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Calculus 1 : Rate of Change

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Example Questions

Example Question #61 : Rate Of Change

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:

$0.1\frac{in}{s}$

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

$0.02\frac{in}{s}$

$0.01\frac{in}{s}$

$0.07\frac{in}{s}$

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 #61 : How To Find Rate Of Change

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:

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

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

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

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

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

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 #61 : How To Find Rate Of Change

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:

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

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

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

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

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

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 #63 : 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:

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

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

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

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

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

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 #62 : How To Find Rate Of Change

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:

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

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

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

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

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

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 #64 : Rate Of Change

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:

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

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

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

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

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

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 #63 : How To Find Rate Of Change

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:

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

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

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

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

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

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 #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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Calculus 1 : Rate of Change

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #61 : Rate Of Change

Example Question #61 : How To Find Rate Of Change

Example Question #61 : How To Find Rate Of Change

Example Question #63 : Rate Of Change

Example Question #62 : How To Find Rate Of Change

Example Question #64 : Rate Of Change

Example Question #63 : How To Find Rate Of Change

Example Question #1952 : Functions

Example Question #64 : How To Find Rate Of Change

Example Question #61 : How To Find Rate Of Change

All Calculus 1 Resources

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