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Calculus 1 : How to find rate of change

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Example Question #71 : How To Find Rate Of Change

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:

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

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

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

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

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

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 #161 : Rate

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:

$12\pi$

$160\pi$

$-2\pi$

$48\pi$

$108\pi$

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 #162 : Rate

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

Possible Answers:

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

$\cos(5x^3)$

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

$\cos(15x^2)$

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 #163 : Rate

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

Possible Answers:

500

750

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 #164 : Rate

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

Possible Answers:

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

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

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

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

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 #165 : Rate

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

Possible Answers:

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

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

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

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

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 #166 : Rate

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:

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

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

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

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

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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Example Question #167 : Rate

The legs of a right triangle are each growing at a rate of $0.6\frac{in}{s}$ . If one leg has a length of five inches and the other a length of twelve inches, what is the rate of growth of hypotenuse?

Possible Answers:

$0.47\frac{in}{s}$

$0.85\frac{in}{s}$

$0.79\frac{in}{s}$

$0.6\frac{in}{s}$

$0.3\frac{in}{s}$

Correct answer:

$0.79\frac{in}{s}$

Explanation:

Begin by finding the hypotenuse using the Pythagorean theorem:

c^2=a^2+b^2

c^2=5^2in^2+12^2in^2

c^2=169in^2

c=13in

Now, to relate the rates of growth, take the time derivative of the Pythagorean Theorem:

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

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

$\frac{dc}{dt}=\frac{5in(0.6\frac{in}{s})+12in(0.6\frac{in}{s})}{13in}$

$\frac{dc}{dt}=0.79\frac{in}{s}$

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Example Question #168 : Rate

The radius of a circle with a square inscribed inside of it (such that each corner touches the edge of the circle) increases at a rate of $0.7\frac{in}{s}$ . What is the rate of growth of the difference in areas of the circle and square if the radius of the circle is $5\sqrt2$ ?

Possible Answers:

$7\sqrt2(2-\pi)\frac{in^2}{s}$

$14\sqrt2\frac{in^2}{s}$

$\pi7\sqrt2\frac{in^2}{s}$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

$7\sqrt2(\pi+2)\frac{in^2}{s}$

Correct answer:

$7\sqrt2(\pi-2)\frac{in^2}{s}$

Explanation:

The areas of the circle and square can be written in terms of the radius of the circle as follows:

$A_{circle}=\pi r^2 = 50\pi in^2$

$A_{square}=(\frac{2r}{\sqrt2})^2=2r^2=50in^2$

The area of the square follows from the fact that the diameter of the circle forms the diagonal of the square, and each side is equal to the diagonal divided by the square root of two. This can be verified via the Pythagorean Theorem (s^2+s^2=d^2) .

Now, the rate of change of areas can be found by taking the derivative of each function with respect to time:

$\frac{dA_{circle}}{dt}=2\pi r \frac{dr}{dt}$

$\frac{dA_{square}}{dt}=4r \frac{dr}{dt}$

Therefore, the rate of growth of the difference in the two areas can be found by finding the difference of these two quantities:

$2r(\pi-2)\frac{dr}{dt}$

$2(5\sqrt2in)(\pi-2)(0.7\frac{in}{s})$

$7\sqrt2(\pi-2)\frac{in^2}{s}$

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Example Question #71 : Rate Of Change

The sides of a square have lengths 5+t^2 . What is the rate of growth of the area at time t=2 ?

Possible Answers:

Correct answer:

Explanation:

The area of a square is the square of its sides:

A=s^2

A(t)=(5+t^2)^2

At time t=2

A(2)=(5+4)^2

A=81

Now to find the rate of change of the area as related to the rate of change of the sides, take the derivative of the area equation with respect to time:

$\frac{dA(t)}{dt}=2s\frac{ds}{dt}$

Since s=5+t^2 , we will use the power rule which states,

$x^n\ dx=nx^{n-1}$

therefore

$\frac{ds}{dt}=2t$ , giving:

$\frac{dA(t)}{dt}=2(5+t^2)(2t)$

$\frac{dA(2)}{dt}=2(5+2^2)(2(2))$

$\frac{dA(2)}{dt}=72$

Note that the same result would be found if you took the derivative of the equation wherein the 5+t^2 was substituted in for :

A(t)=(5+t^2)^2

Use the power rule and chain rule to find the derivative of area with respect to time.

Chain Rule: $f(g(x)) dx=f'(g(x))\cdot g'(x)$

$\frac{dA(t)}{dt}=2(2t)(5+t^2)$

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Calculus 1 : How to find rate of change

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #71 : How To Find Rate Of Change

Example Question #161 : Rate

Example Question #162 : Rate

Example Question #163 : Rate

Example Question #164 : Rate

Example Question #165 : Rate

Example Question #166 : Rate

Example Question #167 : Rate

Example Question #168 : Rate

Example Question #71 : Rate Of Change

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