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Calculus 1 : Functions

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Example Questions

Example Question #297 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the volume is 10.13 times the rate of shrinkage of the surface area?

Possible Answers:

2729.43 $\displaystyle 2729.43$

1289.52 $\displaystyle 1289.52$

6343.81 $\displaystyle 6343.81$

5158.09 $\displaystyle 5158.09$

4596.03 $\displaystyle 4596.03$

Correct answer:

5158.09 $\displaystyle 5158.09$

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 shrinkage of the volume is 10.13 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(10.13)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(10.13)8\pi r \frac{dr}{dt}$

$\displaystyle r =(10.13)2$

r=20.26 $\displaystyle r=20.26$

Then to find the surface area:

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

$\displaystyle A=5158.09$

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Example Question #298 : Rate Of Change

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

Possible Answers:

$\frac{4096}{3}\pi$ $\displaystyle \frac{4096}{3}\pi$

$\frac{256}{3}\pi$ $\displaystyle \frac{256}{3}\pi$

$\frac{2048}{3}\pi$ $\displaystyle \frac{2048}{3}\pi$

$\frac{512}{3}\pi$ $\displaystyle \frac{512}{3}\pi$

$\frac{1024}{3}\pi$ $\displaystyle \frac{1024}{3}\pi$

Correct answer:

$\frac{2048}{3}\pi$ $\displaystyle \frac{2048}{3}\pi$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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 128 times the rate of growth of the circumference, solve for the radius:

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

$r^2=(128)\frac{1}{2}$ $\displaystyle r^2=(128)\frac{1}{2}$

$r =\sqrt{64}$ $\displaystyle r =\sqrt{64}$

r=8 $\displaystyle r=8$

Then to find the volume:

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

$V=\frac{2048}{3}\pi$ $\displaystyle V=\frac{2048}{3}\pi$

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Example Question #299 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the circumference of the sphere at the instance the rate of shrinkage of the volume is seven times the rate of shrinkage of the circumference?

Possible Answers:

$7\pi\sqrt{2}$ $\displaystyle 7\pi\sqrt{2}$

$\pi\sqrt{14}$ $\displaystyle \pi\sqrt{14}$

$2\pi\sqrt{3}$ $\displaystyle 2\pi\sqrt{3}$

$\pi\sqrt{7}$ $\displaystyle \pi\sqrt{7}$

Correct answer:

$\pi\sqrt{14}$ $\displaystyle \pi\sqrt{14}$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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 shrinkage of the volume is seven times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(7)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(7)2\pi \frac{dr}{dt}$

$r^2=(7)\frac{1}{2}$ $\displaystyle r^2=(7)\frac{1}{2}$

$r =\sqrt{\frac{7}{2}}$ $\displaystyle r =\sqrt{\frac{7}{2}}$

The circumference is then

$C=2\pi r$ $\displaystyle C=2\pi r$

C=11.75 $\displaystyle C=11.75$

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Example Question #300 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the volume is fifteen times the rate of shrinkage of the circumference?

Possible Answers:

$225\pi$ $\displaystyle 225\pi$

$15\pi$ $\displaystyle 15\pi$

$25\pi$ $\displaystyle 25\pi$

$30\pi$ $\displaystyle 30\pi$

$700\pi$ $\displaystyle 700\pi$

Correct answer:

$30\pi$ $\displaystyle 30\pi$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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 shrinkage of the volume is fifteen times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(15)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(15)2\pi \frac{dr}{dt}$

$r^2=(15)\frac{1}{2}$ $\displaystyle r^2=(15)\frac{1}{2}$

$r =\sqrt{\frac{15}{2}}$ $\displaystyle r =\sqrt{\frac{15}{2}}$

The surface area is then:

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

$A=30\pi$ $\displaystyle A=30\pi$

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Example Question #301 : 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 0.34 times the rate of growth of the circumference?

Possible Answers:

0.170 $\displaystyle 0.170$

0.825 $\displaystyle 0.825$

0.412 $\displaystyle 0.412$

0.627 $\displaystyle 0.627$

0.340 $\displaystyle 0.340$

Correct answer:

0.825 $\displaystyle 0.825$

Explanation:

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

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

$C=2\pi r^$ $\displaystyle C=2\pi r^$

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

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

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

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 0.34 times the rate of growth of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(0.34)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(0.34)2\pi \frac{dr}{dt}$

$r^2=(0.34)\frac{1}{2}$ $\displaystyle r^2=(0.34)\frac{1}{2}$

$r =\sqrt{\frac{0.34}{2}}=0.4123$ $\displaystyle r =\sqrt{\frac{0.34}{2}}=0.4123$

The diameter is therefore

d=0.825 $\displaystyle d=0.825$

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Example Question #302 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is the volume of the sphere at the instance the rate of growth of the surface area is 24 times the rate of growth of the circumference?

Possible Answers:

$216\pi$ $\displaystyle 216\pi$

$144\pi$ $\displaystyle 144\pi$

$288\pi$ $\displaystyle 288\pi$

$316 \pi$ $\displaystyle 316 \pi$

$108\pi$ $\displaystyle 108\pi$

Correct answer:

$288\pi$ $\displaystyle 288\pi$

Explanation:

Start by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

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

$C=2\pi r$ $\displaystyle C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 24 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(24)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(24)2\pi \frac{dr}{dt}$

$r =\frac{24}{4}$ $\displaystyle r =\frac{24}{4}$

r=6 $\displaystyle r=6$

Now to solve for the volume:

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

$V=288\pi$ $\displaystyle V=288\pi$

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Example Question #303 : 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 surface area is 3.12 times the rate of growth of the circumference?

Possible Answers:

1.04 $\displaystyle 1.04$

2.08 $\displaystyle 2.08$

4.16 $\displaystyle 4.16$

1.56 $\displaystyle 1.56$

0.78 $\displaystyle 0.78$

Correct answer:

1.56 $\displaystyle 1.56$

Explanation:

Start by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

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

$C=2\pi r$ $\displaystyle C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of growth of the surface area is 3.12 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(3.12)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(3.12)2\pi \frac{dr}{dt}$

$r =\frac{3.12}{4}$ $\displaystyle r =\frac{3.12}{4}$

r=0.78 $\displaystyle r=0.78$

The diameter is then

d=1.56 $\displaystyle d=1.56$

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Example Question #304 : How To Find Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the circumference of the sphere at the instance the rate of shrinkage of the surface area is 11.12 times the rate of shrinkage of the circumference?

Possible Answers:

15.1 $\displaystyle 15.1$

19.4 $\displaystyle 19.4$

11.2 $\displaystyle 11.2$

23.5 $\displaystyle 23.5$

17.5 $\displaystyle 17.5$

Correct answer:

17.5 $\displaystyle 17.5$

Explanation:

Start by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

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

$C=2\pi r$ $\displaystyle C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of shrinkage of the surface area is 11.12 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(11.12)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(11.12)2\pi \frac{dr}{dt}$

$r =\frac{11.12}{4}$ $\displaystyle r =\frac{11.12}{4}$

r=2.78 $\displaystyle r=2.78$

The circumference is then

$C=2\pi r$ $\displaystyle C=2\pi r$

C=17.5 $\displaystyle C=17.5$

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Example Question #305 : How To Find Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the surface area of the sphere at the instance the rate of shrinkage of the surface area is 76 times the rate of shrinkage of the circumference?

Possible Answers:

$729 \pi$ $\displaystyle 729 \pi$

$1458 \pi$ $\displaystyle 1458 \pi$

$1444\pi$ $\displaystyle 1444\pi$

$2888 \pi$ $\displaystyle 2888 \pi$

$722\pi$ $\displaystyle 722\pi$

Correct answer:

$1444\pi$ $\displaystyle 1444\pi$

Explanation:

Start by writing the equations for the surface area and circumference of a sphere with respect to the sphere's radius:

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

$C=2\pi r$ $\displaystyle C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. Now we can use the relation given in the problem statement, the rate of shrinkage of the surface area is 76 times the rate of shrinkage of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(76)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(76)2\pi \frac{dr}{dt}$

$r =\frac{76}{4}$ $\displaystyle r =\frac{76}{4}$

r=19 $\displaystyle r=19$

Now all that remains is to find the surface area at this time

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

$A= 1444\pi$ $\displaystyle A= 1444\pi$

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Example Question #306 : How To Find Rate Of Change

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 2.22 times the rate of growth of the surface area?

Possible Answers:

4.44 $\displaystyle 4.44$

4.98 $\displaystyle 4.98$

2.22 $\displaystyle 2.22$

2.49 $\displaystyle 2.49$

6.66 $\displaystyle 6.66$

Correct answer:

4.44 $\displaystyle 4.44$

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

$4\pi r^2 \frac{dr}{dt}=(2.22)8\pi r \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(2.22)8\pi r \frac{dr}{dt}$

$\displaystyle r =(2.22)2$

r=4.44 $\displaystyle r=4.44$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #297 : Rate Of Change

Example Question #298 : Rate Of Change

Example Question #299 : Rate Of Change

Example Question #300 : Rate Of Change

Example Question #301 : How To Find Rate Of Change

Example Question #302 : How To Find Rate Of Change

Example Question #303 : How To Find Rate Of Change

Example Question #304 : How To Find Rate Of Change

Example Question #305 : How To Find Rate Of Change

Example Question #306 : How To Find Rate Of Change

All Calculus 1 Resources

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