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Calculus 1 : Functions

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Example Questions

Example Question #2441 : Functions

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 128 times the rate of growth of the surface area?

Possible Answers:

$\displaystyle 64$

2048 $\displaystyle 2048$

1024 $\displaystyle 1024$

256 $\displaystyle 256$

512 $\displaystyle 512$

Correct answer:

256 $\displaystyle 256$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 128 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(128)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(128)8\pi r \frac{dr}{dt}$

$\displaystyle r =(128)2$

r=256 $\displaystyle r=256$

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Example Question #2442 : Functions

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 81 times the rate of growth of the surface area?

Possible Answers:

243 $\displaystyle 243$

162 $\displaystyle 162$

$\displaystyle 72$

144 $\displaystyle 144$

324 $\displaystyle 324$

Correct answer:

162 $\displaystyle 162$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 81 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(81)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(81)8\pi r \frac{dr}{dt}$

$\displaystyle r =(81)2$

r=162 $\displaystyle r=162$

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Example Question #2443 : Functions

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the volume is 116 times the rate of growth of the surface area?

Possible Answers:

232 $\displaystyle 232$

290 $\displaystyle 290$

$\displaystyle 58$

$\displaystyle 29$

145 $\displaystyle 145$

Correct answer:

232 $\displaystyle 232$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 116 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(116)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(116)8\pi r \frac{dr}{dt}$

$\displaystyle r =(116)2$

r=232 $\displaystyle r=232$

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Example Question #552 : Rate Of Change

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is 54 times the rate of growth of the surface area?

Possible Answers:

$\displaystyle 27$

216 $\displaystyle 216$

$\displaystyle 54$

243 $\displaystyle 243$

108 $\displaystyle 108$

Correct answer:

216 $\displaystyle 216$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 54 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(54)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(54)8\pi r \frac{dr}{dt}$

$\displaystyle r =(54)2$

r=108 $\displaystyle r=108$

The diameter is then

d=2r $\displaystyle d=2r$

d=216 $\displaystyle d=216$

Report an Error

Example Question #561 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is 73 times the rate of growth of the surface area?

Possible Answers:

1168 $\displaystyle 1168$

146 $\displaystyle 146$

$\displaystyle 73$

292 $\displaystyle 292$

584 $\displaystyle 584$

Correct answer:

292 $\displaystyle 292$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 73 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(73)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(73)8\pi r \frac{dr}{dt}$

$\displaystyle r =(73)2$

r=146 $\displaystyle r=146$

The diameter is then

d=2r $\displaystyle d=2r$

d=292 $\displaystyle d=292$

Report an Error

Example Question #2444 : Functions

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is $\frac{1}{4}$ $\displaystyle \frac{1}{4}$ times the rate of growth of the surface area?

Possible Answers:

$\displaystyle 4$

$\displaystyle 1$

$\frac{1}{4}$ $\displaystyle \frac{1}{4}$

$\displaystyle 2$

$\frac{1}{2}$ $\displaystyle \frac{1}{2}$

Correct answer:

$\displaystyle 1$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is $\frac{1}{4}$ $\displaystyle \frac{1}{4}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

$r =(\frac{1}{4})2$ $\displaystyle r =(\frac{1}{4})2$

$r=\frac{1}{2}$ $\displaystyle r=\frac{1}{2}$

The diameter is then

d=2r $\displaystyle d=2r$

d=1 $\displaystyle d=1$

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Example Question #2445 : Functions

A spherical balloon is being filled with air. What is the diameter of the sphere at the instance the rate of growth of the volume is $\frac{1}{96}$ $\displaystyle \frac{1}{96}$ times the rate of growth of the surface area?

Possible Answers:

$\frac{1}{32}$ $\displaystyle \frac{1}{32}$

$\frac{1}{64}$ $\displaystyle \frac{1}{64}$

$\frac{1}{12}$ $\displaystyle \frac{1}{12}$

$\frac{1}{96}$ $\displaystyle \frac{1}{96}$

$\frac{1}{24}$ $\displaystyle \frac{1}{24}$

Correct answer:

$\frac{1}{24}$ $\displaystyle \frac{1}{24}$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is $\frac{1}{96}$ $\displaystyle \frac{1}{96}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{96})8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\frac{1}{96})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{96})2$ $\displaystyle r =(\frac{1}{96})2$

$r=\frac{1}{48}$ $\displaystyle r=\frac{1}{48}$

The diameter is then

d=2r $\displaystyle d=2r$

$d=\frac{1}{24}$ $\displaystyle d=\frac{1}{24}$

Report an Error

Example Question #2446 : Functions

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is 17 times the rate of growth of the surface area?

Possible Answers:

$2312\pi$ $\displaystyle 2312\pi$

$9248\pi$ $\displaystyle 9248\pi$

$1156\pi$ $\displaystyle 1156\pi$

$578\pi$ $\displaystyle 578\pi$

$4624\pi$ $\displaystyle 4624\pi$

Correct answer:

$4624\pi$ $\displaystyle 4624\pi$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is 17 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(17)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(17)8\pi r \frac{dr}{dt}$

$\displaystyle r =(17)2$

r=34 $\displaystyle r=34$

The surface area is then:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$A=4\pi (34)^2=4624\pi$ $\displaystyle A=4\pi (34)^2=4624\pi$

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Example Question #651 : Rate

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is $\frac{1}{26}$ $\displaystyle \frac{1}{26}$ times the rate of growth of the surface area?

Possible Answers:

$\frac{4}{13}\pi$ $\displaystyle \frac{4}{13}\pi$

$\frac{2}{13}\pi$ $\displaystyle \frac{2}{13}\pi$

$\frac{16}{169}\pi$ $\displaystyle \frac{16}{169}\pi$

$\frac{2}{169}\pi$ $\displaystyle \frac{2}{169}\pi$

$\frac{4}{169}\pi$ $\displaystyle \frac{4}{169}\pi$

Correct answer:

$\frac{4}{169}\pi$ $\displaystyle \frac{4}{169}\pi$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is $\frac{1}{26}$ $\displaystyle \frac{1}{26}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{26})8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\frac{1}{26})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{26})2$ $\displaystyle r =(\frac{1}{26})2$

$r=\frac{1}{13}$ $\displaystyle r=\frac{1}{13}$

The surface area is then:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$A=4\pi (\frac{1}{13})^2=\frac{4}{169}\pi$ $\displaystyle A=4\pi (\frac{1}{13})^2=\frac{4}{169}\pi$

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Example Question #655 : Rate

A spherical balloon is being filled with air. What is the surface area of the sphere at the instance the rate of growth of the volume is $\frac{1}{11}$ $\displaystyle \frac{1}{11}$ times the rate of growth of the surface area?

Possible Answers:

$\frac{16}{121}\pi$ $\displaystyle \frac{16}{121}\pi$

$\frac{8}{121}\pi$ $\displaystyle \frac{8}{121}\pi$

$\frac{2}{11}\pi$ $\displaystyle \frac{2}{11}\pi$

$\frac{4}{11}\pi$ $\displaystyle \frac{4}{11}\pi$

$\frac{8}{11}\pi$ $\displaystyle \frac{8}{11}\pi$

Correct answer:

$\frac{16}{121}\pi$ $\displaystyle \frac{16}{121}\pi$

Explanation:

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

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

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

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

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

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

The rate of change of the radius is going to be the same for the sphere. So given our problem conditions, the rate of growth of the volume is $\frac{1}{11}$ $\displaystyle \frac{1}{11}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{11})8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(\frac{1}{11})8\pi r \frac{dr}{dt}$

$r =(\frac{1}{11})2$ $\displaystyle r =(\frac{1}{11})2$

$r=\frac{2}{11}$ $\displaystyle r=\frac{2}{11}$

The surface area is then:

$A=4\pi r^2$ $\displaystyle A=4\pi r^2$

$A=4\pi (\frac{2}{11})^2=\frac{16}{121}\pi$ $\displaystyle A=4\pi (\frac{2}{11})^2=\frac{16}{121}\pi$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2441 : Functions

Example Question #2442 : Functions

Example Question #2443 : Functions

Example Question #552 : Rate Of Change

Example Question #561 : How To Find Rate Of Change

Example Question #2444 : Functions

Example Question #2445 : Functions

Example Question #2446 : Functions

Example Question #651 : Rate

Example Question #655 : Rate

All Calculus 1 Resources

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