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Calculus 1 : Calculus

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Example Questions

Example Question #53 : How To Find Rate Of Change

A circle's circumference is increasing at a rate of 0.6 feet per second. If the diameter of the circle is 80 feet at a moment, how fast is the area of the circle increasing at the same moment?

Possible Answers:

$24 \frac{ft}{sec}$

None of the above.

$25 \frac{ft}{sec}$

$27\frac{ft}{sec}$

$26 \frac{ft}{sec}$

Correct answer:

$24 \frac{ft}{sec}$

Explanation:

In order to solve this problem, we must know that the circumfrence of a circle is equivalent to $C=2\pi r$ .

Therefore by taking the derivative with respect to time, we obtain

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

Given that we know that

$\frac{dC}{dt}=0.6$ , we can solve for $\frac{dr}{dt}$ .

$0.6=2\pi\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{3}{10\pi}$

In order to solve for how fast the area of the circle changes, we must know that the area of a circle is defined as $A=\pi r^2$ .

Taking the derivative with respect to time, we obtain

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

We know that $r=40,\frac{dr}{dt}=\frac{3}{10\pi}$ , so by plugging in all the variables, we can solve for $\frac{dA}{dt}$ .

$\frac{dA}{dt}=2\pi(40)\left(\frac{3}{10\pi}\right)$

$\frac{dA}{dt}=24 \frac{ft}{sec}$ .

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Example Question #51 : How To Find Rate Of Change

Find the rate of change of the function f(t) = tan(t) + 4t^2 - 6 .

Possible Answers:

sec^2(t) + 4t^2

sin^2(t) + 4t^2

cos(t) - 8t

sec^2(t) + 8t

tan(t) +4t

Correct answer:

sec^2(t) + 8t

Explanation:

The rate of change of a function is its derivative.

Recall the following rules of differentiation to help solve this problem.

Power Rule:

$\frac{\mathrm{d} }{\mathrm{d} t}(at^{n}) = nat^{n-1}$

Differentiation rule for tangent:

$\frac{\mathrm{d} }{\mathrm{d} t}(atan(bt)) = absec^2(bt)$

Therefore, the rate of change of f(t) , by the power rule and the differentiation rule for tangent, is

f'(t) = sec^2(t) + 8t .

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Example Question #51 : How To Find Rate Of Change

The sides of a square shrink at a rate of $0.01\frac{in}{s}$ . What is the rate of growth of the square if its sides have lengths of 10in ?

Possible Answers:

$0.1\frac{in^2}{s}$

$-0.1\frac{in^2}{s}$

$0.0002\frac{in^2}{s}$

$-0.2\frac{in^2}{s}$

$-0.0001\frac{in^2}{s}$

Correct answer:

$-0.2\frac{in^2}{s}$

Explanation:

The area of a square is given by the formula:

A=l^2

The rate of growth of the area can be related to the rate of growth of sides by differentiating each side with respect to time:

$\frac{dA}{dt}=\frac{d}{dt}l^2=2l\frac{dl}{dt}$

Therefore, the rate of growth of the square is:

$\frac{dA}{dt}=2(10in)(-0.01\frac{in}{s})=-0.2\frac{in^2}{s}$

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Example Question #141 : Rate

The area of a circle is increasing at a rate of $0.14\pi \frac{in^2}{s}$ . If the area of the circle is $49\pi$ , what is the rate of increase of the radius?

Possible Answers:

$0.02\frac{in}{s}$

$0.01\frac{in}{s}$

$0.07\frac{in}{s}$

$0.1\frac{in}{s}$

$\sqrt{0.07}\frac{in}{s}$

Correct answer:

$0.01\frac{in}{s}$

Explanation:

The area of circle in terms of its radius is given as:

$A=\pi r^2$

This can be rewritten to find the radius:

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

In this problem, the radius can be found to be:

$r=\sqrt{\frac{49\pi}{\pi}}=7$

Now, relate rates of change by deriving each side of the area equation with respect to time:

$\frac{dA}{dt}=\frac{d}{dt}(\pi r^2)=2\pi r \frac{dr}{dt}$

Solve then for $\frac{dr}{dt}$ :

$\frac{dr}{dt}=\frac{1}{2\pi r}\frac{dA}{dt}$

$\frac{dr}{dt}=\frac{1}{14\pi in}(0.14\pi\frac{in^2}{s})$

$\frac{dr}{dt}=0.01\frac{in}{s}$

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Example Question #2971 : Calculus

A triangle is growing taller at a rate of $1 \frac{m}{s}$ . If it has a base of 6 m and height of , how fast is the area of the triangle changing?

Possible Answers:

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

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

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

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

$0.5\frac{m^2}{s}$

Correct answer:

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

Explanation:

The area of a triangle in terms of its base and height is given by the formula:

$A=\frac{1}{2}bh$

To find how the rates of change of each term relate, derive each side of the equation with respect to time:

$\frac{dA}{dt}=\frac{d}{dt}(\frac{1}{2}bh)$

$\frac{dA}{dt}=\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}$

The base isn't widening, so $\frac{db}{dt}=0$

Therefore:

$\frac{dA}{dt}=\frac{1}{2}(6m)(1\frac{m}{s})+\frac{1}{2}(1m)(0)$

$\frac{dA}{dt}=3\frac{m^2}{s}$

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Example Question #2972 : Calculus

A rectangle is inexplicably changing in shape. Its length is growing at a rate of $3\frac{in}{s}$ and its width is shrinking at a rate of $2\frac{in}{s}$ . If the length is 20 in and its width is 15 in , what is the rate of change of the area?

Possible Answers:

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

$85\frac{in^2}{s}$

$45\frac{in^2}{s}$

$6\frac{in^2}{s}$

$-6\frac{in^2}{s}$

Correct answer:

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

Explanation:

The area of a rectangle is given in terms of its length and width as:

A=lw .

The rates of change of each parameter with respect to time can be found by deriving each side of the equation with respect to time:

$\frac{dA}{dt}=l\frac{dw}{dt}+w\frac{dl}{dt}$

Therefore, with our known values, it's possible to find the inexplicable rate of change of the rectangle. Remember that the width is shrinking, so its rate should be treated as negative:

$\frac{dA}{dt}=20in(-2\frac{in}{s})+15in(3\frac{in}{s})$

$\frac{dA}{dt}=5\frac{in^2}{s}$

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Example Question #61 : How To Find Rate Of Change

The sides of a square are shrinking at an increasing rate. If the rate of rate increase is $0.2\frac{in}{s^2}$ , the rate of increase is $0.01\frac{in}{s}$ and the sides of a length of 4in , what is the rate at which the rate of growth of the area of the square is changing?

Possible Answers:

$1.62\frac{in^2}{s^2}$

$1.6\frac{in^2}{s^2}$

$0.2\frac{in^2}{s^2}$

$0.02\frac{in^2}{s^2}$

$1.2\frac{in^2}{s^2}$

Correct answer:

$1.62\frac{in^2}{s^2}$

Explanation:

Note for this problem, we're looking for a change in a rate, so this is dealing with second time derivatives:

Relating the area of a square to length:

A=l^2

Leads to

$\frac{d^2A}{dt^2}=\frac{d^2}{dt^2}l^2$

$\frac{d^2A}{dt^2}=\frac{d}{dt}(2l\frac{dl}{dt})$

Now, it's important to note that the $\frac{dl}{dt}$ term cannot be ignored in this second derivation; treat it like another variable!

$\frac{d^2A}{dt^2}=2l\frac{d^2l}{dt^2}+2\frac{dl}{dt}\frac{dl}{dt}$

$\frac{d^2A}{dt^2}=2l\frac{d^2l}{dt^2}+2(\frac{dl}{dt})^2$

Plugging in the given values, this gives:

$\frac{d^2A}{dt^2}=2(4in)(0.2\frac{in}{s^2})+2(0.01\frac{in}{s})^2$

$\frac{d^2A}{dt^2}=1.6\frac{in^2}{s^2}+0.02\frac{in^2}{s^2}$

$\frac{d^2A}{dt^2}=1.62\frac{in^2}{s^2}$

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Example Question #2973 : Calculus

A child is breathing into a bubble wand to create a soap bubble. Treating the bubble as an expanding sphere, If the sphere has a volume of $\frac{32}{3}\pi cm^3$ and is growing at a rate of $\pi\frac{cm^3}{s}$ , what is the rate of growth of the sphere's radius?

Possible Answers:

$\frac{1}{4}\frac{cm}{s}$

$\frac{\pi}{16}\frac{cm}{s}$

$\frac{\pi}{4}\frac{cm}{s}$

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

$\frac{1}{16}\frac{cm}{s}$

Correct answer:

$\frac{1}{16}\frac{cm}{s}$

Explanation:

Begin this problem by solving for the radius. The volume of a sphere is given as

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

Solving for the radius:

$r=\sqrt[3]{\frac{3}{4\pi}\frac{32}{3}\pi cm^3}=2cm$

Now time rate of change between quantities can be found by deriving each side with respect to time:

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

Solving for the radius rate of time then gives:

$\frac{dr}{dt}=\frac{1}{4\pi r^2}\frac{dV}{dt}$

$\frac{dr}{dt}=\frac{1}{16\pi cm^2}(\pi \frac{cm^3}{s})$

$\frac{dr}{dt}=\frac {1}{16}\frac{cm}{s}$

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Example Question #2974 : Calculus

A triangle is being stretched in a peculiar way. The base is shrinking at a rate of $1\frac{m}{s}$ while the height is increasing at a rate of $2\frac{m}{s}$ . If the height of the triangle is 20m and the base of the triangle is , what is the rate of change of the triangle's area?

Possible Answers:

$16\frac{m^2}{s}$

$17\frac{m^2}{s}$

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

$23\frac{m^2}{s}$

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

Correct answer:

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

Explanation:

The area of a triangle is given by the function:

$A=\frac{1}{2}bh$

The relationship between the rate of change of each parameter can be found by deriving each side of the equation with respect to time:

$\frac{dA}{dt}=\frac{1}{2}b\frac{dh}{dt}+\frac{1}{2}h\frac{db}{dt}$

$\frac{dA}{dt}=\frac{1}{2}\left(6m(2\frac{m}{s})+20m(-1\frac{m}{s})\right)$

$\frac{dA}{dt}=-4\frac{m^2}{s}$

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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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #53 : How To Find Rate Of Change

Example Question #51 : How To Find Rate Of Change

Example Question #51 : How To Find Rate Of Change

Example Question #141 : Rate

Example Question #2971 : Calculus

Example Question #2972 : Calculus

Example Question #61 : How To Find Rate Of Change

Example Question #2973 : Calculus

Example Question #2974 : Calculus

Example Question #2975 : Calculus

All Calculus 1 Resources

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