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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 #1952 : Functions

A bar of length 10ft is resting against a wall, the ground end 6ft from the base of the wall. If the end resting on the ground is moved forward at a rate of $0.2\frac{ft}{s}$ , how fast does the end resting on the wall rise? The plane of the ground is perpendicular to the wall.

Possible Answers:

$-0.2\frac{ft}{s}$

$1\frac{ft}{s}$

$0.15\frac{ft}{s}$

$0.2\frac{ft}{s}$

$-0.15\frac{ft}{s}$

Correct answer:

$0.15\frac{ft}{s}$

Explanation:

The bar resting against the wall forms a right triangle, its length serving as the hypotenuse. This can be used to find the height of the end resting on the wall via the Pythagorean Theorem:

a^2+b^2=c^2

36ft^2+b^2=100ft^2

b^2=64ft^2

b=8ft

The rates of change can also be related using the Pythagorean Theorem:

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}$

Since the length of the bar isn't changing, $\frac{dc}{dt}=0$

$2b\frac{db}{dt}=-2a\frac{da}{dt}$

$\frac{db}{dt}=-\frac{a}{b}\frac{da}{dt}$

$\frac{db}{dt}=-\frac{6ft}{8ft}(-0.2)\frac{ft}{s}$

Note that the negative value is used because the end on the ground is closing the distance to the wall.

$\frac{db}{dt}=0.15\frac{ft}{s}$

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Example Question #64 : How To Find Rate Of Change

A circle is stretching into an ellipse, its horizontal radius expanding at a rate of 0.1m . If the circle has an area of $\pi m^2$ , what is the rate of growth?

Possible Answers:

$0.01\pi \frac{m^2}{s}$

$0.2\pi \frac{m^2}{s}$

$0.02\pi \frac{m^2}{s}$

$0.1\pi \frac{m^2}{s}$

$1\pi \frac{m^2}{s}$

Correct answer:

$0.1\pi \frac{m^2}{s}$

Explanation:

The area of a circle is given by the formula:

$A=\pi r^2$

This can be used to find the original radii:

$\pi r^2 =\pi m^2$

r=1m

Now, the area of an ellipse in terms of its vertical and horizontal radii is:

$A=\pi ab$

To relate rates of change, derive each side with respect to time:

$\frac{dA}{dt}=\pi a \frac{db}{dt}+\pi b \frac{da}{dt}$

Since the vertical radius does not change, let's designate it as , $\frac{db}{dt}=0$ ;

$\frac{dA}{dt}=\pi b \frac{da}{dt}$

$\frac{dA}{dt}=\pi(1m)(0.1\frac{m}{s})$

$\frac{dA}{dt}=0.1\pi \frac{m^2}{s}$

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Example Question #61 : How To Find Rate Of Change

The shorter leg of a right triangle is growing at a rate of $0.2\frac{in}{s}$ . If the shorter leg has a length of 5in and the hypotenuse has a length of 10in , what is the rate of growth of the angle across from the shorter leg? The hypotenuse is growing, but the longer leg is not.

Possible Answers:

$0.06\frac{rad}{s}$

$0.03\frac{rad}{s}$

$0.0346\frac{rad}{s}$

$0.0173\frac{rad}{s}$

$0.8127\frac{rad}{s}$

Correct answer:

$0.0173\frac{rad}{s}$

Explanation:

The hypotenuse and the shorter leg can be related in terms of the angle across from the leg:

$a=csin(\alpha )$

This angle can be found as:

$\alpha=sin^{-1}(\frac{a}{c})$

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

Now, rates of change can be related by deriving each side of the original equation with respect to time:

$\frac{da}{dt}=sin(\alpha)\frac{dc}{dt}+ccos(\alpha))\frac{d\alpha}{dt}$

However, we do not know what $\frac{dc}{dt}$ is. We can find this by using the Pythagorean theorem and once again deriving:

a^2+b^2=c^2

$2a\frac{da}{dt}+2b\frac{db}{dt}=2c\frac{dc}{dt}$

Since we're told the longer side does not change in length, $\frac{db}{dt}=0$ , leaving

$a\frac{da}{dt}=c\frac{dc}{dt}$

$\frac{dc}{dt}=\frac{a}{c}\frac{da}{dt}$

$\frac{dc}{dt}=\frac{5in}{10in}(0.2\frac{in}{s})$

$\frac{dc}{dt}=0.1\frac{in}{s}$

Now, let's return to this previous derivative using our known values and solve for $\frac{d\alpha}{dt}$ :

$\frac{da}{dt}=sin(\alpha)\frac{dc}{dt}+ccos(\alpha))\frac{d\alpha}{dt}$

$\frac{d\alpha}{dt}=\frac{\frac{da}{dt}-sin(\alpha)\frac{dc}{dt}}{ccos(\alpha)}$

$\frac{d\alpha}{dt}=\frac{0.2\frac{in}{s}-0.5(0.1\frac{in}{s})}{10in(\frac{\sqrt3}{2})}$

$\frac{d\alpha}{dt}=0.0173\frac{rad}{s}$

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Example Question #1955 : Functions

The circumference of a circle is increasing at a rate of $0.5\pi\frac{in}{s}$ . If the circle has an area of $9\pi in^2$ , what is the rate of growth of the area?

Possible Answers:

$3\frac{in^2}{s}$

$3\pi^2\frac{in^2}{s}$

$2.25\pi\frac{in^2}{s}$

$1.5\pi\frac{in^2}{s}$

$1.5\frac{in^2}{s}$

Correct answer:

$1.5\pi\frac{in^2}{s}$

Explanation:

Begin by finding the radius of the circle:

$A=\pi r^2$

$r=\sqrt{\frac{A}{\pi}}$

$r=\sqrt{\frac{9\pi in^2}{\pi}}$

r=3in

Now, rates of change can be related by taking the time derivative of both sides of an equation:

$\frac{dA}{dt}=2\pi r \frac{dr}{dt}$

However, the rate of change of the radius, $\frac{dr}{dt}$ , is still unknown. It can be found by relating it to the circumference's rate of change:

$C=2\pi r$

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

$\frac{dr}{dt}=\frac{1}{2\pi}0.5\pi\frac{in}{s}$

$\frac{dr}{dt}=0.25\frac{in}{s}$

Going back to our earlier equation, we can solve for the rate of change of the area:

$\frac{dA}{dt}=2 \pi (3in)(0.25\frac{in}{s})$

$\frac{dA}{dt}=1.5\pi\frac{in^2}{s}$

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Example Question #2981 : Calculus

The radius of a cylinder is given by the function 3+t , while the height at any time is given by the function 12-2t . What is the rate of growth of the cylinder at time t=1 ?

Possible Answers:

$12\pi$

$108\pi$

$48\pi$

$-2\pi$

$160\pi$

Correct answer:

$48\pi$

Explanation:

The volume of a cylinder is given by the equation:

$V=\pi r^2 h$

$V(t)=\pi (3+t)^2(12-2t)$

$V(t)=\pi(t^2+6t+9)(12-2t)$

$V(t)=\pi (108+54t -2t^3)$

From this, the rate of change can be found by taking the derivative with respect to time:

$\frac{dV(t)}{dt}=\pi(54-6t^2)$

$\frac{dV(1)}{dt}=\pi(54-6(1)^2)$

$\frac{dV(1)}{dt}=48\pi$

It may be worth noting that at time t=3 the cylinder actually begins to shrink.

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Example Question #2982 : Calculus

If $h(x)=\sin(5x^3)$ , what is h'(x) ?

Possible Answers:

$15x^2\cos(5x^3)$

$\cos(15x^2)$

$-5x^3\cos(5x^3)$

$\cos(5x^3)$

Correct answer:

$15x^2\cos(5x^3)$

Explanation:

The given function consists of a function, g(x)=5x^3 , inside another function, $f(x)=\sin(x)$ , such that h(x)=sin(5x^3)=f(g(x)) .

Thus we can use the Chain Rule to find h'(x) .

The Chain Rule says that if

h(x)=f(g(x)) ,

then

$h'(x)=f'(g(x))\cdot g'(x)$ .

Recall (or look up) that the derivative of sine is cosine, so $f'(x)=\cos(x)$ , and use the Power Rule to get g'(x)=15x^2 .

Combining the three functions f'(x) , g(x) , and g'(x) , we have $h'(x)=15x^2\cos(5x^3)$ .

Note: The Power Rule says that for a function

p(x)=kx^n , $p'(x)=knx^{n-1}$ .

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Example Question #2983 : Calculus

What is the rate of change of the function $g(x)=4x^{\frac{3}{2}}$ when x=25 ?

Possible Answers:

750

500

Correct answer:

Explanation:

The rate of change of the function $g(x)=4x^{\frac{3}{2}}$ at is the value of the derivative

$g'(x)=\frac{dg(x)}{dx}$ at x=25 .

Use the Power Rule to find that

$g'(x)=4\cdot\frac{3}{2}\cdot x^{\frac{3}{2}-1}=6x^{\frac{1}{2}}=6\sqrt{x}$ .

The rate of change at is

$g'(25)=6\sqrt{25}=30$ .

Note: The Power Rule says that for a function

p(x)=kx^n , $p'(x)=knx^{n-1}$ .

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Example Question #2984 : Calculus

What is $\frac{dy}{dx}$ if $y(x)=\frac{\sin x}{e^{x^2}}$ ?

Possible Answers:

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

$\frac{\sin x-2x\cos x}{4x^2e^{x^2}}$

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

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

Correct answer:

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

Explanation:

Since y(x) is a quotient of two functions $\sin x$ and $e^{x^2}$ , we can use the Quotient Rule, which says that for a function

$h(x)=\frac{f(x)}{g(x)}$ ,

$h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ .

$\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(e^{x^2})=2xe^{x^2}$ by the Chain Rule.

Applying the Quotient Rule,

$\\ \frac{dy}{dx}=\frac{e^{x^2}\cdot \frac{d}{dx}(\sin x)-\sin x\cdot \frac{d}{dx}(e^{x^2})}{(e^{x^2})^2}\\ \\=\frac{e^{x^2}\cdot \cos x-\sin x\cdot 2xe^{x^2}}{e^{2x^2}}\\ \\=\frac{e^{x^2}(\cos x-2x\sin x)}{e^{2x^2}}\\ \\=\frac{\cos x-2x\sin x}{e^{x^2}}$ .

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Example Question #2985 : Calculus

What is the rate of change of $f(x)=\sin^3 x$ when $x=\tfrac{\pi}{4}$ ?

Possible Answers:

$\tfrac{3\sqrt{2}}{4}$

$\tfrac{\sqrt{2}}{4}$

$\tfrac{\sqrt{2}}{2}$

$\tfrac{3\sqrt{2}}{2}$

Correct answer:

$\tfrac{3\sqrt{2}}{4}$

Explanation:

We are looking for $f'(\tfrac{\pi}{4})$ .

The Chain Rule says that for

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

Applying the Chain Rule,

$f'(x)=\cos x\cdot3(\sin x)^2=3\cos x \sin^2x$ .

So,

$f'(\tfrac{\pi}{4})=3\cos (\tfrac{\pi}{4}) \sin^2 (\tfrac{\pi}{4})=3(\tfrac{\sqrt{2}}{2})(\tfrac{\sqrt2}{2})^2=\tfrac{3\sqrt{2}}{4}$ .

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Example Question #71 : How To Find Rate Of Change

A spherical water balloon is filled at a rate of $0.2 \frac{m^3}{s}$ . What is the rate of change of the surface area of the water balloon when the balloon's radius is one meter?

Possible Answers:

$\frac{4}{5}\frac{m^2}{s}$

$\frac{6}{5}\frac{m^2}{s}$

$\frac{1}{20\pi} \frac{m^2}{s}$

$\frac{2}{5}\frac{m^2}{s}$

Correct answer:

$\frac{2}{5}\frac{m^2}{s}$

Explanation:

The rate of change of the surface area is found by relating the volume to the surface area:

$V=\frac{4}{3} \pi r^3$ , $\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

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

We must solve for the rate of change of the radius using the volume equation, and then solve for the rate of change of the surface area by plugging in the given radius and rate of change of the radius:

$\frac{dr}{dt}=\frac{1}{20\pi}=\frac{0.2}{4\pi (1)}$

$\frac{dA}{dt}=8\pi (1)\left(\frac{1}{20 \pi}\right)=\frac{2}{5}$

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1952 : Functions

Example Question #64 : How To Find Rate Of Change

Example Question #61 : How To Find Rate Of Change

Example Question #1955 : Functions

Example Question #2981 : Calculus

Example Question #2982 : Calculus

Example Question #2983 : Calculus

Example Question #2984 : Calculus

Example Question #2985 : Calculus

Example Question #71 : How To Find Rate Of Change

All Calculus 1 Resources

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