Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1081 : Rate

Find the direct constant of proportionality of f(x)=e^t-cos(2t) from $t:(0,\pi)$ .

Possible Answers:

$\frac{e^\pi-1}{\pi}$

$\frac{e^\pi+1}{\pi}$

$\frac{-e^\pi+1}{\pi}$

$\frac{e^\pi}{\pi}$

Correct answer:

$\frac{e^\pi-1}{\pi}$

Explanation:

To find direct constant of proportionality , you need to find the slope of the line between the two points at t=0 and $t=\pi$ . To do so we use the formula:

f(x)=e^t-cos(2t)

$k=\frac{f(\pi)-f(0)}{\pi -0}=\frac{(e^\pi-cos(2*\pi))-(e^0-cos(0))}{\pi}=\frac{e^\pi-1}{\pi}$

Report an Error

Example Question #2 : Asymptotic And Unbounded Behavior

The rate of growth of the population of Reindeer in Norway is proportional to the population. The population increased from 9876 to 10381 between 2013 and 2015. What is the expected population in 2030?

Possible Answers:

42581

38914

15150

49362

53880

Correct answer:

15150

Explanation:

We're told that the rate of growth of the population is proportional to the population itself, meaning that this problem deals with exponential growth/decay. The population can be modeled thusly:

$p(t)=p_0e^{kt}$

Where p_0 is an initial population value, and is the constant of proportionality.

Since the population increased from 9876 to 10381 between 2013 and 2015, we can solve for this constant of proportionality:

$10381=9876e^{k(2015-2013)}$

$\frac{10381}{9876}=e^{2k}$

$2k=ln(\frac{10381}{9876})$

$k=\frac{ln(\frac{10381}{9876})}{2}=0.0252$

Using this, we can calculate the expected value from 2015 to 2030:

$P=10381e^{(2030-2015)(0.0252)} \approx 15150$

Report an Error

Example Question #1 : Volume

Suppose I want to construct a cylindrical container. It costs 5 dollars per square foot to construct the two circular ends and 2 dollars per square foot for the rounded side. If I have a budget of 100 dollars, what's the maximum volume possible for this container?

Possible Answers:

$\sqrt{\frac{10\pi}{3}}$

$\frac{70}{6}\sqrt{33\pi}$

$\frac{70}{3}\sqrt{\frac{\pi}{3}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

$\frac{15}{2}\sqrt{\frac{10\pi}{3}}$

Correct answer:

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

Explanation:

Write down equations representing the volume(V) and cost of the cylinder.

$V=\pi r^2h$

$100=5(2\pi r^2)+2(2\pi rh)=10\pi r^2 +4\pi rh$

We want to find the values of and that will maximize the volume. Before taking the derivative of the volume equation, let's eliminate by using the cost equation.

$h=\frac{100-10\pi r^2}{4\pi r}$

Plug this into the volume equation

$V=\pi r^2\left (\frac{100-10\pi r^2}{4\pi r} \right )=r\left (\frac{100-10\pi r^2}{4} \right )$

$V=\frac{100r-10\pi r^3}{4}$

So now is eliminated in the volume equation. Take the derivative with respect to , set it equal to zero, and solve for .

$\frac{dV}{dr}=\frac{100-30\pi r^2}{4}=0$

$r=\sqrt{\frac{10}{3\pi}}$

We can use this value to find

$h=\frac{100-10\pi \left (\sqrt{\frac{10}{3\pi}} \right )^2}{4\pi \sqrt{\frac{10}{3\pi}}}=\frac{100-\frac{100}{3}}{4\sqrt{\frac{10\pi}{3}}}=\sqrt{\frac{3}{10\pi}} \frac{200}{12}$

$h=\frac{50}{3}\sqrt{\frac{3}{10\pi}}$

Now that we found AND that maximizes the volume, we can find that maximum volume

$V=\pi r^2h=\pi\left (\sqrt{\frac{10}{3\pi}} \right )^2 \frac{50}{3}\sqrt{\frac{3}{10\pi}}$

$V=\frac{50}{3}\sqrt{\frac{10}{3\pi}}$

Report an Error

Example Question #2 : Volume

We have the function $f(x)=\sqrt{x}$ and it is used to form a three dimensional figure by rotating it about the line $y=4$ . Find the volume of that figure from $x=0$ to $x=5$ .

Possible Answers:

100

Volume is infinite

432.129

103.269

43.874

Correct answer:

103.269

Explanation:

Imagine a test rectangle with a length of $4-\sqrt{x}$ and it is rotated around $y=4$ to form a cicular disk with area $\pi (4-\sqrt{x})^2$ . The disk has a thickness $\Delta x$ so that its volume is $\pi(4-\sqrt{x})^2 \Delta x$ . To find the total volume of the figure, turn this into an integral.

$Volume=\pi\int_{0}^{5}(4-\sqrt{x})^2 dx$

Perform the integration

$Volume=\pi\int_{0}^{5}(x-8\sqrt{x}+16) dx=\pi\left ( \frac{1}{2}(5)^2-\frac{16}{3}(5)^{\frac{3}{2}}+16(5) \right )$

$Volume=103.269$

Report an Error

Example Question #3 : Volume

Consider a volume V of height H along some axis, which we will simply call h, such that 0 < h < H . We can express the figure as cross-sectional areas A(h) perpendicular to this h-axis. For example for a cone whose base has radius r, we can choose the axis to go through the point of the cone at h = 0 and then the cone is a stack of little circles of radius r h/H , so $A(h) = \pi r^2 h^2/H^2$ .

What is in terms of A(h) ?

Possible Answers:

$V = H \cdot \left( \lim_{h \to H} A(h) - \lim_{h \to 0} A(h) \right )$

$V = \int_0^H A(h) ~ dh$

$V = \frac{1}{2H} \int_{-H}^H ~A(h) ~ dh$

$V = \int \frac{dh}{dx} ~ A(h) ~ dx + C$

$V = H ~ \cdot ~\frac{A(H) + A(0)}{2}$

Correct answer:

$V = \int_0^H A(h) ~ dh$

Explanation:

The volume is approximated by the Riemann sum formed when we stack n layers atop each other with $\Delta h = H/n$ , h_k = H k / n

$V \approx \sum_{k=1}^n A(h_k) ~ \Delta h$

In the limit as n gets large, this expression becomes the integral:

$V = \int_0^H A(h) ~dh$

Report an Error

Example Question #4 : Volume

What is the volume inside the bowl z(x, y) = x^2 + y^2 , 0 < z < 1 ? Hint: This is a solid of revolution about the z-axis with a radius of $\sqrt{x^2 + y^2}$ .

Possible Answers:

$\pi/2$

3/2

$\pi/3$

$\pi/4$

Correct answer:

$\pi/2$

Explanation:

Since z = r^2 we know that the volume element

$dV = A ~ dz = \pi ~ r^2 ~ dz = \pi ~z ~dz$ ,

and integrating gives:

$V = \int_0^1 \pi ~z~ dz = \pi \left[ \frac{z^2}{2} \right ]_0^1 = \pi \left( \frac{1}{2} - 0 \right ) = \frac{\pi}{2}$

Report an Error

Example Question #5 : Regions

The following function:

f(x) = x + x^2

Is rotated around the -axis to create a three-dimensional shape. What is the volume of this object within the interval x = 0 to ?

Possible Answers:

$\frac{196}{9}\pi$

$\frac{136}{15}\pi$

$\frac{256}{15}\pi$

$\frac{128}{7}\pi$

Correct answer:

$\frac{256}{15}\pi$

Explanation:

Note that for a given value of x in the function:

f(x) = x + x^2

The value of f(x) is the distance between the corresponding point on the curve and the x-axis. If the curve is rotated around the x-axis to create a three-dimensional object, f(x) can be seen as the radius of a circular cross-section of the object for any value of x.

To define the volume of this object, we can view it as a sum of infinitely thin disks stacked along the x-axis:

$V = \int \pi(f(x)^{2})dx$

Note how this follows the formula for the volume of a cylinder or disk

$( \pi \cdot height \cdot radius ^2 )$ ,

where f(x) is the radius and is the height.

Plugging in our function and range, we can rewrite this as:

$V = \int_{0}^{2} \pi(x+x^{2})^{2}dx$

$V = \int_{0}^{2}\pi(x^{4} + 2x^{3} + x^{2})$

Integrating this yields

$V = \pi \left(\frac{x^{5}}{5} + \frac{x^{4}}{2} + \frac{x^{3}}{3}\right)$

Using the upper bound of 2 and lower bound of 0 we find,

$V = \pi\left(\frac{2^{5}}{5}+\frac{2^{4}}{2} +\frac{2^{3}}{3}\right) - \pi\left(\frac{0^{5}}{5}+\frac{0^{4}}{2} +\frac{0^{3}}{3}\right) =\pi \frac{256}{15}$ .

Report an Error

Example Question #4 : Volume

Determine the volume for the region bounded by y=x and y=x^2 if the region is revolved about the -axis.

Possible Answers:

$\frac{\pi}{30}$

$\frac{\pi}{5}$

$\frac{2\pi}{15}$

$\frac{2\pi}{11}$

$\frac{4\pi}{3}$

Correct answer:

$\frac{2\pi}{15}$

Explanation:

In the region bounded by y=x and y=x^2 , write the formula to determine the volume of a solid revolved about an axis. This is the washer method.

$V=\int_{a}^{b} A(x)\:dx= \pi \int_{a}^{b} (R(x)^2-r(x)^2)\:dx$

Determine which function is the top function. This will be the big radius, and the small radius is represented by the bottom curve. The bounds of the integral are at the intersections of the two bounded functions. Evaluate the integral.

$V=\pi \int_{0}^{1} ((x)^2-(x^2)^2)\:dx$

$V=\pi \int_{0}^{1} x^2-x^4\:dx$

$V=\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_{0}^{1}=\pi\left[\frac{1}{3}-\frac{1}{5}\right] = \frac{2\pi}{15}$

Report an Error

Example Question #6 : How To Find Volume Of A Region

The unit circle is described by $y=\sqrt{1-x^2}$ and is graphed below. Find the volume of the shaded region rotated around the -axis. The shaded region starts at $x=-\sqrt{2}/2$ and ends at $x=\sqrt{2}/2$ .

Screen shot 2015 05 31 at 12.26.17 pm

Possible Answers:

$\frac{5\pi}{6}\sqrt{8}$

$\pi$

$\frac{5\pi}{6}\sqrt{2}$

$2\sqrt{2}\pi$

$\sqrt{2}\pi$

Correct answer:

$\frac{5\pi}{6}\sqrt{2}$

Explanation:

To find the volume of a region rotated around the -axis, we must find:

$V=\pi\int_a^b{f^2(x)}dx$

In our case, this is:

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{(\sqrt{1-x^2})^2}dx$

$V=\pi\int_{\sqrt{2}/2}^{\sqrt{2}/2}{1-x^2} \; dx$

$V =\pi \left[x-\frac{1}{3}x^3\right]_{-\sqrt{2}/2}^{\sqrt{2}/2}$

$V =\pi \left(\sqrt{2}/2-\frac{1}{3}(\sqrt{2}/2)^3\right) - \pi\left(-\sqrt{2}/2-\frac{1}{3}(-\sqrt{2}/2)^3\right)$

$V = \pi\left(\frac{\sqrt{2}}{2}-\frac{1}{3}\cdot \frac{2\sqrt{2}}{8}\right) - \pi\left(-\frac{\sqrt{2}}{2}-\frac{1}{3}\left(-\frac{2\sqrt{2}}{8}\right) \right )\right)$

$V = \pi \left(\frac{\sqrt{2}}{2}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{\sqrt{2}}{2} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{12\sqrt{2}}{24}-\frac{2\sqrt{2}}{24} \right )-\pi\left(-\frac{12\sqrt{2}}{24} +\frac{2\sqrt{2}}{24}\right )$

$V = \pi \left(\frac{10\sqrt{2}}{24} \right )-\pi\left(-\frac{10\sqrt{2}}{24} \right )$

$V = \frac{20\pi}{24}\sqrt{2}$

$V = \frac{5\pi}{6}\sqrt{2}$

Report an Error

Example Question #7 : Regions

Using the method of cylindrical disks, find the volume of the region of the graph of

f(x)=x^2

revolved around the -axis on the interval [0,10] .

Possible Answers:

$20000\pi$ units cubed

$\frac{1000}{3}\pi$ units cubed

$10000\pi$ units cubed

$40000\pi$ units cubed

Correct answer:

$20000\pi$ units cubed

Explanation:

The formula for the volume is given as

$V=\pi\int_{a}^{b}r^2 dx$

where r=f(x) .

As such,

$V=\pi\int_{0}^{10}(x^2)^2 dx=\pi\int_{0}^{10}x^4 dx$ .

When taking the integral, we will use the inverse power rule which states,

$\int x^n=\frac{x^{n+1}}{n+1}$ .

Applying this rule we get

$\pi\int_{0}^{10}x^4 dx=\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{10}$ .

And by the corollary of the first Fundamental Theorem of Calculus,

$\pi\left(\frac{1}{5}x^5\right)\left.\begin{matrix} \, \end{matrix}\right|_{0}^{10}=\pi \left(\left[\frac{1}{5}(10)^5\right]-\left[\frac{1}{5}(0)^5\right]\right)=20000\pi$ .

As such, the volume is

$20000\pi$ units cubed.

Report an Error

View Calculus Tutors

Pushpendra
Certified Tutor

university of allahabad, Bachelor of Science, Mathematics.

View Calculus Tutors

Jeremy
Certified Tutor

Florida Southern College, Bachelor of Science, Chemical Engineering.

View Calculus Tutors

Siavash
Certified Tutor

Ryerson University, Bachelor of Engineering, Civil Engineering. University of Toronto, Master of Engineering, Biomedical Engi...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Calculus Tutors in Miami, Reading Tutors in Atlanta, Spanish Tutors in Philadelphia, Spanish Tutors in San Francisco-Bay Area, English Tutors in Houston, Calculus Tutors in Los Angeles, Biology Tutors in Chicago, Statistics Tutors in Boston, Reading Tutors in Los Angeles, LSAT Tutors in Seattle

Popular Courses & Classes

ACT Courses & Classes in Atlanta, ACT Courses & Classes in San Diego, LSAT Courses & Classes in Seattle, GMAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Seattle, SSAT Courses & Classes in Seattle, MCAT Courses & Classes in Washington DC, MCAT Courses & Classes in New York City, SAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

SAT Test Prep in Los Angeles, SAT Test Prep in Seattle, SSAT Test Prep in Dallas Fort Worth, ACT Test Prep in Los Angeles, ACT Test Prep in Miami, GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in New York City, SSAT Test Prep in Boston, GMAT Test Prep in Los Angeles, SAT Test Prep in San Diego

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1081 : Rate

Example Question #2 : Asymptotic And Unbounded Behavior

Example Question #1 : Volume

Example Question #2 : Volume

Example Question #3 : Volume

Example Question #4 : Volume

Example Question #5 : Regions

Example Question #4 : Volume

Example Question #6 : How To Find Volume Of A Region

Example Question #7 : Regions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: