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Calculus 1 : Calculus

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Example Questions

Example Question #31 : How To Find Rate Of Change

A cup with a shape of an inverted cone that is being filled up with water has a base radius of 3 inches and height 5 inches. If the water is being poured into the cup at a rate of $1\frac{in^{3}}{second}$ find the rate at which the water level of the cup is rising when the water is 4 inches deep.

Possible Answers:

$\frac{1}{2.14\pi}\frac{in^3}{sec}$

$\frac{1}{2.12\pi}\frac{inc^3}{sec}$

Not enough information is given.

$\frac{1}{2\pi}\frac{in^3}{sec}$

$\frac{1}{9.54\pi}\frac{in^3}{sec}$

Correct answer:

$\frac{1}{2.12\pi}\frac{inc^3}{sec}$

Explanation:

To solve this equation, we must first know that the equation for the volume of a cone is

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

where is the base radius of the cone and is the height of the cone.

We are asked to find the rate of which the water level of the cup changes after the water height in the cone is equal to 4 inches, therefore we take the derivative of the volume equation with respect to time.

To take the derivative of the volume equation, we must use the power rule

$\frac{d}{dx}x^n=nx^{n-1}$ as well as the product rule (uv)'=u'v+uv' .

The derivative comes out to be

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

We know that

$\frac{dV}{dt}=1\frac{in^3}{sec},r=2.4 in, h=4in,\frac{dr}{dt}=\frac{3}{5}\frac{dh}{dt}$

based off of the information given to us in the beginning; now we must just solve for $\frac{dh}{dt}$ .

Note that $r,\frac{dr}{dt}$ different then expected due to the ratio of the cone.

This ratio is

$\frac{h}{2r}=\frac{5}{6}$ or $r=\frac{3}{5}h$ .

$1\frac{in^3}{sec}=\frac{\pi}{3}\left(2.4^2\frac{dh}{dt}+\frac{3}{5}\frac{dh}{dt}\right)$

$\frac{3}{6.36\pi}\frac{in^3}{sec}=\frac{dh}{dt}=\frac{1}{2.12\pi}\frac{inc^3}{sec}$

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Example Question #1923 : Functions

A spherical snowball stuck into an oven and is melting at a rate such that its volume is decreasing by $2\frac{cm^3}{sec}$ . When the snowball has a radius of 10 centimeters, how fast is the diameter of the snowball changing? Assume only the surface of the snowball is melting at a uniform rate.

Possible Answers:

Not enough information is given.

$\frac{1}{300\pi}\frac{cm}{sec}$

$\frac{1}{100\pi}\frac{cm}{sec}$

$\frac{1}{400\pi}\frac{cm}{sec}$

$\frac{1}{200\pi}\frac{cm}{sec}$

Correct answer:

$\frac{1}{100\pi}\frac{cm}{sec}$

Explanation:

In order to solve this question, we must first know that the volume of a sphere is defined as

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

We are asked to find the rate of which the diameter of the snowball changes when it is melting. Therefore we must take the derivative of the volume equation with respect to time. To do this we will need to utilize the power rule

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

The derivative becomes

$\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$ and since we know that $r=10 cm, \frac{dV}{dt}=2\frac{cm^3}{sec}$ , we simply plug the variables into the equation and solve for $\frac{dr}{dt}$ .

$2\frac{cm^3}{sec}=4\pi (10cm)^2\frac{dr}{dt}$

$\frac{dr}{dt}=\frac{1}{200\pi}\frac{cm}{sec}$

However we are asked to find the rate of which the diamater of the snowball changes, therefore we multiply the change of rate of the radius by 2 and obtain that the diameter changes at a rate of $\frac{1}{100\pi}\frac{cm}{sec}$ .

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Example Question #1924 : Functions

A car 40 miles east of point A starts driving north at a rate of 30 miles per hour. What is the rate at which the distance between the car and point A is changing after one hour?

Possible Answers:

None of the above.

18pmh

19mph

17mph

20mph

Correct answer:

18pmh

Explanation:

In order to solve this question, we must first see that a triangle is being formed between the car and point A. Therefore a Pythagorean theorem derivative can be used to solve this problem. The Pythagorean therorem

x^2+y^2=z^2 can be derived with respect to time in using the power rule

$\frac{d}{dx}x^n=nx^{n-1}$ to obtain

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$ .

We know from the information given to us that

$x=40 miles,y=30miles, \frac{dy}{dt}=30 mph,\frac{dx}{dt}=0$ ,

therefore simply plugging in and solving for $\frac{dz}{dt}$ will give us the answer to this problem. However, we must still solve for . can be solved by plugging in the values of and back into the Pythagorean theorem. is equal to 30 miles because after one hour that is how far the car has traveled.

(30)^2+(40)^2=z^2

z=50 miles

Now we simply plug everything into the derivative equation and solve for $\frac{dz}{dt}$

$2(40miles)(0)+2(30miles)(30mph)=2(50miles)\frac{dz}{dt}$

$\frac{dz}{dt}=\frac{1800}{100}mph=18 mph$

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Example Question #1925 : Functions

Varsity tutors clock

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:

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

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

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

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

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

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

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

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

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

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

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:

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

Varsity tutors clock answre

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 #41 : 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:

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

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

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

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

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

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 #1925 : Functions

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:

$0.34\frac{in}{s}$

$1.7\frac{in}{s}$

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

$0.68\frac{in}{s}$

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

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:

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

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

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

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

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

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #31 : How To Find Rate Of Change

Example Question #1923 : Functions

Example Question #1924 : Functions

Example Question #1925 : Functions

Example Question #41 : How To Find Rate Of Change

Example Question #1925 : Functions

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

All Calculus 1 Resources

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