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Calculus 1 : Calculus

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

Example Questions

Example Question #406 : How To Find Rate Of Change

Find the slope at x=3 :

f(x)= 2(x+2)^2 + 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 of that function.

In this case, we must find the derivative of the following: f(x)= 2(x+2)^2 + 3x

That is done by doing the following: f'(x)= 4x+11

Then plug x=3 into the derivative: $f'(3)= 4({\color{Blue} 3})+11$

Therefore, the answer is:

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Example Question #407 : How To Find Rate Of Change

Find the derivative of the following function:
f(x)=e^xSin(x)

Possible Answers:

$e^{Cosx}$

e^xCos(x)

-e^x(Cos(x)+Sin(x))

e^x(-Cos(x)+Sin(x))

e^x(Cos(x)+Sin(x))

Correct answer:

e^x(Cos(x)+Sin(x))

Explanation:

We notice that this function,

f(x)=e^xSin(x) , is of the form f(x)=g(x)h(x)

such that we must use the Product Rule to find the derivative.

f'(x)=g(x)h'(x)+h(x)g'(x) .

Doing so we find the derivative to be:

f'(x)=e^xCos(x)+e^xSin(x)=e^x(Cos(x)+Sin(x))

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Example Question #408 : Rate Of Change

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameter of the sphere at the instance the rate of shrinkage of the volume is 18.5 times the rate of shrinkage of the surface area?

Possible Answers:

$\frac{37}{\pi}$

$\frac{74}{\pi}$

$37\pi$

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 shrinkage of the volume is 18.5 times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(18.5)8\pi r \frac{dr}{dt}$

r =(18.5)2

r=37

The diameter is then:

d=2r

d=74

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Example Question #411 : 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 volume is $\frac{3}{\pi}$ times the rate of shrinkage of the surface area?

Possible Answers:

$\frac{12}{\pi}$

$12\pi^2$

$\frac{\pi}{12}$

$\frac{12}{\pi^2}$

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 shrinkage of the volume is $\frac{3}{\pi}$ times the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

$4\pi r^2 \frac{dr}{dt}=(\frac{3}{\pi})8\pi r \frac{dr}{dt}$

$r =(\frac{3}{\pi})2$

$r=\frac{6}{\pi}$

The circumference of a sphere is:

$C=2\pi r$

$C=2\pi(\frac{6}{\pi})$

C=12

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Example Question #501 : Rate

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 0.16 times the rate of growth of the surface area?

Possible Answers:

0.282

0.137

0.314

0.070

0.039

Correct answer:

0.137

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

$4\pi r^2 \frac{dr}{dt}=(0.16)8\pi r \frac{dr}{dt}$

r =(0.16)2

r=0.32

Finding the volume:

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

$V=\frac{4}{3}\pi (0.32)^3=0.137$

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Example Question #502 : 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 volume is 12.5 times the rate of growth of the surface area?

Possible Answers:

$50\pi$

$75\pi$

$25\pi$

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

$4\pi r^2 \frac{dr}{dt}=(12.5)8\pi r \frac{dr}{dt}$

r =(12.5)2

r=25

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Example Question #503 : 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 volume is 98 times the rate of growth of the circumference?

Possible Answers:

$\sqrt{7}$

$\sqrt{\frac{7}{2}}$

$\frac{7}{2}$

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

$4\pi r^2 \frac{dr}{dt}=(98)2\pi \frac{dr}{dt}$

$r^2=(98)\frac{1}{2}$

$r =\sqrt{\frac{98}{2}}=\sqrt{49}=7$

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Example Question #504 : Rate

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 72 times the rate of growth of the circumference?

Possible Answers:

144

$576\pi$

$288\pi$

288

$144\pi$

Correct answer:

$288\pi$

Explanation:

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

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

$C=2\pi r^$

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

$4\pi r^2 \frac{dr}{dt}=(72)2\pi \frac{dr}{dt}$

$r^2=(72)\frac{1}{2}$

$r =\sqrt{\frac{72}{2}}=\sqrt{36}=6$

Finding the volume at this time:

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

$V=\frac{4}{3}\pi (6)^3=288\pi$

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Example Question #505 : Rate

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 338 times the rate of shrinkage of the circumference?

Possible Answers:

$26\pi$

$13\pi$

Correct answer:

$26\pi$

Explanation:

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

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

$C=2\pi r^$

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 shrinkage of the volume is 338 times the rate of shrinkage of the circumference, solve for the radius:

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

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

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

The circumference is then:

$C=2\pi r$

$C=2\pi (13)=26\pi$

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Example Question #506 : Rate

A spherical balloon is deflating, while maintaining its spherical shape. What is the diameterof the sphere at the instance the rate of shrinkage of the volume is 450 times the rate of shrinkage of the circumference?

Possible Answers:

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 shrinkage of the volume is 450 times the rate of shrinkage of the circumference, solve for the radius:

$4\pi r^2 \frac{dr}{dt}=(450)2\pi \frac{dr}{dt}$

$r^2=(450)\frac{1}{2}$

$r =\sqrt{\frac{450}{2}}=\sqrt{225}=15$

The diameter is then:

d=2r

d=30

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #406 : How To Find Rate Of Change

Example Question #407 : How To Find Rate Of Change

Example Question #408 : Rate Of Change

Example Question #411 : How To Find Rate Of Change

Example Question #501 : Rate

Example Question #502 : Rate

Example Question #503 : Rate

Example Question #504 : Rate

Example Question #505 : Rate

Example Question #506 : Rate

All Calculus 1 Resources

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