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Calculus 1 : Calculus

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Example Questions

Example Question #191 : Rate Of Change

A cube is growing in size. What is the area of one of the cube's faces at the time that the rate of growth of the cube's volume is equal to twice the rate of growth of its sides?

Possible Answers:

$\frac{2}{3}$

$\sqrt{\frac{2}{3} }$

$6\sqrt{\frac{2}{3}}$

Correct answer:

$\frac{2}{3}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

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

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of growth of the cube's volume is equal to twice the rate of growth of its sides

$3s^2\frac{ds}{dt}=(2)\frac{ds}{dt}$

$s=\sqrt{\frac{2}{3}}$

The area of one of those cube's faces is the length of its sides squared:

$A=s^2=\frac{2}{3}$

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Example Question #288 : Rate

A square is growing in area. What is the length of the square's sides at the moment that the rate of growth of the area is the same as the rate of growth of the square's diagonal?

Possible Answers:

$\sqrt{2}$

$\frac{1}{2}$

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

Correct answer:

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

Explanation:

Start by writing the equations for a square's dimensions.

A=s^2

$d=s\sqrt{2}$

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

$\frac{dA}{dt}=2s\frac{ds}{dt}$

$\frac{dd}{dt}=\sqrt{2}\frac{ds}{dt}$

Now, solve for the length of the side of the square to satisfy the problem condition, the rate of growth of the area is the same as the rate of growth of the square's diagonal:

$2s\frac{ds}{dt}=\sqrt{2}\frac{ds}{dt}$

$s=\frac{\sqrt{2}}{2}$

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Example Question #198 : Rate Of Change

Use normal rules of differentiation to find the derrivative f'(0) of the function

$f(x)=2^{5x}+(2x^2+3)^3$

Possible Answers:

$-2 \ln 5$

$5\ln 2$

$2 \ln 5$

$-5 \ln 2$

Correct answer:

$5\ln 2$

Explanation:

Taking the derrivative,

$f'(x)=\frac{d}{dx}[2^{5x}+(2x^2+3)^3]$

We separate the summed terms:

$=\frac{d}{dx}[2^{5x}]+\frac{d}{dx}[(2x^2+3)^3]$

Using the chain rule:

$=2^{5x}\ln 2\frac{d}{dx}[5x]+2(2x^2+3)^2\frac{d}{dx}[(2x^2+3)]$

$=2^{5x}\ln 2*5+3(2x^2+3)^2* 4x$

Simplifying

$=2^{5x}5\ln 2+12x(2x^2+3)^2$

Now, evaluating at f'(0) :

$=2^{5(0)}5\ln 2+12(0)(2(0)^2+3)^2$

$=(1)5 \ln 2+(0)$

$=5 \ln 2$

which is our answer.

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Example Question #201 : How To Find Rate Of Change

Use normal rules of differentiation to find the derrivative $f'(\frac{2 \pi}{3})$ of the function

$f(x)= \tan(\frac{1}{2}x^2)-e^{\cos x}$ .

Possible Answers:

$\approx 5.2456$

$\approx 10.4568$

$\approx 4.3333$

$\approx -2.5783$

$\approx 6.6867$

Correct answer:

$\approx 6.6867$

Explanation:

Taking the derivative

$f'(x)= \frac{d}{dx}[\tan(\frac{1}{2}x^2)-e^{\cos x}]$

Separating the summed terms:

$= \frac{d}{dx}[\tan(\frac{1}{2}x^2)]-\frac{d}{dx}[e^{\cos x}]$

Applying the chain rule:

$= \frac{d}{dx}[\tan(\frac{1}{2}x^2)]-\frac{d}{dx}[e^{\cos x}]$

$= \sec^2 (\frac{1}{2}x^2)\frac{d}{dx}[\frac{1}{2}x^2]-e^{cos x}\frac{d}{dx}[\cos x]$

$= \sec^2 (\frac{1}{2}x^2)*x-e^{cos x}(-\sin x)$

Siimplifying:

$= x\sec^2 (\frac{1}{2}x^2)+\sin x *e^{cos x}$

Evaluating at $x=\frac{2\pi}{3}$ :

$= (\frac{2\pi}{3})\sec^2 (\frac{1}{2}(\frac{2\pi}{3})^2)+\sin (\frac{2\pi}{3}) *e^{cos (\frac{2\pi}{3})}$

$= (\frac{2\pi}{3})\sec^2 (\frac{2\pi^2}{9})+\sin (\frac{2\pi}{3})*e^{cos (\frac{2\pi}{3})}$

$= (\frac{2\pi}{3})\sec^2 (\frac{2\pi^2}{9})+(\frac{\sqrt{3}}{2})*e^{cos (-\frac{1}{2})}$

$\approx 6.6867$

Which is our answer.

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Example Question #201 : Rate Of Change

Use implicit differentiation to find the derrivative $\frac{dy}{dx}$ .

Find $\frac{dy}{dx}$ given the equation $\ln (y^2+1)=2xy$ .

Possible Answers:

$\frac{y^3+y}{y-xy^2-x}$

$\frac{1}{y^2+1}-2xy$

$\frac{(y+2)^2}{y+xy^2-xy}$

$\frac{x^3+x}{xy+xy^2-y^2}$

$2x-\frac{2y}{y^2+1}$

Correct answer:

$\frac{y^3+y}{y-xy^2-x}$

Explanation:

We take the derivative of both sides of the equation

$\frac{d}{dx}[\ln (y^2+1)]=\frac{d}{dx}[2xy]$

We apply the chain rule on the left side and the product rule on the right side:

$\frac{1}{(y^2+1)}*\frac{d}{dx}[y^2+1]=\frac{d}{dx}[2x]y+2x\frac{d}{dx}[y]$

$\frac{2y}{(y^2+1)}*\frac{dy}{dx}=2y+2x\frac{dy}{dx}$

$\frac{2y}{(y^2+1)}*\frac{dy}{dx}-2x\frac{dy}{dx}=2y$

$\frac{dy}{dx}[\frac{2y}{(y^2+1)}-2x]=2y$

$\frac{dy}{dx}=\frac{2y}{\frac{2y}{(y^2+1)}-2x}$

Simplifying:

$\frac{dy}{dx}=\frac{2y}{(\frac{2y-2x(y^2+1)}{(y^2+1)})}$

$\frac{dy}{dx}=\frac{2y(y^2+1)}{2y-2x(y^2+1)}$

$\frac{dy}{dx}=\frac{2y(y^2+1)}{2[y-x(y^2+1)]}$

$\frac{dy}{dx}=\frac{y(y^2+1)}{y-x(y^2+1)}$

$\frac{dy}{dx}=\frac{y^3+y}{y-xy^2-x}$

Which is our answer.

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Example Question #202 : How To Find Rate Of Change

Use implicit differentiation to find the derrivative $\frac{dy}{dx}$ .

Find $\frac{dy}{dx}$ given (x-1)^2+(y+2)^2=36 .

Possible Answers:

$\frac{-x+1}{y+2}$

$\frac{(x+1)^2}{(y-4)^2}$

$\frac{2x+1}{2y-1}$

$\frac{-2x^2+1}{2y^2+4}-36$

$-\frac{(x+1)^2}{(y-4)^2}$

Correct answer:

$\frac{-x+1}{y+2}$

Explanation:

Taking the derivative of both sides of the equation

$\frac{d}{dx}[(x-1)^2+(y+2)^2]=\frac{d}{dx}[36]$

We apply normal rules of differentiation and solve for dy/dx

$2(x-1)+2(y+2)\frac{dy}{dx}=0$

$2(y+2)\frac{dy}{dx}=-2(x-1)$

$\frac{dy}{dx}=\frac{-2(x-1)}{2(y+2)}$

$\frac{dy}{dx}=\frac{-x+1}{y+2}$

Which is our answer.

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Example Question #3111 : Calculus

Use normal rules of differentiation to find the derrivative $f'(\frac{5\pi}{6})$ of the function $f(x)= \ln(\cos^2(x))$

Possible Answers:

$\frac{1}{2}$

$\frac{2\sqrt{3}}{3}$

$\frac{ -2 \sqrt{3}}{2}$

$-\frac{2\sqrt{3}}{3}$

Correct answer:

$\frac{2\sqrt{3}}{3}$

Explanation:

Taking the derivative

$f'(x)=\frac{d}{dx} [\ln(\cos^2(x))]$

Applying the chain rule:

$=\frac{1}{\cos^2 x}*\frac{d}{dx}[\cos^2 x]$

$=\frac{1}{\cos^2 x}*2 \cos x * (- \sin x)$

Simplifying:

$=\frac{-2 \cos x \sin x}{\cos^2 x}$

$=\frac{-2 \sin x}{\cos x}$

$=-2 \tan x$

Evaluating at $x= \frac{5 \pi}{6}$

$=-2 \tan (\frac{5 \pi }{6})$

$= \frac{2\sqrt{3}}{3}$

Which is our answer.

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Example Question #204 : How To Find Rate Of Change

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

Possible Answers:

$196\pi$

$28\pi$

$14\pi$

Correct answer:

$28\pi$

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

$4\pi r^2 \frac{dr}{dt}=(7)8\pi r \frac{dr}{dt}$

r =(7)2

r=14

Circumference of a sphere is given by the function:

$C=2\pi r$

$C=28\pi$

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Example Question #205 : Rate Of Change

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

Possible Answers:

$3\pi$

$\frac{1}{3}\pi$

$\frac{2}{3}\pi$

$4\pi$

$\frac{4}{3}\pi$

Correct answer:

$\frac{4}{3}\pi$

Explanation:

Let's 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. So given our problem conditions, the rate of shrinkage of the volume is twice the rate of shrinkage of the circumference, let's solve for a radius that satisfies it.

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

r^2 =1

r=1

Then to find the volume at this time:

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

$V=\frac{4}{3}\pi (1^3)$

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

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Example Question #3112 : Calculus

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 a seventh of the rate of growth of the surface area?

Possible Answers:

$\frac{8}{7}$

$\frac{1}{7}$

$\frac{16}{7}$

$\frac{2}{7}$

$\frac{4}{7}$

Correct answer:

$\frac{4}{7}$

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

$4\pi r^2 \frac{dr}{dt}=(\frac{1}{7})8\pi r \frac{dr}{dt}$

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

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

The diameter is given by the formula:

D=2r

$D=\frac{4}{7}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #191 : Rate Of Change

Example Question #288 : Rate

Example Question #198 : Rate Of Change

Example Question #201 : How To Find Rate Of Change

Example Question #201 : Rate Of Change

Example Question #202 : How To Find Rate Of Change

Example Question #3111 : Calculus

Example Question #204 : How To Find Rate Of Change

Example Question #205 : Rate Of Change

Example Question #3112 : Calculus

All Calculus 1 Resources

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