We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Length of Curve, Distance traveled, Accumulated Change, Motion of Curve

Study concepts, example questions & explanations for AP Calculus BC

Create An Account

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Determine the length of the following function between $1\leq t\leq 4$

$v(t)=\frac{2}{3}(t-1)^{\frac{3}{2}}$

Possible Answers:

$-\frac{7}{3}$

$\frac{16}{9}$

$\frac{14}{3}$

$-\frac{14}{3}$

$\frac{7}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

In order to begin the problem, we must first remember the formula for finding the arc length of a function along any given interval:

$L=\int ds$

where ds is given by the equation below:

$ds=\sqrt{1+(\frac{dy}{dx})^{2}}dx$

We can see from our equation for ds that we must find the derivative of our function, which in our case is dv/dt instead of dy/dx, so we begin by differentiating our function v(t) with respect to t:

$\frac{dv}{dt}=(t-1)^{\frac{1}{2}}=\sqrt{t-1}$

Now we can plug this into the given equation to find ds:

$ds=\sqrt{1+(\frac{dv}{dt})^{2}}dt=\sqrt{1+(\sqrt{t-1})^{2}}=\sqrt{1+t-1}=\sqrt{t}$

Our last step is to plug our value for ds into the equation for arc length, which we can see only involves integrating ds. The interval along which the problem asks for the length of the function gives us our limits of integration, so we simply integrate ds from t=1 to t=4:

$L=\int ds=\int_{1}^{4}\sqrt{t}dt=[\frac{2}{3}t^{\frac{3}{2}}]_{1}^{4}$

$=\frac{2}{3}(4)^{\frac{3}{2}}-\frac{2}{3}(1)^{\frac{3}{2}}=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

Report an Error

Example Question #1 : Applications In Physics

In physics, the work done on an object is equal to the integral of the force on that object dotted with its displacent.

This looks like $\small W=\int(F\cdot ds)$ ( $\small W$ is work, $\small F$ is force, and $\small ds$ is the infinitesimally small displacement vector). For a force whose direction is the line of motion, the equation becomes $\small \small W=\int (Fdx)$ .

If the force on an object as a function of displacement is $\small F(x)=3x^{2}+x$ , what is the work as a function of displacement $\small W(x)$ ? Assume $\small W(0)=0$ and the force is in the direction of the object's motion.

Possible Answers:

$\small W(x)=x^{3}+x^{2}/2$

$\small \small W(x)=6x+1$

Not enough information

$\small \small W(x)=3x^{2}+x$

$\small \small W(x)=3x^{3}/2+x^{2}$

Correct answer:

$\small W(x)=x^{3}+x^{2}/2$

Explanation:

$\small \small W=\int (Fdx)$ , so $\small W=\int(3x^{2}+x)dx$ .

Both the terms of the force are power terms in the form $\small ax^{n}$ , which have the integral $\small \frac{a}{n+1}x^{n+1}$ , so the integral of the force is $\small x^{3}+x^{2}/2+c$ .

We know

$\small W(0)=0\rightarrow 0^{3}+0^{2}/2+c=0 \therefore c=0$ .

This means

$\small W(x)=x^{3}+x^{2}/2$ .

Report an Error

Example Question #11 : Application Of Integrals

Give the arclength of the graph of the function $f(x) = \frac{2}{3} x \sqrt{x}$ on the interval $\left [ 0,3 \right ]$ .

Possible Answers:

$\frac{14}{3}$

$\frac{16}{3}$

$\frac{11}{3}$

Correct answer:

$\frac{14}{3}$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{2}{3} x \sqrt{x} = \frac{2}{3} x^{\frac{3}{2}}$

$f(x) = \frac{2}{3} \cdot \frac{3}{2} x^{\frac{1}{2}} = \sqrt{x}$ .

The above integral becomes

$\int_{0}^{3} \sqrt{1 + (\sqrt{x}) ^{2}} \; \mathrm{d}x$

$\int_{0}^{3} \sqrt{x+1} \; \mathrm{d}x$

Substitute u = x + 1 . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 1$ , $\mathrm{d} u = \mathrm{d} x$ , and the integral becomes

$\int_{1}^{4} \sqrt{u} \; \mathrm{d}u = \int_{1}^{4} u^{\frac{1}{2}} \; \mathrm{d}u$

$= \left.\begin{matrix} \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \left.\begin{matrix} \frac{2u \sqrt{u}}{3} \\ \\ \end{matrix}\right| \begin{matrix} 4\\ \\ 1 \end{matrix}$

$= \frac{2 \cdot 4 \sqrt{4}}{3} - \frac{2 \cdot 1 \sqrt{1}}{3}$

$= \frac{2 \cdot 4 \cdot 2 }{3} - \frac{2 \cdot 1 \cdot 1}{3} = \frac{16}{3} - \frac{2}{3} =\frac{14}{3}$

Report an Error

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Give the arclength of the graph of the function $f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ on the interval $\left [ 0, \frac{\pi}{8} \right ]$ .

Possible Answers:

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

$\ln \left ( 1 + \sqrt{2} \right )$

$1+\frac{1 }{2} \ln 2$

$\frac{1 }{2}+\frac{1 }{2} \sqrt{2}$

$1 + \sqrt{2}$

Correct answer:

$\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right )$

Explanation:

The length of the curve of f(x) on the interval [a,b] can be determined by evaluating the integral

$\int_{a}^{b} \sqrt{1 + f'(x) ^{2}} \; \mathrm{d}x$ .

$f(x) = \frac{1}{2} \ln \left ( \cos 2 x \right )$ , so

$f'(x) = \frac{1}{2} \cdot \frac{\frac{\mathrm{d} }{\mathrm{d} x}\cos 2 x}{\cos 2 x} = \frac{1}{2} \cdot \frac{-2 \sin 2x}{\cos 2 x} = - \tan 2x$

The integral becomes

$\int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \left ( - \tan 2x \right )^{2}} \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{1 + \tan ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sqrt{\sec ^{2} 2x } \; \mathrm{d}x$

$= \int_{0}^{ \frac{\pi}{8}} \sec 2x \; \mathrm{d}x$

Use substitution - set u = 2x . Then $\frac{\mathrm{d} u}{\mathrm{d} x} = 2$ , and $\mathrm{d} x = \frac{1}{2}\mathrm{d} u$ . The bounds of integration become $2 \cdot 0 = 0$ and $2 \cdot \frac{\pi}{8} = \frac{\pi}{4}$ , and the integral becomes

$\frac{1}{2} \int_{0}^{ \frac{\pi}{4}} \sec u \; \mathrm{d}u$

$=\left.\begin{matrix}\frac{1}{2} \ln | \sec u + \tan u | \\ \\ \end{matrix}\right|\begin{matrix} \frac{\pi}{4}\\ \\ 0 \end{matrix}$

$=\frac{1}{2} \ln \left | \sec \frac{\pi}{4} + \tan \frac{\pi}{4} \right | - \frac{1}{2} \ln | \sec 0 + \tan 0 |$

$=\frac{1}{2} \ln \left | \sqrt{2} + 1 \right | -\frac{1}{2} \ln | 1 + 0 |$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) -\frac{1}{2} \ln 1$

$=\frac{1}{2} \ln \left ( 1 + \sqrt{2} \right ) - 0$

$=\frac{1}{2} \ln \left (1 + \sqrt{2} \right )$

Report an Error

Example Question #1 : Average Values And Lengths Of Functions

What is the length of the curve $g(x)=x^{\frac{3}{2}}$ over the interval $0\le x\le4$ ?

Possible Answers:

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

$\frac{8}{27}(10\sqrt{10}-1)$

$4\sqrt2-1$

$\frac{56}{27}$

Correct answer:

$\frac{8}{27}(10\sqrt{10}-1)$

Explanation:

The general formula for finding the length of a curve f(x) over an interval [a,b] is $\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx$

In this example, the arc length can be found by computing the integral

$\int_{0}^{4}\sqrt{1+(\frac{d}{dx}x^{\frac{3}{2}})^2}\ dx$ .

The derivative of $x^{\frac{3}{2}}$ can be found using the power rule, $\frac{d}{dx} x^n=nx^{n-1}$ , which leads to

$\int_{0}^{4}\sqrt{1+\left(\frac{3}{2}x^{\frac{1}{2}}\right)^2}\ dx=\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx$ .

At this point, a substitution is useful.

Let

$u=1+\frac{9}{4}x,\ du=\frac{9}{4}dx,\frac{4}{9}du=dx$ .

We can also express the limits of integration in terms of to simplify computation. When x=0, u=1 , and when x = 4, u=10 .

Making these substitutions leads to

$\int_{0}^{4}\sqrt{1+\frac{9}{4}x}\ dx=\int_{1}^{10}\frac{4}{9}\sqrt{u}du=\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du$ .

Now use the power rule, which in general is $\int x^n=\frac{x^{n+1}}{n+1}+c$ , to evaluate the integral.

$\frac{4}{9}\int_{1}^{10}u^{\frac{1}{2}}du=\frac{4}{9}\frac{u^{\frac{3}{2}}}{\frac{3}{2}}|_{1}^{10}$

$=\frac{8}{27}(10^{\frac{3}{2}}-1^\frac{3}{2})=\frac{8}{27}(10\sqrt{10}-1)$

Report an Error

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Find the total distance traveled by a particle along the curve y=x^2 from x=0 to x=2 .

Possible Answers:

$\int_0^2\sqrt{1+2x^2}dx$

$\int_0^2\sqrt{1+2x}dx$

$\int_0^2\sqrt{1+4x^2}dx$

$\int_0^2\sqrt{1+x^2}dx$

Correct answer:

$\int_0^2\sqrt{1+4x^2}dx$

Explanation:

To find the required distance, we can use the arc length expression given by $\int_{x_0}^{x_1}\sqrt{1+f'(x)^2}dx$ .

Taking the derivative of our function, we have y'=f'(x)=2x . Plugging in our values for our integral bounds, we have

$\int_0^2\sqrt{1+(2x)^2}dx = \int_0^2 \sqrt{1+4x^2}dx$ .

As with most arc length integrals, this integral is too difficult (if not, outright impossible) to evaluate explicitly by hand. So we will just leave it this form, or evaluate it with some computer software.

Report an Error

View AP Calculus BC Tutors

Kayley
Certified Tutor

New Mexico State University-Main Campus, Bachelor of Science, Biology, General.

View AP Calculus BC Tutors

Sree
Certified Tutor

Saint Louis University-Main Campus, Bachelor in Arts, Chemistry.

View AP Calculus BC Tutors

Ian
Certified Tutor

Williams College, Bachelor in Arts, Physics.

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in San Francisco-Bay Area, SAT Tutors in Phoenix, GRE Tutors in New York City, English Tutors in Dallas Fort Worth, ISEE Tutors in Denver, Spanish Tutors in Boston, GRE Tutors in San Francisco-Bay Area, MCAT Tutors in Phoenix, English Tutors in Chicago, ACT Tutors in San Diego

Popular Courses & Classes

GRE Courses & Classes in Miami, GRE Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in New York City, SAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Seattle, LSAT Courses & Classes in San Diego, ACT Courses & Classes in Los Angeles, Spanish Courses & Classes in Philadelphia, ACT Courses & Classes in San Diego, Spanish Courses & Classes in Boston

Popular Test Prep

SAT Test Prep in Seattle, ACT Test Prep in Philadelphia, SAT Test Prep in Dallas Fort Worth, SSAT Test Prep in Los Angeles, GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in San Diego, LSAT Test Prep in San Diego, SAT Test Prep in Houston, ISEE Test Prep in Chicago, ACT 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

AP Calculus BC : Length of Curve, Distance traveled, Accumulated Change, Motion of Curve

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #1 : Applications In Physics

Example Question #11 : Application Of Integrals

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

Example Question #1 : Average Values And Lengths Of Functions

Example Question #1 : Length Of Curve, Distance Traveled, Accumulated Change, Motion Of Curve

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: