Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Parametric Calculations

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

Example Question #3 : Parametric, Polar, And Vector Functions

Find the length of the following parametric curve

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

Possible Answers:

$e^{2 \pi}$

sin(t)+cos(t)

$e^{t}cos(t)$

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

Correct answer:

$\sqrt{2}\left ( e^{2 \pi}-1}\right )$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the product rule,

$\frac{d}{dt}(a*b)=a\frac{db}{dt}+b\frac{da}{dt}$ , when and are functions of ,

the trigonometric rule,

$\frac{d}{dt}[cos(t)]=-sin(t)$ and $\frac{d}{dt}[sin(t)]=cos(t)$

and exponential rule,

$\frac{d}{dt}[e^{t}]=e^{t}$ to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

In this case

$x=e^{t}cos(t)$ , $y=e^{t}sin(t)$

$\frac{dx}{dt}=\frac{d}{dt}e^{t}cos(t)=e^{t}\frac{d}{dt}cos(t)+cos(t)\frac{d}{dt}e^{t}$

$=e^{t}*(-sin(t))+cos(t)*e^{t}=e^{t}cos(t)-e^{t}sin(t)$

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

$=e^{t}cos(t)+sin(t)*e^{t}=e^{t}cos(t)+e^{t}sin(t)$

$\alpha=0, \beta=2\pi$

The length of this curve is

$L=\int_{0}^{2\pi }\sqrt{\left (e^{t}cos(t)-e^{t}sin(t)\right )^{2}+\left (e^{t}cos(t)+e^{t}sin(t)\right )^{2}}dt$

Using the identity $(a\pm b)^{2}=a^{2}\pm2ab+b^{2}$

$L=\int_{0}^{2\pi }\sqrt{\binom{\left (e^{2t}cos^{2}(t)-2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right )}{\left +(e^{2t}cos^{2}(t)+2e^{t}cos(t)e^{t}sin(t)+e^{2t}sin^{2}(t)\right)}dt}$

$=\int_{0}^{2\pi }\sqrt{(2e^{2t}cos^{2}(t)+2e^{2t}sin^{2}(t))dt}$

$=\int_{0}^{2\pi }\sqrt{2e^{2t}(cos^{2}(t)+sin^{2}(t))dt}$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

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

Using the trigonometric identity $sin^{2}(at)+cos^{2}(at)=1$ where is a constant and $e^{2t}=(e^{t})^2$

$=\int_{0}^{2\pi }\sqrt{2}e^{t}dt}=\sqrt{2}\int_{0}^{2\pi }e^{t}dt}$

Using the exponential rule, $\int e^{t}dt}=e^{t}$

$\sqrt{2}\int_{0}^{2\pi }e^{t}dt}=\sqrt{2}e^{t}|_{0}^{2\pi}=\sqrt{2}\left ( e^{2 \pi}-e^{0}}\right )$

Using the exponential rule, $e^{0}=1$ , gives us the final solution

$L=\sqrt{2}\left ( e^{2 \pi}-1}\right )$

Report an Error

Example Question #21 : Parametric Calculations

Find the length of the following parametric curve

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

Possible Answers:

$2\sqrt{3}$

$\frac{1}{5}\left ( 6\sqrt{6}-1\right )$

$\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

$\frac{1}{6}$

Correct answer:

$\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We use the power rule $\frac{d}{dt}t^{c}=ct^{c-1}$ , where is a constant, to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$ .

$x=\frac{1}{3}t^{3}$ , $y=\frac{1}{2}t^{2}$

In this case

$\frac{dx}{dt}=\frac{d}{dt}(\frac{1}{3}t^{3})=t^{2}$

$\frac{dy}{dt}=\frac{d}{dt}(\frac{1}{2}t^{2})=t$

$\alpha=0, \beta=1$

The length of this curve is

$L=\int_{0}^{1 }\sqrt{(t^{2})^{2}+(t)^{2}}dt$

$=\int_{0}^{1 }\sqrt{t^{4}+t^{2}}dt$

$=\int_{0}^{1 }\sqrt{t^{2}(t^{2}+1)}dt$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$=\int_{0}^{1 }\sqrt{t^{2}}\sqrt{(t^{2}+1)}dt$

$=\int_{0}^{1 }t\sqrt{(t^{2}+1)}dt$

Using a u-substitution

Let $u=t^{2}+1$

$du=2tdt\Rightarrow dt=2t/du$

and changing the bounds

$u=t^{2}+1=(1)^{2}+1=2$

$u=t^{2}+1=(0)^{2}+1=1$

$L=\frac{1}{2}\int_{1}^{2}\sqrt{u}du$

$L=\frac{1}{2}*\frac{2}{3}\left ( 2^{\frac{3}{2}}-1^{\frac{3}{2}}\right )$

$L=\frac{1}{3}\left ( 2\sqrt{2}-1\right )$

Report an Error

Example Question #22 : Parametric Calculations

Find the length of the following parametric curve

$x=6t^{2}$ , $y=4t^{^{3}}$ , $1\leq t \leq 8$ .

Possible Answers:

$4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

$\sqrt{122}-8$

$\sqrt{130}-5\sqrt{7}$

Correct answer:

$4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

$x=6t^{2}$ , $y=4t^{^{3}}$ , $1\leq t \leq 8$ .

We use the power rule $\frac{d}{dt}t^{c}=ct^{c-1}$ , where is a constant, to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

$\frac{dx}{dt}=\frac{d}{dt}(6t^{2})=12t$

$\frac{dx}{dt}=\frac{d}{dt}(4t^{3})=4t^{2}$

In this case, the length of this curve is

$L=\int_{1}^{8 }\sqrt{\left (12t \right )^{2}+\left (12t^{2} \right )^{2}}dt$

$=\int_{1}^{8 }\sqrt{144t^{2}+144t^{4}}dt$

$=\int_{1}^{8 }\sqrt{144t^{2}(1+t^{2})}dt$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$\int_{1}^{8 }\sqrt{144t^{2}(1+t^{2})}dt=\int_{1}^{8 }\sqrt{144t^{2}}\sqrt{1+t^{2}}dt=\int_{1}^{8 }\12t\sqrt{1+t^{2}}dt$

using a u-substitution

$u=1+t^{2}$

$du=2tdt\Rightarrow dt=2t/du$

and changing the bounds

$u=1+t^{2}=1+8^{2}=65$

$u=1+t^{2}=1+1^{2}=2$

$L=\frac{12}{2}\int_{2}^{65 }\sqrt{u}du$

$=\frac{12}{2}\left [\frac{2}{3} \right ]\left [ 65^{\frac{3}{2}}-2^{\frac{3}{2}}\right ]$

$=4\left [ \sqrt{65^{3}}-\sqrt{2^{3}} \right ]$

$=4\left [ \sqrt{65^{2}*65}-\sqrt{2^{2}*2} \right ]$

$=4\left [ \sqrt{65^{2}}\sqrt{65}-\sqrt{2^{2}}\sqrt{2} \right ]$

$=4\left [ 65\sqrt{65}-2\sqrt{2} \right ]$

Report an Error

Example Question #23 : Parametric Calculations

Find the length of the following parametric curve

x=cos(5t) , y=sin(5t) , $0\leq t \leq 2\pi$ .

Possible Answers:

$5\pi$

$10\pi$

$5\pi-5$

Correct answer:

$10\pi$

Explanation:

The length of a curve is found using the equation $L=\int_{\alpha }^{\beta }\sqrt{\left ( \frac{dx}{dt} \right )^{2}+\left ( \frac{dy}{dt} \right )^{2}}dt$

We then use the following trigonometric rules,

$\frac{d}{dt}[cos(at)]=-asin(at)$ and $\frac{d}{dt}[sin(bt)]=bcos(bt)$ ,

where and are constants.

In this case

$\frac{dx}{dt}=\frac{d}{dt}(cos(5t))=-5sin(5t)$ ,

$\frac{dy}{dt}=\frac{d}{dt}(sin(5t))=5cos(5t)$ , $\alpha=0, \beta=2\pi$

The length of this curve is

$L=\int_{0}^{2\pi }\sqrt{\left (-5sin(5t) \right )^{2}+\left (5cos(5t) \right )^{2}}dt$

$=\int_{0}^{2\pi }\sqrt{25sin^{2}(5t)+25cos^{2}(5t)}dt$

$=\int_{0}^{2\pi }\sqrt{25(sin^{2}(5t)+cos^{2}(5t))}dt$

Using the identity $\sqrt{a*b}=\sqrt{a}\sqrt{b}$

$\int_{0}^{2\pi }\sqrt{25(sin^{2}(5t)+cos^{2}(5t))}dt=\int_{0}^{2\pi }\sqrt{25}\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt$

$=\int_{0}^{2\pi }5\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt$

Using the trigonometric identity $sin^{2}(at)+cos^{2}(at)=1$ where is a constant

$\int_{0}^{2\pi }5\sqrt{sin^{2}(5t)+cos^{2}(5t)}dt=5\int_{0}^{2\pi }dt$

Using the rule of integration for constants

$\int_{a}^{b}dt=t|_{a}^{b}=(b-a)$

$5\int_{0}^{2\pi }dt=5(2\pi -0)=10\pi$

Report an Error

Example Question #22 : Parametric Calculations

Given x=-t+3 and y=3-2y , what is the arc length between $0\leq t\leq 2$

Possible Answers:

$L=2\sqrt{5}$

$L=\sqrt{5}$

$L=4\sqrt{5}$

$L=3\sqrt{5}$

$L=5\sqrt{5}$

Correct answer:

$L=2\sqrt{5}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given x=-t+3 and y=3-2y , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(1)(-1)t^{1-1}+(0)3=-1$ and

$\frac{dy}{dt}=(0)3-(1)2y^{1-1}=-2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{2}(\sqrt{(-1)^{2}+(-2)^{2}})dt$

$L=\int_{0}^{2}(\sqrt{1+4})dt$

$L=\int_{0}^{2}(\sqrt{5})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{5}]_{0}^{2}\textrm{}$

$L=(2)\sqrt{5}-(0)\sqrt{5}$

$L=2\sqrt{5}$

Report an Error

Example Question #23 : Parametric Calculations

Given x=2t+15 and y=4+9y , what is the arc length between $0\leq t\leq 3$ ?

Possible Answers:

$\sqrt{85}$

$5\sqrt{85}$

$2\sqrt{85}$

$4\sqrt{85}$

$3\sqrt{85}$

Correct answer:

$3\sqrt{85}$

Explanation:

In order to find the arc length, we must use the arc length formula for parametric curves:

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

Given x=2t+15 and y=4+9y , wwe can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(1)2t^{1-1}+(0)15=2$ and

$\frac{dy}{dt}=(0)4+(1)9y^{1-1}=9$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{0}^{3}(\sqrt{(2)^{2}+(9)^{2}})dt$

$L=\int_{0}^{3}(\sqrt{4+81})dt$

$L=\int_{0}^{3}(\sqrt{85})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{85}]_{0}^{3}\textrm{}$

$L=(3)\sqrt{85}-(0)\sqrt{85}$

$L=3\sqrt{85}$

Report an Error

Example Question #24 : Parametric Calculations

Given x=2-3t and y=9+2t , what is the arc length between $1\leq t\leq 3$ ?

Possible Answers:

$5\sqrt{13}$

$\sqrt{13}$

$2\sqrt{13}$

$3\sqrt{13}$

$4\sqrt{13}$

Correct answer:

$2\sqrt{13}$

Explanation:

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

Given x=2-3t and y=9+2t , we can use using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ ,

to derive

$\frac{dx}{dt}=(0)2-(1)3t^{1-1}=-3$ and

$\frac{dy}{dt}=(0)9+(1)2t^{1-1}=2$ .

Plugging these values and our boundary values for into the arc length equation, we get:

$L=\int_{1}^{3}(\sqrt{(-3)^{2}+(2)^{2}})dt$

$L=\int_{1}^{3}(\sqrt{9+4})dt$

$L=\int_{1}^{3}(\sqrt{13})dt$

Now, using the Power Rule for Integrals

$\int x^{n}dx=\frac{x^{n+1}}{n+1}$ for all $n\neq0$ ,

we can determine that:

$L=[t\sqrt{13}]_{1}^{3}\textrm{}$

$L=(3)\sqrt{13}-(1)\sqrt{13}$

$L=2\sqrt{13}$

Report an Error

View Calculus 2 Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Calculus 2 Tutors

Dawn
Certified Tutor

University of Oregon, Bachelor in Arts, Biochemistry and Molecular Biology. Johns Hopkins University, Doctor of Philosophy, E...

View Calculus 2 Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

All Calculus 2 Resources

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

Popular Subjects

GMAT Tutors in New York City, Statistics Tutors in Washington DC, MCAT Tutors in Chicago, Reading Tutors in Seattle, Biology Tutors in Phoenix, English Tutors in San Diego, ISEE Tutors in Denver, GMAT Tutors in Atlanta, Chemistry Tutors in Atlanta, French Tutors in San Diego

Popular Courses & Classes

ISEE Courses & Classes in Houston, LSAT Courses & Classes in San Diego, GMAT Courses & Classes in Washington DC, GMAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in Atlanta, Spanish Courses & Classes in Atlanta, SAT Courses & Classes in Miami, Spanish Courses & Classes in Phoenix, MCAT Courses & Classes in New York City, SSAT Courses & Classes in Phoenix

Popular Test Prep

ACT Test Prep in Boston, GRE Test Prep in Dallas Fort Worth, ISEE Test Prep in Houston, ACT Test Prep in Washington DC, MCAT Test Prep in Washington DC, GRE Test Prep in Chicago, MCAT Test Prep in Miami, MCAT Test Prep in Philadelphia, SSAT Test Prep in Boston, SAT Test Prep in Chicago

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 Calculations

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Parametric, Polar, And Vector Functions

Example Question #21 : Parametric Calculations

Example Question #22 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #22 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #24 : Parametric Calculations

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: