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Calculus 1 : Rate of Change

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Example Question #41 : How To Find Rate Of Change

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The seconds hand in the clock shown has a length of five inches. Find the velocity of its tip in the and directions at the depicted moment in time.

Possible Answers:

$x:-\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi) \frac{in}{s}$

$y:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$y:-\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$x:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

$y:\frac{\pi}{3}sin(\frac{5}{6}\pi)\frac{in}{s}$

Correct answer:

$x:\frac{\pi}{3}sin(\frac{5}{6}\pi) \frac{in}{s}$

$y:\frac{-\pi}{3}cos(\frac{5}{6}\pi)\frac{in}{s}$

Explanation:

For this problem, it will be useful to work in radians. The angle of the secondshand, between the positive x-axis and itself, is given as:

$\theta=\frac{5}{12}(2\pi)=\frac{5}{6}\pi$

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Now, to find the x and y velocities, first write out the form of their positions:

$x=rcos(\theta)$

$y=rsin(\theta)$

Therefore, their respective velocities are given by their respective derivatives:

$\frac{dx}{dt}=\frac{d}{dt}(rcos(\theta))=\frac{dr}{dt}cos(\theta)-r\frac{d\theta}{dt}sin(\theta)$

$\frac{dy}{dt}=\frac{d}{dt}(rsin(\theta))=\frac{dr}{dt}sin(\theta)+r\frac{d\theta}{dt}cos(\theta)$

The secondshand doesn't change in length, so $\frac{dr}{dt}=0$ . As for $\frac{d\theta}{dt}$ , that can be found knowing how long it takes the secondshand to complete the circuit around the clock:

$\frac{d\theta}{dt}=-\frac{2\pi}{180}\frac{360}{60}\frac{rad}{s}=-\frac{\pi}{15}\frac{rad}{s}$

Note that movement in the clockwise direction, when dealing with angular motion, is treated as negative.

Knowing this, we can find the velocities:

$\frac{dx}{dt}=-r\frac{d\theta}{dt}sin(\theta)=\frac{\pi}{3}sin(\frac{5}{6}\pi)$

$\frac{dy}{dt}=r\frac{d\theta}{dt}cos(\theta)=\frac{-\pi}{3}cos(\frac{5}{6}\pi)$

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Example Question #42 : How To Find Rate Of Change

A metal cylinder with a radius of five inches and a height of twenty inches is being heated, causing it to expand. If the radius grows at a rate of $0.001\frac{in}{s}$ and the height grows at a rate of $0.002\frac{in}{s}$ , what is the rate of expansion of the cylinder's volume?

Possible Answers:

$0.125\frac{in^3}{s}$

$0.425\pi \frac{in^3}{s}$

$0.25\pi \frac{in^3}{s}$

$0.625\frac{in^3}{s}$

$0.5\pi \frac{in^3}{s}$

Correct answer:

$0.25\pi \frac{in^3}{s}$

Explanation:

The volume of a cylinder is given by the formula:

$V=\pi r^2h$ wherein represents the cylinder's radius, and its height.

Therefore, to find the rate of expansion, derive each side of the equation with respect to time:

$\frac{dV}{dt}=\frac{d}{dt}(\pi r^2h)=2\pi r h\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}$

Therefore, for the values given in the problem statement, the rate of expansion of the volume can be found:

$\frac{dV}{dt}=2\pi r h\frac{dr}{dt}+\pi r^2 \frac{dh}{dt}=2\pi(5in)(20in)\left(0.001\frac{in}{s}\right)+\pi(5in)^2\left(0.002\frac{in}{s}\right)$

$\frac{dV}{dt}=0.25\pi \frac{in^3}{s}$

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Example Question #41 : How To Find Rate Of Change

The sides of a right triangle are increasing in length. If the shorter side, which has a length of 3in , increases at a rate of $0.3\frac{in}{s}$ , and the longer side, which has a length of 4in , increases at a rate of $0.2\frac{in}{s}$ , what is the rate of growth of the hypotenuse?

Possible Answers:

$\sqrt{0.13}\frac{in}{s}$

$0.68\frac{in}{s}$

$\sqrt{0.65}\frac{in}{s}$

$0.34\frac{in}{s}$

$1.7\frac{in}{s}$

Correct answer:

$0.34\frac{in}{s}$

Explanation:

Since this is a right triangle, the hypotenuse can be related to the sides utilizing the pythagorean theorem:

c^2=a^2+b^2

Although the root of each side may be taken, it would be simpler to take the derivative of each side with respect to time as is:

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

Therefore, the rate of change of the hypotenuse is:

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

Using the Pythagorean Theorem, the hypotenuse is:

$c=\sqrt{3^2+4^2}=5$

Thus:

$\frac{dc}{dt}=\frac{3in(0.3\frac{in}{s})+4in(0.2\frac{in}{s})}{5in}=0.34\frac{in}{s}$

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Example Question #1 : Comparing Relative Magnitudes (Exponential Growth,Logarithmic Growth, Polynomial Growth)

A cylinder of height 4in and radius 1in is expanding. The radius increases at a rate of $0.1\frac{in}{s}$ and its height increases at a rate of $0.2\frac{in}{s}$ . What is the rate of growth of its surface area?

Possible Answers:

$2.8\pi\frac{in^2}{s}$

$1.4\pi\frac{in^2}{s}$

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

$0.4\pi\frac{in^2}{s}$

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

Correct answer:

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

Explanation:

The surface area of a cylinder is given by the formula:

$A=2\pi r^2+2\pi rh$

To find the rate of growth over time, take the derivative of each side with respect to time:

$\frac{dA}{dt}=4\pi r \frac{dr}{dt}+2\pi h \frac{dr}{dt} + 2\pi r \frac{dh}{dt}$

Therefore, the rate of growth of surface area is:

$\frac{dA}{dt}=4\pi (1in)\left(0.1\frac{in}{s}\right)+2\pi(1in)(4in)\left(0.1\frac{in}{s}\right)+2\pi(1in)\left(0.2\frac{in}{s}\right)$

$\frac{dA}{dt}=(0.4\pi+0.8\pi+0.4\pi)\frac{in^2}{s}=1.6\pi\frac{in^2}{s}$

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Example Question #44 : How To Find Rate Of Change

Water is being poured into a cylindrical glass at a rate of $4\frac{in^3}{s}$ . If the cylinder has a diameter of 4in and a height of 8in , what is the rate at which the water rises?

Possible Answers:

$\frac{1}{2\pi}\frac{in}{s}$

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

$\frac{1}{\pi}\frac{in}{s}$

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

$\pi\frac{in}{s}$

Correct answer:

$\frac{1}{\pi}\frac{in}{s}$

Explanation:

The volume equation of a cylinder is

$V=\pi r ^2h$

Therefore, the rate of change of each term can be related by taking the derivative of each side of the equation:

$\frac{dV}{dt}=2\pi r h \frac{dr}{dt} + \pi r^2\frac{dh}{dt}$

Treating the glass as solid, the radius of liquid in it will not change, only the height, so $\frac{dr}{dt}=0$ . This simplifies the equation:

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

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

Since the radius is half the diameter:

$\frac{dh}{dt}=\frac{1}{\pi}\frac{in}{s}$

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Example Question #45 : How To Find Rate Of Change

The position in meters of a particle after seconds is modeled by the equation

$f(t)=2te^{t}$ , where $t\ge0$ .

At what rate, in meters per second, is the position of the particle changing at t = 3 seconds?

Possible Answers:

$8e^{3}$

$2e^{3}$

$2+e^{3}$

$6e^{3}$

Correct answer:

$8e^{3}$

Explanation:

The derivative of a function gives us a new function that describes the rate of change of the original function at every point in its domain. Therefore the derivative of f(t) will produce a new function that describes the rate of change in position with respect to time. Using the product rule, the derivative of f(t) is:

$f'(t)=2e^{t}+2te^{t}$

Next substitute 3 for t to find the rate of change of the particle at exactly 3 seconds.

$f'(3)=2e^{3}+2(3)e^{3} = 8e^{3}$ .

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Example Question #46 : How To Find Rate Of Change

What is the rate of change of the function

$y=\frac{x^2}{sin(x)}$ when $x=\frac{\pi}{6}$ ?

Possible Answers:

$\frac{12\pi-\pi^2\sqrt3}{18}$

$\frac{2\pi\sqrt3}{9}$

$\frac{\pi^2-\pi\sqrt2}{2}$

$\frac{\pi\sqrt2}{2}-\frac{\pi^2\sqrt2}{16}$

Correct answer:

$\frac{12\pi-\pi^2\sqrt3}{18}$

Explanation:

The find the rate of change of a function at a particular point, find the derivative of the function, then substitute the point into the function. Using the quotient rule, the derivative is:

$y'=\frac{2xsin(x)-x^2cos(x)}{sin^2(x)}$

Next substitute $\frac{\pi}{6}$ for x.

$y'(\frac{\pi}{6})=\frac{2(\frac{\pi}{6})sin(\frac{\pi}{6})-(\frac{\pi}{6})^2cos(\frac{\pi}{6})}{sin^2(\frac{\pi}{6})} =\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{(\frac{1}{2})^2}$

Now we need to simplify the right hand side.

$\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{(\frac{1}{2})^2} =\frac{(\frac{\pi}{3})(\frac{1}{2})-(\frac{\pi^2}{36})\frac{\sqrt3}{2}}{\frac{1}{4}}$

$=4(\frac{\pi}{3})(\frac{1}{2})-4(\frac{\pi^2}{36})\frac{\sqrt3}{2}$

$=\frac{2\pi}{3}-(\frac{\pi^2}{9})\frac{\sqrt3}{2}$

$=\frac{2\pi}{3}-\frac{\pi^2\sqrt3}{18}$

$=\frac{12\pi-\pi^2\sqrt3}{18}$

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Example Question #47 : How To Find Rate Of Change

What is the rate of change of a square's sides if it has an area of 144in^2 which is growing at a rate of $1\frac{in^2}{s}$ ?

Possible Answers:

$\frac{1}{12}\frac{in}{s}$

$\frac{1}{24}\frac{in}{s}$

$12\frac{in}{s}$

$\frac{1}{6}\frac{in}{s}$

$24\frac{in}{s}$

Correct answer:

$\frac{1}{24}\frac{in}{s}$

Explanation:

The area of a square in terms of the lengths of its sides is given as:

A=l^2

Therefore the length of each side can be found to be

l^2=144in^2

l=12in

Rates of change can also be related by taking the time derivative of each side:

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

The rate of change of the sides, $\frac{dl}{dt}$ , is what is being solved for:

$\frac{dl}{dt}=\frac{1}{2l}\frac{dA}{dt}=\frac{1}{2(12in)}(1\frac{in^2}{s})=\frac{1}{24}\frac{in}{s}$

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Example Question #48 : How To Find Rate Of Change

Find the derivative of the following function through implicit differentiation:

3y^2+2y=x^2+2sin(x)

Possible Answers:

x+cos(x)

$\frac{-x+cos(x)}{3y+1}$

$\frac{x+cos(x)}{3y+1}$

{3y+1}

$\frac{x+cos(x)}{3y+1}$

Correct answer:

$\frac{x+cos(x)}{3y+1}$

Explanation:

3y^2+2y=x^2+2sin(x)

By differentiating both sides:

6ydy+2dy=2xdx+2cos(x)dx

By solving for $\frac{dy}{dx}$ :

$\frac{dy}{dx}=\frac{x+cos(x))}{3y+1}$

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Example Question #49 : How To Find Rate Of Change

Air flows out of a spherical balloon at a rate of -12 units^3/min . What is the rate of change of the circumference when the radius of the balloon is units?

Possible Answers:

$\frac{6}{25} units/min$

$\frac{-3}{25\pi} units/min$

$\frac{-6}{25} units/min$

$\frac{3}{25\pi} units/min$

$0\, units/min$

Correct answer:

$\frac{-6}{25} units/min$

Explanation:

It is given that V'=-12 units^3/min .

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

By differentiating both sides with respect to time we can solve this problem:

$V'=4\pi r^2r'$

$r'=\frac{V'}{4\pi r^2}=\frac{-12}{4\pi 5^2} units/min=\frac{-3}{25\pi} units/min$

$C'=2\pi r'=\frac{-2\pi \cdot 3}{25 \pi} unit/min=\frac{-6}{25} units/min$

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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 #41 : How To Find Rate Of Change

Example Question #42 : How To Find Rate Of Change

Example Question #41 : How To Find Rate Of Change

Example Question #1 : Comparing Relative Magnitudes (Exponential Growth,Logarithmic Growth, Polynomial Growth)

Example Question #44 : How To Find Rate Of Change

Example Question #45 : How To Find Rate Of Change

Example Question #46 : How To Find Rate Of Change

Example Question #47 : How To Find Rate Of Change

Example Question #48 : How To Find Rate Of Change

Example Question #49 : How To Find Rate Of Change

All Calculus 1 Resources

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