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Calculus 1 : Rate of Change

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Example Questions

Example Question #611 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $\frac{31}{36}$ ?

Possible Answers:

$\frac{31}{18}$

$\frac{31}{3}$

$\frac{31}{9}$

$\frac{31}{6}$

$\frac{31}{24}$

Correct answer:

$\frac{31}{9}$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

The rate of change of the radius is going to be the same for the sphere regardless of the considered parameter. To find the ratio of the rates of changes of the surface area and circumference, divide:

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

$\phi =4r$

$\phi =4(\frac{31}{36})=\frac{31}{9}$

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Example Question #612 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $\frac{77}{78}$ ?

Possible Answers:

$\frac{308}{39}$

$\frac{154}{39}$

$\frac{77}{39}$

$\frac{77}{13}$

$\frac{77}{312}$

Correct answer:

$\frac{154}{39}$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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

$\phi =4r$

$\phi =4(\frac{77}{78})=\frac{154}{39}$

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Example Question #613 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $\frac{13}{12}$ ?

Possible Answers:

$\frac{26}{3}$

$\frac{13}{3}$

$\frac{13}{6}$

$\frac{52}{3}$

Correct answer:

$\frac{13}{3}$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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

$\phi =4r$

$\phi =4(\frac{13}{12})=\frac{13}{3}$

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Example Question #614 : How To Find Rate Of Change

A spherical balloon is being filled with air. What is ratio of the rate of growth of the surface area of the sphere to the rate of growth of the circumference when the radius is $\frac{11}{128}$ ?

Possible Answers:

$\frac{11}{8}$

$\frac{11}{32}$

$\frac{11}{512}$

$\frac{11}{64}$

$\frac{11}{16}$

Correct answer:

$\frac{11}{32}$

Explanation:

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

$A=4\pi r^2$

$C=2\pi r$

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

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

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

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

$\phi =4r$

$\phi =4(\frac{11}{128})=\frac{11}{32}$

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Example Question #615 : Rate Of Change

A spherical balloon is being inflated with air. Determine the rate as a function of at which the volume is changing over time, $\frac{dV}{dt}$ , if $\frac{dr}{dt}=5\,cm/sec$ .

The volume of a sphere is given as

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

Possible Answers:

$\frac{dV}{dt}=\pi r^2$ $\frac{cm^3}{sec}$

$\frac{dV}{dt}=4\pi r^2$ $\frac{cm^3}{sec}$

$\frac{dV}{dt}=20\pi r^2$ $\frac{cm^3}{sec}$

$\frac{dV}{dt}=\frac{4}{3}\pi r^2$ $\frac{cm^3}{sec}$

Correct answer:

$\frac{dV}{dt}=20\pi r^2$ $\frac{cm^3}{sec}$

Explanation:

In order to find the rate at which the volume is changing over time we take the derivative with respect to .

$\frac{d}{dt}[V]=\frac{dV}{dt}$

As such

$\frac{d}{dt}[\frac{4}{3}\pi r^3]=\frac{4}{3}\pi *3r^{3-1}*\frac{dr}{dt}$

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

Substituting in $\frac{dr}{dt}=5\,cm/sec$ we get

$=4\pi r^2*5$

Hence the function for the rate at which the volume is changing over time is

$\frac{dV}{dt}=20\pi r^2$

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Example Question #615 : How To Find Rate Of Change

Find the slope at x=0 given the following function:

f(x)= (9x^2 - 2x + 5)^4

Possible Answers:

500

1000

Answer not listed

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

In this case the derivative is: $f'(x)= 4(9x^2 - 2x + 5)^{3} (18x - 2)$

Then plug x=0 into the derivative function: $f'(0)= 4(9({\color{Blue} 0})^2 - 2({\color{Blue} 0}) + 5)^{3} (18({\color{Blue} 0}) - 2)$

Therefore, the slope is:

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Example Question #616 : How To Find Rate Of Change

Find the slope at x=8 given the following function:

f(x)= 23 + e

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

In this case the derivative is: f'(x)= 0

Then plug x=8 into the derivative function: f'(x)= 0

Therefore, the slope is:

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Example Question #3531 : Calculus

Find the slope at x=2 given the following function:

f(x)= ln(2x) + ln(3x)

Possible Answers:

Answer not listed

Correct answer:

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

In this case the derivative is: $f'(x)= \frac{1}{x} + \frac{1}{x}$

Then plug x=2 into the derivative function: $f'(2)= \frac{1}{{\color{Blue} 2}} + \frac{1}{{\color{Blue} 2}}$

Therefore, the slope is:

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Example Question #3532 : Calculus

Find the slope at x=2 given the following function:

$f(x)= \sin (4x^4 - x)$

Possible Answers:

-14.8

34.7

-18.9

Answer not listed

7..9

Correct answer:

-14.8

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

In this case the derivative is: $f'(x)= \cos(4x^4 - x) (16x^{3} -1)$

Then plug x=1 into the derivative function: $f'(1)= \cos(4({\color{Blue} 1})^4 - {\color{Blue} 1}) (16({\color{Blue} 1})^{3} -1)$

Therefore, the slope is: -14.8

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Example Question #3533 : Calculus

Find the slope at x=0 given the following function:

$f(x)= e^{3x^6 - e^x}$

Possible Answers:

1.4

-0.4

5.8

Answer notl listed

-7.3

Correct answer:

-0.4

Explanation:

In order to find the slope of a certain point given a function, you must first find the derivative.

In this case the derivative is: $f'(x)= e^{3x^6 - e^x} (18x^{5} - e^x)$

Then plug x=0 into the derivative function: $f'(0)= e^{3({\color{Blue} 0})^6 - e^{{\color{Blue} 0}}} (18({\color{Blue} 0})^{5} - e^{\color{Blue} 0})$

Therefore, the slope is: -0.4

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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #611 : How To Find Rate Of Change

Example Question #612 : How To Find Rate Of Change

Example Question #613 : How To Find Rate Of Change

Example Question #614 : How To Find Rate Of Change

Example Question #615 : Rate Of Change

Example Question #615 : How To Find Rate Of Change

Example Question #616 : How To Find Rate Of Change

Example Question #3531 : Calculus

Example Question #3532 : Calculus

Example Question #3533 : Calculus

All Calculus 1 Resources

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