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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

324

144

162

243

Correct answer:

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:

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

$A=4\pi r^2$

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 growth of the volume is 81 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

r =(81)2

r=162

Report an Error

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:

290

232

145

Correct answer:

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:

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

$A=4\pi r^2$

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 growth of the volume is 116 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

r =(116)2

r=232

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 54 times the rate of growth of the surface area?

Possible Answers:

108

243

216

Correct answer:

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:

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

$A=4\pi r^2$

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 growth of the volume is 54 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

r =(54)2

r=108

The diameter is then

d=2r

d=216

Report an Error

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 73 times the rate of growth of the surface area?

Possible Answers:

584

292

146

1168

Correct answer:

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:

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

$A=4\pi r^2$

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 growth of the volume is 73 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

r =(73)2

r=146

The diameter is then

d=2r

d=292

Report an Error

Example Question #2446 : 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 $\frac{1}{4}$ times the rate of growth of the surface area?

Possible Answers:

$\frac{1}{2}$

$\frac{1}{4}$

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 growth of the volume is $\frac{1}{4}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

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

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

The diameter is then

d=2r

d=1

Report an Error

Example Question #2447 : 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 $\frac{1}{96}$ times the rate of growth of the surface area?

Possible Answers:

$\frac{1}{64}$

$\frac{1}{32}$

$\frac{1}{24}$

$\frac{1}{96}$

$\frac{1}{12}$

Correct answer:

$\frac{1}{24}$

Explanation:

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

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

$A=4\pi r^2$

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 growth of the volume is $\frac{1}{96}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

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

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

The diameter is then

d=2r

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

Report an Error

Example Question #2448 : 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:

$4624\pi$

$1156\pi$

$578\pi$

$2312\pi$

$9248\pi$

Correct answer:

$4624\pi$

Explanation:

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

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

$A=4\pi r^2$

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 growth of the volume is 17 times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

r =(17)2

r=34

The surface area is then:

$A=4\pi r^2$

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

Report an Error

Example Question #654 : 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 $\frac{1}{26}$ times the rate of growth of the surface area?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

$A=4\pi r^2$

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 growth of the volume is $\frac{1}{26}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

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

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

The surface area is then:

$A=4\pi r^2$

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

Report an Error

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 $\frac{1}{11}$ times the rate of growth of the surface area?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

$A=4\pi r^2$

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 growth of the volume is $\frac{1}{11}$ times the rate of growth of the surface area, let's solve for a radius that satisfies it.

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

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

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

The surface area is then:

$A=4\pi r^2$

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

Report an Error

Example Question #3471 : Calculus

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 338 times the rate of growth of the circumference?

Possible Answers:

$26\sqrt{3}$

$13\sqrt{2}$

$26\sqrt{2}$

Correct answer:

Explanation:

Begin by writing the equations for the volume and circumference of a sphere with respect to the sphere's radius:

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

$C=2\pi r^$

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

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

$\frac{dC}{dt}=2\pi \frac{dr}{dt}$

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now given the problem information, the rate of growth of the volume is 338 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(338)2\pi \frac{dr}{dt}$

$r^2=(338)\frac{1}{2}$

$r =\sqrt{\frac{338}{2}}=\sqrt{169}=13$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2442 : Functions

Example Question #2443 : Functions

Example Question #2444 : Functions

Example Question #2445 : Functions

Example Question #2446 : Functions

Example Question #2447 : Functions

Example Question #2448 : Functions

Example Question #654 : Rate

Example Question #655 : Rate

Example Question #3471 : Calculus

All Calculus 1 Resources

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