frac{\sqrt{3}}{12})\sqrt{2}=\frac{\sqrt{6}}{4}$

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Example Question #3437 : Calculus

Possible Answers:

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

$\frac{1}{6}$

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

$\frac{\sqrt{6}}{12}$

$\frac{1}{12}$

Correct answer:

$\frac{1}{6}$

Explanation:

To solve this problem, define a regular tetrahedron's dimensions, its surface area and height, in terms of the length of its sides:

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

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

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

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

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

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering; $\frac{ds}{dt}$ is $\frac{ds}{dt}$ . Find the ratio by dividing quantities:

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

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

$\phi=3(\frac{\sqrt{2}}{36})\sqrt{2}=\frac{1}{6}$

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Example Question #3438 : Calculus

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 200?

Possible Answers:

400

100

200

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{200}{4}=50$

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Example Question #3439 : Calculus

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 164?

Possible Answers:

10.25

20.5

5.125

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{164}{4}=41$

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Example Question #3431 : Calculus

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 316?

Possible Answers:

39.5

316

632

158

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{316}{4}=79$

Report an Error

Example Question #522 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 432?

Possible Answers:

108

360

180

216

Correct answer:

108

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{432}{4}=108$

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Example Question #521 : How To Find Rate Of Change

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 520?

Possible Answers:

260

130

325

650

Correct answer:

130

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{520}{4}=130$

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Example Question #2413 : Functions

A cube is growing in size. What is the ratio of the rate of growth of the cube's volume to the rate of growth of its surface area when its sides have length 924?

Possible Answers:

204

231

308

616

Correct answer:

231

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, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and surface area:

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

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

$\phi =\frac{924}{4}=231$

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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 #521 : Rate Of Change

Example Question #3435 : Calculus

Example Question #521 : How To Find Rate Of Change

Example Question #3437 : Calculus

Example Question #3438 : Calculus

Example Question #3439 : Calculus

Example Question #3431 : Calculus

Example Question #522 : How To Find Rate Of Change

Example Question #521 : How To Find Rate Of Change

Example Question #2413 : Functions

All Calculus 1 Resources

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