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Calculus 1 : Functions

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Example Questions

Example Question #2975 : Calculus

The radius of the base of a cone is increasing at a rate of $0.2\frac{m}{s}$ . If the current radius is and its height is 20m , what is the rate of growth of its volume?

Possible Answers:

$\frac{40\pi}{3}\frac{m^3}{s}$

$\frac{20\pi}{3}\frac{m^3}{s}$

$\frac{8\pi}{3}\frac{m^3}{s}$

$\frac{4\pi}{15}\frac{m^3}{s}$

$\frac{2\pi}{15}\frac{m^3}{s}$

Correct answer:

$\frac{40\pi}{3}\frac{m^3}{s}$

Explanation:

The volume of a cone is given by the formula:

$V=\frac{1}{3}\pi r^2h$

To relate rate of changes over time, derive each side of the equation with respect to time:

$\frac{dV}{dt}=\frac{1}{3}\pi(2rh\frac{dr}{dt}+r^2\frac{dh}{dt})$

Since height isn't changing, $\frac{dh}{dt}=0$ , which leaves:

$\frac{dV}{dt}=\frac{2}{3}\pi(rh\frac{dr}{dt})$

$\frac{dV}{dt}=\frac{2}{3}\pi(5m(20m)(0.2)\frac{m}{s})$

$\frac{dV}{dt}=\frac{40\pi}{3}\frac{m^3}{s}$

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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 #61 : 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:

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

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

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

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

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

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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Calculus 1 : Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2975 : Calculus

Example Question #1952 : Functions

Example Question #61 : 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

All Calculus 1 Resources

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