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Calculus 1 : How to find rate of change

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Example Questions

Example Question #721 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length $\frac{14}{9}$ $\displaystyle \frac{14}{9}$ ?

Possible Answers:

$\frac{7}{3}$ $\displaystyle \frac{7}{3}$

$\frac{7}{6}$ $\displaystyle \frac{7}{6}$

$\displaystyle 7$

$\frac{14}{3}$ $\displaystyle \frac{14}{3}$

$\frac{7}{9}$ $\displaystyle \frac{7}{9}$

Correct answer:

$\frac{7}{3}$ $\displaystyle \frac{7}{3}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

$\phi =\frac{3(\frac{14}{9})}{2}=\frac{7}{3}$ $\displaystyle \phi =\frac{3(\frac{14}{9})}{2}=\frac{7}{3}$

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Example Question #731 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length $\frac{16}{3}$ $\displaystyle \frac{16}{3}$ ?

Possible Answers:

$\frac{2}{9}$ $\displaystyle \frac{2}{9}$

$\frac{16}{9}$ $\displaystyle \frac{16}{9}$

$\frac{4}{3}$ $\displaystyle \frac{4}{3}$

$\frac{8}{3}$ $\displaystyle \frac{8}{3}$

$\displaystyle 8$

Correct answer:

$\displaystyle 8$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

$\phi =\frac{3(\frac{16}{3})}{2}=8$ $\displaystyle \phi =\frac{3(\frac{16}{3})}{2}=8$

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Example Question #732 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length $\frac{31}{6}$ $\displaystyle \frac{31}{6}$ ?

Possible Answers:

$\frac{31}{24}$ $\displaystyle \frac{31}{24}$

$\frac{93}{4}$ $\displaystyle \frac{93}{4}$

$\frac{31}{4}$ $\displaystyle \frac{31}{4}$

$\frac{31}{2}$ $\displaystyle \frac{31}{2}$

$\frac{93}{8}$ $\displaystyle \frac{93}{8}$

Correct answer:

$\frac{31}{4}$ $\displaystyle \frac{31}{4}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

$\phi =\frac{3(\frac{31}{6})}{2}=\frac{31}{4}$ $\displaystyle \phi =\frac{3(\frac{31}{6})}{2}=\frac{31}{4}$

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Example Question #733 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length $\frac{64}{27}$ $\displaystyle \frac{64}{27}$ ?

Possible Answers:

$\frac{16}{9}$ $\displaystyle \frac{16}{9}$

$\frac{32}{9}$ $\displaystyle \frac{32}{9}$

$\frac{8}{9}$ $\displaystyle \frac{8}{9}$

$\frac{4}{3}$ $\displaystyle \frac{4}{3}$

$\frac{16}{3}$ $\displaystyle \frac{16}{3}$

Correct answer:

$\frac{32}{9}$ $\displaystyle \frac{32}{9}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

$\phi =\frac{3(\frac{64}{27})}{2}=\frac{32}{9}$ $\displaystyle \phi =\frac{3(\frac{64}{27})}{2}=\frac{32}{9}$

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Example Question #734 : Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of the area of a single face when its sides have length $\frac{22}{45}$ $\displaystyle \frac{22}{45}$ ?

Possible Answers:

$\frac{13}{15}$ $\displaystyle \frac{13}{15}$

$\frac{11}{90}$ $\displaystyle \frac{11}{90}$

$\frac{3}{5}$ $\displaystyle \frac{3}{5}$

$\frac{13}{90}$ $\displaystyle \frac{13}{90}$

$\frac{11}{15}$ $\displaystyle \frac{11}{15}$

Correct answer:

$\frac{11}{15}$ $\displaystyle \frac{11}{15}$

Explanation:

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

V=s^3 $\displaystyle V=s^3$

a=s^2 $\displaystyle a=s^2$

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

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

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

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

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

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

$\phi =\frac{3(\frac{22}{45})}{2}=\frac{11}{15}$ $\displaystyle \phi =\frac{3(\frac{22}{45})}{2}=\frac{11}{15}$

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Example Question #735 : Rate

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 diagonal when its sides have length $3\sqrt{5}$ $\displaystyle 3\sqrt{5}$ ?

Possible Answers:

$15\sqrt{15}$ $\displaystyle 15\sqrt{15}$

$15\sqrt{3}$ $\displaystyle 15\sqrt{3}$

$3\sqrt{15}$ $\displaystyle 3\sqrt{15}$

$45\sqrt{3}$ $\displaystyle 45\sqrt{3}$

$9\sqrt{15}$ $\displaystyle 9\sqrt{15}$

Correct answer:

$45\sqrt{3}$ $\displaystyle 45\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(3\sqrt{5})^2\sqrt{3}=45\sqrt{3}$ $\displaystyle \phi =(3\sqrt{5})^2\sqrt{3}=45\sqrt{3}$

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Example Question #736 : Rate

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 diagonal when its sides have length 17?

Possible Answers:

289 $\displaystyle 289$

$289\sqrt{3}$ $\displaystyle 289\sqrt{3}$

867 $\displaystyle 867$

$17\sqrt{3}$ $\displaystyle 17\sqrt{3}$

$\displaystyle 51$

Correct answer:

$289\sqrt{3}$ $\displaystyle 289\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(17)^2\sqrt{3}=289\sqrt{3}$ $\displaystyle \phi =(17)^2\sqrt{3}=289\sqrt{3}$

Report an Error

Example Question #737 : Rate

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 diagonal when its sides have length $4\sqrt{2}$ $\displaystyle 4\sqrt{2}$ ?

Possible Answers:

$16\sqrt{6}$ $\displaystyle 16\sqrt{6}$

$32\sqrt{3}$ $\displaystyle 32\sqrt{3}$

$16\sqrt{3}$ $\displaystyle 16\sqrt{3}$

$8\sqrt{3}$ $\displaystyle 8\sqrt{3}$

$32\sqrt{6}$ $\displaystyle 32\sqrt{6}$

Correct answer:

$32\sqrt{3}$ $\displaystyle 32\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(4\sqrt{2})^2\sqrt{3}=32\sqrt{3}$ $\displaystyle \phi =(4\sqrt{2})^2\sqrt{3}=32\sqrt{3}$

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Example Question #738 : Rate

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 diagonal when its sides have length $5\sqrt{7}$ $\displaystyle 5\sqrt{7}$ ?

Possible Answers:

$25\sqrt{21}$ $\displaystyle 25\sqrt{21}$

$175\sqrt{21}$ $\displaystyle 175\sqrt{21}$

$35\sqrt{3}$ $\displaystyle 35\sqrt{3}$

$175\sqrt{3}$ $\displaystyle 175\sqrt{3}$

$25\sqrt{3}$ $\displaystyle 25\sqrt{3}$

Correct answer:

$175\sqrt{3}$ $\displaystyle 175\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(5\sqrt{7})^2\sqrt{3}=175\sqrt{3}$ $\displaystyle \phi =(5\sqrt{7})^2\sqrt{3}=175\sqrt{3}$

Report an Error

Example Question #739 : Rate

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 diagonal when its sides have length $4\sqrt{11}$ $\displaystyle 4\sqrt{11}$ ?

Possible Answers:

$22\sqrt{3}$ $\displaystyle 22\sqrt{3}$

$16\sqrt{33}$ $\displaystyle 16\sqrt{33}$

$176\sqrt{33}$ $\displaystyle 176\sqrt{33}$

$176\sqrt{3}$ $\displaystyle 176\sqrt{3}$

$44\sqrt{6}$ $\displaystyle 44\sqrt{6}$

Correct answer:

$176\sqrt{3}$ $\displaystyle 176\sqrt{3}$

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 $\displaystyle V=s^3$

$d=s\sqrt{3}$ $\displaystyle d=s\sqrt{3}$

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

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

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

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

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

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

$\phi =(4\sqrt{11})^2\sqrt{3}=176\sqrt{3}$ $\displaystyle \phi =(4\sqrt{11})^2\sqrt{3}=176\sqrt{3}$

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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 #721 : Rate

Example Question #731 : Rate

Example Question #732 : Rate

Example Question #733 : Rate

Example Question #734 : Rate

Example Question #735 : Rate

Example Question #736 : Rate

Example Question #737 : Rate

Example Question #738 : Rate

Example Question #739 : Rate

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