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Calculus 1 : Functions

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Example Questions

Example Question #3490 : 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 726 times the rate of growth of the circumference?

Possible Answers:

$11\sqrt{6}$ $\displaystyle 11\sqrt{6}$

$11\sqrt{3}$ $\displaystyle 11\sqrt{3}$

$6\sqrt{11}$ $\displaystyle 6\sqrt{11}$

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

$22\sqrt{6}$ $\displaystyle 22\sqrt{6}$

Correct answer:

$11\sqrt{3}$ $\displaystyle 11\sqrt{3}$

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

$4\pi r^2 \frac{dr}{dt}=(726)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(726)2\pi \frac{dr}{dt}$

$r^2=(726)\frac{1}{2}$ $\displaystyle r^2=(726)\frac{1}{2}$

$r =\sqrt{363}=11\sqrt{3}$ $\displaystyle r =\sqrt{363}=11\sqrt{3}$

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Example Question #3491 : 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 676 times the rate of growth of the circumference?

Possible Answers:

$13\sqrt{2}$ $\displaystyle 13\sqrt{2}$

$2\sqrt{13}$ $\displaystyle 2\sqrt{13}$

$26\sqrt{2}$ $\displaystyle 26\sqrt{2}$

$13\sqrt{26}$ $\displaystyle 13\sqrt{26}$

$\sqrt{13}$ $\displaystyle \sqrt{13}$

Correct answer:

$13\sqrt{2}$ $\displaystyle 13\sqrt{2}$

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

$4\pi r^2 \frac{dr}{dt}=(676)2\pi \frac{dr}{dt}$ $\displaystyle 4\pi r^2 \frac{dr}{dt}=(676)2\pi \frac{dr}{dt}$

$r^2=(676)\frac{1}{2}$ $\displaystyle r^2=(676)\frac{1}{2}$

$r =\sqrt{338}=13\sqrt{2}$ $\displaystyle r =\sqrt{338}=13\sqrt{2}$

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Example Question #3492 : 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 surface area is 804 times the rate of growth of the circumference?

Possible Answers:

402 $\displaystyle 402$

$\displaystyle 67$

134 $\displaystyle 134$

536 $\displaystyle 536$

201 $\displaystyle 201$

Correct answer:

201 $\displaystyle 201$

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 804 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(804)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(804)2\pi \frac{dr}{dt}$

$r =\frac{804}{4}=201$ $\displaystyle r =\frac{804}{4}=201$

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Example Question #3493 : 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 surface area is 728 times the rate of growth of the circumference?

Possible Answers:

$\displaystyle 91$

$\displaystyle 52$

364 $\displaystyle 364$

416 $\displaystyle 416$

182 $\displaystyle 182$

Correct answer:

182 $\displaystyle 182$

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 728 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(728)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(728)2\pi \frac{dr}{dt}$

$r =\frac{728}{4}=182$ $\displaystyle r =\frac{728}{4}=182$

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Example Question #3494 : 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 surface area is 716 times the rate of growth of the circumference?

Possible Answers:

716 $\displaystyle 716$

1432 $\displaystyle 1432$

2864 $\displaystyle 2864$

358 $\displaystyle 358$

179 $\displaystyle 179$

Correct answer:

179 $\displaystyle 179$

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 716 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(316)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(316)2\pi \frac{dr}{dt}$

$r =\frac{716}{4}=179$ $\displaystyle r =\frac{716}{4}=179$

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Example Question #671 : Rate

A spherical balloon is being filled with air. What is the radius of the sphere at the instance the rate of growth of the surface area is 92 times the rate of growth of the circumference?

Possible Answers:

$\displaystyle 48$

$\displaystyle 96$

$\displaystyle 23$

184 $\displaystyle 184$

$\displaystyle 46$

Correct answer:

$\displaystyle 23$

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 92 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(92)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(92)2\pi \frac{dr}{dt}$

$r =\frac{92}{4}=23$ $\displaystyle r =\frac{92}{4}=23$

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Example Question #672 : Rate

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 1352 times the rate of growth of the circumference?

Possible Answers:

338 $\displaystyle 338$

845 $\displaystyle 845$

169 $\displaystyle 169$

676 $\displaystyle 676$

1690 $\displaystyle 1690$

Correct answer:

676 $\displaystyle 676$

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 1352 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(1352)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(1352)2\pi \frac{dr}{dt}$

$r =\frac{1352}{4}=338$ $\displaystyle r =\frac{1352}{4}=338$

The diameter is then:

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

d=676 $\displaystyle d=676$

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Example Question #673 : Rate

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 56 times the rate of growth of the circumference?

Possible Answers:

$\displaystyle 42$

$\displaystyle 35$

$\displaystyle 28$

$\displaystyle 14$

$\displaystyle 21$

Correct answer:

$\displaystyle 28$

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 56 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(56)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(56)2\pi \frac{dr}{dt}$

$r =\frac{56}{4}=14$ $\displaystyle r =\frac{56}{4}=14$

The diameter is then:

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

d=28 $\displaystyle d=28$

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Example Question #674 : Rate

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 344 times the rate of growth of the circumference?

Possible Answers:

$\displaystyle 43$

172 $\displaystyle 172$

$\displaystyle 86$

252 $\displaystyle 252$

126 $\displaystyle 126$

Correct answer:

172 $\displaystyle 172$

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 344 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(344)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(344)2\pi \frac{dr}{dt}$

$r =\frac{344}{4}=86$ $\displaystyle r =\frac{344}{4}=86$

The diameter can then be found:

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

d=172 $\displaystyle d=172$

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Example Question #675 : Rate

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 1348 times the rate of growth of the circumference?

Possible Answers:

185 $\displaystyle 185$

2696 $\displaystyle 2696$

674 $\displaystyle 674$

370 $\displaystyle 370$

337 $\displaystyle 337$

Correct answer:

674 $\displaystyle 674$

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 1348 times the rate of growth of the circumference, to solve for the length of the radius at that instant:

$8\pi r \frac{dr}{dt}=(1348)2\pi \frac{dr}{dt}$ $\displaystyle 8\pi r \frac{dr}{dt}=(1348)2\pi \frac{dr}{dt}$

$r =\frac{1348}{4}=337$ $\displaystyle r =\frac{1348}{4}=337$

The diameter is then:

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

d=674 $\displaystyle d=674$

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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #3490 : Calculus

Example Question #3491 : Calculus

Example Question #3492 : Calculus

Example Question #3493 : Calculus

Example Question #3494 : Calculus

Example Question #671 : Rate

Example Question #672 : Rate

Example Question #673 : Rate

Example Question #674 : Rate

Example Question #675 : Rate

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