Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Parametric Form

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 3 4 5 6 7 8 9 10 Next →

Example Question #91 : Parametric Form

Find the arc length of the curve (Round to three significant digits):

$x=\frac{1}{4}t^3,\,y=\frac{1}{10}t^2,\,0\leq t\leq 1$

Possible Answers:

0.306

0.422

0.460

0.409

0.272

Correct answer:

0.272

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

a=0, b =1

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[\frac{1}{4}t^3]=\frac{3}{4}t^2$

$\frac{dy}{dt}=\frac{d}{dt}[\frac{1}{10}t^2]=\frac{1}{5}t$

Applying these into the above formula results in:

$L=\int_{0}^{1}\sqrt{(\frac{3}{4}t^2)^2+(\frac{1}{5}t)^2}\,dt=\int_{0}^{1}\sqrt{\frac{9}{16}t^4+\frac{1}{25}t^2}\,dt$

Looking at this integral, it seems pretty intimidating at first, but we can do some algebraic manipulation to make it look like a simple u-substitution problem. First we notice that there is a t^2 term that is common within the two separate functions. If we factor it out then the integral becomes:

$\int_{0}^{1}\sqrt{t^2[\frac{9}{16}t^2+\frac{1}{25}]}\,dt=\int_{0}^{1}t\sqrt{\frac{9}{16}t^2+\frac{1}{25}}\,dt$

Now we can apply the rules of u-substitution:

$u=\frac{9}{16}t^2+\frac{1}{25},\,du=\frac{18}{16}t\,dt=\frac{9}{8}t\,dt$

Therefore:

$\frac{8}{9}du=t\,dt$

Now we must deal with the new bounds of our integral:

$u(0)=\frac{9}{16}(0)^2+\frac{1}{25}=\frac{1}{25}$

$u(1)=\frac{9}{16}(1)^2+\frac{1}{25}=\frac{241}{400}$

Our new integral becomes:

$\frac{8}{9}\int_{\frac{1}{25}}^{\frac{241}{400}}\sqrt{u}\,du=\frac{8}{9}[\frac{2}{3}u^{\frac{3}{2}}]_{\frac{1}{25}}^{\frac{241}{400}}=\frac{16}{27}[(\frac{241}{400})^\frac{3}{2}-(\frac{1}{25})^\frac{3}{2}]$

Plugging this result into a calculator results in:

0.2723945261

Because the problem statement requires us to round this to three significant figures, the final result is:

0.272

This is one of the answer choices.

Report an Error

Example Question #92 : Parametric Form

Find the arc length of the curve:

$x=8\cos(3t),\,y=8\sin(3t),\,0\leq t\leq \frac{\pi}{2}$

Possible Answers:

$\frac{3\pi}{2}$

$4\pi$

$8\pi$

$24\pi$

$12\pi$

Correct answer:

$12\pi$

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

$a=0, b =\frac{\pi}{2}$

Since we are also given and , we can easily compute the derivatives of each:

$\frac{dx}{dt}=\frac{d}{dt}[8\cos(3t)]=-8\sin(3t)\cdot3=-24\sin(3t)$

$\frac{dy}{dt}=\frac{d}{dt}[8\sin(3t)]=8\cos(3t)\cdot3=24\cos(3t)$

Applying these into the above formula results in:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(-24\sin(3t))^2+(24\cos(3t))^2}\,dt=\int_{0}^{\frac{\pi}{2}}\sqrt{24^2\sin^2(3t)+24^2\cos^2(3t)}\,dt$ We can factor out the common 24^2 , and pull it outside of the square-root, and we will notice one of the most common trigonometric identities:

$\int_{0}^{\frac{\pi}{2}}\sqrt{24^2[\sin^2(3t)+\cos^2(3t)]}\,dt=\int_{0}^{\frac{\pi}{2}}24\sqrt{\sin^2(3t)+\cos^2(3t)}\,dt$

The term inside the square-root symbol can be simplified to .

$24\int_{0}^{\frac{\pi}{2}}\sqrt{1}\,dt=24\int_{0}^{\frac{\pi}{2}}dt=24[t]_{0}^{\frac{\pi}{2}}=24[\frac{\pi}{2}-0]=\frac{24\pi}{2}=12\pi$

This is one of the answer choices.

Report an Error

Example Question #93 : Parametric Form

Find the arc length of the curve:

$x=e^{3t}\cos(3t),\,y=e^{3t}\sin(3t),\,0 \leq t\leq \frac{\pi}{2}$

Possible Answers:

$\sqrt{2}e^{\frac{3\pi}{2}}$

$3\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

$\frac{1}{3}(e^{\frac{3\pi}{2}}-1)$

$3\sqrt{2}e^{\frac{3\pi}{2}}$

$\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

Correct answer:

$\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

Explanation:

Finding the length of the curve requires simply applying the formula:

$L=\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}dt$

Where:

$a=0, b =\frac{\pi}{2}$

Since we are also given and , we can easily compute the derivatives of each, using the Product Rule:

$\frac{dx}{dt}=\frac{d}{dt}[e^{3t}\cos(3t)]=-3e^{3t}\sin(3t)+3e^{3t}\cos(3t)=3e^{3t}[\cos(3t)-\sin(3t)]$

$\frac{dy}{dt}=\frac{d}{dt}[e^{3t}\sin(3t)]=3e^{3t}\cos(3t)+3e^{3t}\sin(3t)=3e^{3t}[\cos(3t)+\sin(3t)]$

Applying these into the above formula results in:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t}[\cos(3t)-\sin(3t)])^2+(3e^{3t}[\cos(3t)+\sin(3t)])^2}\,dt$

Simplifying the above will require these two formulas:

$(a+b)^2=a^2+2ab+b^2 \rightarrow \,(1)$

$(a-b)^2=a^2-2ab+b^2\rightarrow\,(2)$

It may also be useful to know this formula:

$(x^a)^b=x^{ab}\rightarrow\,(3)$

We can factor out the common $(3e^{3t})^2$ to make the above expression easier to look at:

$L=\int_{0}^{\frac{\pi}{2}}\sqrt{(3e^{3t})^2([\cos(3t)-\sin(3t)]^2+[\cos(3t)+\sin(3t)])^2}\,dt$

We can take the $(3e^{3t})^2$ outside of the square-root by cancelling out the representing the "square". Then we can apply formulas (1) & (2) to the trigonometric expressions:

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{[\cos^2(3t)-2\cos(3t)\sin(3t)+\sin^2(3t)]+[\cos^2(3t)+2\cos(3t)\sin(3t)+\sin^2(t)]}\,dt$

We can now simplify the terms inside the square-root to get:

$L=3e^{3t}\int_{0}^{\frac{\pi}{2}}\sqrt{[2\cos^2(3t)+2\sin^2(3t)]}\,dt$

If we factor out the common "2" above, we are left with the trigonometric identity, which simplifies to , since:

$\sin^2(t)+\cos^2(t)=1$

Therefore the integral now becomes:

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2[\cos^2(3t)+\sin^2(3t)]}\,dt$

$L=\int_{0}^{\frac{\pi}{2}}3e^{3t}\sqrt{2(1)}\,dt=\int_{0}^{\frac{\pi}{2}}3\sqrt{2}e^{3t}\,dt$

This is a simple integral which can be solved using u-substitution. But first, we can factor out the constant term $3\sqrt{2}$ , outside of the integral:

$L=3\sqrt{2}\int_{0}^{\frac{\pi}{2}}e^{3t}\,dt$

We will make our substitutions:

$u=3t,\,du=3\,dt\rightarrow\frac{1}{3}du=dt$

We also need to change the bounds of the new integral:

u(0)=3(0)=0

$u(\frac{\pi}{2})=3(\frac{\pi}{2})=\frac{3\pi}{2}$

Our new integral becomes:

$L=3\sqrt{2}\int_{0}^{\frac{3\pi}{2}}\frac{1}{3}e^u\,du=\sqrt{2}[e^u]_{0}^{\frac{3\pi}{2}}=\sqrt{2}(e^{\frac{3\pi}{2}}-e^{0})=\sqrt{2}(e^{\frac{3\pi}{2}}-1)$

This is one of the answer choices.

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 10 Next →

View Calculus 2 Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View Calculus 2 Tutors

Kurosh
Certified Tutor

University of tehran, Bachelor of Engineering, Electrical Engineering.

View Calculus 2 Tutors

Filiberto
Certified Tutor

University of Arizona, Bachelor of Science, Electrical Engineering.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in Los Angeles, Computer Science Tutors in Atlanta, Math Tutors in Phoenix, Computer Science Tutors in San Diego, Math Tutors in Houston, SSAT Tutors in Los Angeles, SSAT Tutors in Miami, French Tutors in Miami, Chemistry Tutors in Seattle, English Tutors in Chicago

Popular Courses & Classes

GRE Courses & Classes in Seattle, ACT Courses & Classes in Miami, LSAT Courses & Classes in Miami, ACT Courses & Classes in Washington DC, SAT Courses & Classes in Chicago, ACT Courses & Classes in Atlanta, SSAT Courses & Classes in Seattle, LSAT Courses & Classes in Denver, LSAT Courses & Classes in Chicago, Spanish Courses & Classes in Dallas Fort Worth

Popular Test Prep

MCAT Test Prep in Philadelphia, LSAT Test Prep in Phoenix, GRE Test Prep in Miami, MCAT Test Prep in Washington DC, MCAT Test Prep in Denver, LSAT Test Prep in Miami, SAT Test Prep in Los Angeles, ACT Test Prep in Washington DC, MCAT Test Prep in Atlanta, ACT Test Prep in New York City

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 2 : Parametric Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : Parametric Form

Find the arc length of the curve (Round to three significant digits):

Example Question #92 : Parametric Form

Find the arc length of the curve:

Example Question #93 : Parametric Form

Find the arc length of the curve:

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: