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Calculus 1 : Calculus

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Example Questions

Example Question #188 : How To Find Rate Of Change

Use normal rules of differentiation to find the derivative of the function $f(x)=\frac{1}{3} x^{-3} \arctan x$ .

Possible Answers:

$- \frac{\arctan x}{x^4}+\frac{1}{3x(1+x^2)}$

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

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

This function is differentiable at no point in its domain.

$-x^{-2}+\frac{1}{3}x^{-3}\sec^2x$

Correct answer:

$- \frac{\arctan x}{x^4}+\frac{1}{3x(1+x^2)}$

Explanation:

We take the derivative of the function:

$f'(x)=\frac{d}{dx}[\frac{1}{3} x^{-3} \arctan x]$

We notice that this function is the product of two functions, so we use the product rule:

$=\frac{d}{dx}[\frac{1}{3}x^{-3}]\arctan x+\frac{1}{3}x^{-3}\frac{d}{dx}[\arctan x]$

$=\frac{1}{3}\cdot (-3)x^{-4}\arctan x+\frac{1}{3}x^{-3}\frac{1}{1+x^2}$

Simplifying, we get:

$=-x^{-4}\arctan x+\frac{1}{3x^3(1+x^2)}$ .

Which is equivalent to:

$=-\frac{\arctan x}{x^4} +\frac{1}{3x^3(1+x^2)}$ , our answer

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Example Question #181 : How To Find Rate Of Change

Using normal rules of differentiation, find the derrivative of the function,

$f(x)=\frac{\sin\alpha \cdot \log_3 y}{4^y}$ with respect to

Possible Answers:

This function is not differentiable at any point on its domain.

$\cos \alpha 4^{-y}+\log_3 y \ln 4$

$\sin \alpha\cdot 4^{-y}(\ln4 \log_3 y-\frac{1}{y \ln 3})$

$-\cos \alpha 4^{-y}-\log_3 y \ln 4$

$\sin \alpha\cdot 4^{-y}(\frac{1}{y \ln 3}-\ln4 \log_3 y)$

Correct answer:

$\sin \alpha\cdot 4^{-y}(\frac{1}{y \ln 3}-\ln4 \log_3 y)$

Explanation:

taking the derivative

$f'(x)=\frac{d}{dx}[\frac{\sin\alpha \cdot \log_3 y}{4^y}]$

we notice we can factor out a coeficient:

$=\sin\alpha \cdot\frac{d}{dx}[\frac{\log_3 y}{4^y}]$

then, we apply the quotient rule:

$=\sin\alpha \cdot(\frac{\frac{d}{dx}[\log_3 y]\cdot 4^y- \log_3 y\cdot \frac{d}{dx}[4y]}{(4^y)^2})$

$=\sin\alpha \cdot (\frac{\frac{1}{y\ln3}\cdot 4^y- \log_3 y\cdot 4^y\ln 4}{4^{2y}})$

taking out a common factor:

$=\sin\alpha \cdot 4^y(\frac{\frac{1}{y\ln3}-\ln 4 \log_3 y}{4^{2y}})$

and simplifying,

$=\sin\alpha \cdot 4^{-y}(\frac{1}{y\ln3}-\ln 4 \log_3 y)$

we have our answer.

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Example Question #190 : How To Find Rate Of Change

The radius of a sphere and the sides of a cube begin to grow at the same rate. What must the ratio of the radius' length to the sides' length be such that the rate of growth of the two shapes' volumes is equal?

Possible Answers:

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

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

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

$\sqrt[3]{\frac{3}{4\pi}}$

$\sqrt[3]{\frac{4\pi}{3}}$

Correct answer:

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

Explanation:

Begin by writing the volume equations for a sphere and a cube:

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

V_c = s^3

The rate of change of these volumes can be found by taking the derivatives of these equation with respect to time:

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

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

We're given the relationship between the rate of expansion of the pertinent lengths for these shapes, namely $\frac{dr}{dt}=\frac{ds}{dt}$ .

To find the necessary ratios of lengths for the volumes to have equal rates of expansion, set the volume rate expressions equal to each other:

$\frac{dV_s}{dt}=\frac{dV_c}{dt}$

$4\pi r^2 \frac{dr}{dt}=3s^2 \frac{ds}{dt}$

$4\pi r^2 =3s^2$

$\frac{r^2}{s^2}=\frac{3}{4\pi}$

$\frac{r}{s}=\sqrt{\frac{3}{4\pi}}$

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Example Question #191 : Rate Of Change

If the position of an object at time is represented by the function p(t) = 3t^2-12t+9 , when does the object stop moving (i.e. The velocity is zero)?

Possible Answers:

The velocity is never .

t=2

t=1 or t=3

t=3

t=1

Correct answer:

t=2

Explanation:

When the velocity is , that means p'(t)=0 . That is, 6t-12=0 . So t=2 .

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Example Question #192 : Rate Of Change

The width of a rectangular prism increases half as fast as its length and a third as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the length, height, and width are equal?

Possible Answers:

$\frac{1}{4}w^2$

3w^2

$\frac{1}{6}w^2$

6w^2

5w^2

Correct answer:

6w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

The rate of change of the volume can be found by taking the derivative of the equation with respect to time:

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases half as fast as its length and a third as fast as its height: $\frac{dw}{dt}=\frac{1}{2}\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{3}\frac{dh}{dt}$

The width when the length, height, and width are equal: w=l=h

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w^2(2\frac{dw}{dt})+w^2(3\frac{dh}{dt})+w^2\frac{dw}{dt}$

$\frac{dV}{dt}=6w^2\frac{dw}{dt}$

The rate of change of the volume is 6w^2 times the rate of change of the rate of change of the width.

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Example Question #193 : Rate Of Change

The width of a rectangular prism increases twice as fast as its length and half as fast as its height. How does the rate of change of the prism's volume compare to that of the rate of change of the width when the width is half the length, which is half the height?

Possible Answers:

9w^2

14w^2

7w^2

22w^2

20w^2

Correct answer:

14w^2

Explanation:

Begin by writing the expression for the volume of a rectangular prism:

V=whl

The rate of change of the volume can be found by taking the derivative of the equation with respect to time:

$\frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}$

Now, we're given some information:

The width of a rectangular prism increases twice as fast as its length and half as fast as its height: $\frac{dw}{dt}=2\frac{dl}{dt},\frac{dw}{dt}=\frac{1}{2}\frac{dh}{dt}$

The width is half the length, which is half the height: $w=\frac{1}{2}l,l=\frac{1}{2}h\rightarrow w=\frac{1}{4}h$

Using this, rewrite the volume equation in terms of width:

$\frac{dV}{dt}=w(4w)(\frac{1}{2}\frac{dw}{dt})+w(2w)(2\frac{dw}{dt})+(4w)(2w)\frac{dw}{dt}$

$\frac{dV}{dt}=2w^2\frac{dw}{dt}+4w^2\frac{dw}{dt}+8w^2\frac{dw}{dt}$

$\frac{dV}{dt}=14w^2\frac{dw}{dt}$

The rate of change of the volume is 14w^2 times the rate of change of the rate of change of the width.

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Example Question #194 : 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 equal to the rate of shrinkage of the surface area?

Possible Answers:

$8\pi$

$4\pi$

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 equal to the rate of shrinkage of the surface area, let's solve for a radius that satisfies it.

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

r =2

Since the diameter is twice this, d=4

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Example Question #195 : Rate Of Change

A regular tetrahedron is growing in size. What is the height of the tetrahedron at the time the rate of growth of its sides is equal to the rate of growth of its volume?

Possible Answers:

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

$\frac{6\sqrt[4]{2}}{\sqrt{3}}$

$\frac{2\sqrt[4]{2}}{{3}}$

$\frac{\sqrt{3}}{\sqrt[4]{2}}$

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

Correct answer:

$\frac{2\sqrt[4]{2}}{\sqrt{3}}$

Explanation:

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

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

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

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

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 sides is equal to the rate of growth of its volume, solve for the corresponding length of the tetrahedron's sides:

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

$s^2=2\sqrt{2}=\sqrt{8}$

$s=\sqrt[4]{8}$

The height of a tetrahedron is given by the equation:

$h=\frac{\sqrt{6}}{3}s$

$h=\frac{\sqrt{6}}{3}\sqrt[4]{8}=\frac{2\sqrt[4]{2}}{\sqrt{3}}$

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Example Question #196 : Rate Of Change

A cube is growing in size. What is the length of the sides of the cube at the time that the rate of growth of the cube's surface is equal to 1.5 times the rate of growth of its volume?

Possible Answers:

$\frac{3}{4}$

$\frac{4}{3}$

$\frac{6}{5}$

Correct answer:

$\frac{4}{3}$

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 growth of the cube's surface is equal to 1.5 times the rate of growth of its volume:

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

$s=\frac{12}{9}=\frac{4}{3}$

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Example Question #197 : Rate Of Change

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

Possible Answers:

$\frac{1}{\sqrt[4]{3}}$

$\frac{\sqrt{3}}{4}$

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

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

$\frac{\sqrt[4]{3}}{4}$

Correct answer:

$\frac{\sqrt{2}}{\sqrt[4]{3}}$

Explanation:

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

V=s^3

$d=s\sqrt{3}$

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

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

$\frac{dd}{dt}=\sqrt{3}\frac{ds}{dt}$

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 diagonal:

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

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

$s=\frac{\sqrt{2}}{\sqrt[4]{3}}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #188 : How To Find Rate Of Change

Example Question #181 : How To Find Rate Of Change

Example Question #190 : How To Find Rate Of Change

Example Question #191 : Rate Of Change

Example Question #192 : Rate Of Change

Example Question #193 : Rate Of Change

Example Question #194 : Rate Of Change

Example Question #195 : Rate Of Change

Example Question #196 : Rate Of Change

Example Question #197 : Rate Of Change

All Calculus 1 Resources

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