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Calculus 1 : Rate of Change

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Example Questions

Example Question #201 : 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:

$\approx 10.4568$

$\approx 6.6867$

$\approx 5.2456$

$\approx 4.3333$

$\approx -2.5783$

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 : 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:

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

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

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

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

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

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 #203 : Rate Of Change

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:

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

$\frac{1}{2}$

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

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

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 : 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:

$14\pi$

$28\pi$

$196\pi$

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 #206 : 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 a seventh of the rate of growth of the surface area?

Possible Answers:

$\frac{8}{7}$

$\frac{1}{7}$

$\frac{2}{7}$

$\frac{16}{7}$

$\frac{4}{7}$

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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Example Question #207 : Rate Of Change

A cube is growing in size. What is the length of the diagonal of the cube at the time that the rate of growth of the cube's volume is equal to an eigth of the rate of growth of its surface area?

Possible Answers:

$\frac{1}{2}$

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

$2\sqrt{3}$

$\sqrt{3}$

Correct answer:

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

Explanation:

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

V=s^3

A=6s^2

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

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the cube's volume is equal to an eigth of the rate of growth of its surface area:

$3s^2\frac{ds}{dt}=(\frac{1}{8})12s\frac{ds}{dt}$

$s=(\frac{1}{8})4$

$s=\frac{1}{2}$

The diagonal of a cube is given by the equation:

$d=\sqrt{3s^2}$

$d=s\sqrt{3}$

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

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Example Question #208 : Rate Of Change

A cube is diminishing in size into nothingness. What is the area of one of the cube's sides at the time that the rate of shrinkage of the cube's volume is equal to thirteen times the rate of shrinkage of its surface area?

Possible Answers:

16224

$\frac{16}{169}$

2704

$\frac{96}{169}$

104

Correct answer:

2704

Explanation:

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

V=s^3

A=6s^2

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

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

$\frac{dA}{dt}=12s\frac{ds}{dt}$

Now, solve for the length of the side of the cube to satisfy the problem condition, the rate of shrinkage of the cube's volume is equal to thirteen times the rate of shrinkage of its surface area:

$3s^2\frac{ds}{dt}=(13)12s\frac{ds}{dt}$

s=(13)4

s=52

The area of one of a cube's is given by the equation:

a=s^2

a=52^2

a=2704

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Example Question #209 : Rate Of Change

A regular tetrahedron is growing in size. What is the surface area of the tetrahedron at the time the rate of growth of its volume is twice the rate of growth of its surface area?

Possible Answers:

$96\sqrt{3}$

$48\sqrt{3}$

$384\sqrt{3}$

$84\sqrt{3}$

$192\sqrt{3}$

Correct answer:

$384\sqrt{3}$

Explanation:

To tackle this problem, define a regular tetrahedron's dimensions in terms of the length of its sides:

$V=\frac{s^3}{6\sqrt{2}}$

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

Rates of change can then be found by taking the derivative of each property with respect to time:

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

$\frac{dA}{dt}=2\sqrt{3}s\frac{ds}{dt}$

The rate of change of the sides isn't going to vary no matter what dimension of the tetrahedron we're considering, so given our problem condition, the rate of growth of its volume is twice the rate of growth of its surface area, solve for the corresponding length of the tetrahedron's sides:

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

$s^2=(2)4\sqrt{6}s$

$s=8\sqrt{6}$

The surface area of the tetrahedron:

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

$A=\sqrt{3}(8\sqrt{6})^2$

$A=384\sqrt{3}$

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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 #201 : Rate Of Change

Example Question #201 : Rate Of Change

Example Question #202 : Rate Of Change

Example Question #203 : Rate Of Change

Example Question #204 : Rate Of Change

Example Question #205 : Rate Of Change

Example Question #206 : Rate Of Change

Example Question #207 : Rate Of Change

Example Question #208 : Rate Of Change

Example Question #209 : Rate Of Change

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