Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Parametric, Polar, and Vector

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 #21 : Parametric Calculations

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

Possible Answers:

$L=3\sqrt{5}$

$L=4\sqrt{5}$

$L=5\sqrt{5}$

$L=2\sqrt{5}$

$L=\sqrt{5}$

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 #21 : Parametric Calculations

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

Possible Answers:

$5\sqrt{85}$

$4\sqrt{85}$

$2\sqrt{85}$

$\sqrt{85}$

$3\sqrt{85}$

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 #23 : Parametric Calculations

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

Possible Answers:

$3\sqrt{13}$

$4\sqrt{13}$

$5\sqrt{13}$

$\sqrt{13}$

$2\sqrt{13}$

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

Example Question #1 : Polar

Rewrite the polar equation

$r = \sin ^{2} \theta$

in rectangular form.

Possible Answers:

$y ^{3} = \left (x^{2} +y^{2} \right )^{4}$

$y ^{3} = \left (x^{2} +y^{2} \right )^{2}$

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

$y ^{4} = \left (x^{3} +y^{3} \right )^{2}$

$y ^{4} = \left (x^{2} +y^{2} \right )^{2}$

Correct answer:

$y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

Explanation:

$r = \sin ^{2} \theta$

$r^{2} \cdot r = r^{2} \sin ^{2} \theta$

$r^{3} =\left ( r \sin \theta \right ) ^{2}$

$\left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}$

$\left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{3} = y ^{4}$

or $y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

Report an Error

Example Question #1 : Polar Form

Rewrite in polar form:

$x^{2} = 3xy + 1$

Possible Answers:

$r =\sqrt{ \frac{1}{\cos \theta + 3 \sin \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta + 3 \cos \theta\; \sin \theta }}$

$r =\sqrt{ \frac{1}{\sin \theta + 3 \cos \theta\; } }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

$r = \sqrt{\frac{1}{\sin ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Correct answer:

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Explanation:

$x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta\; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta\; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta\; \sin \theta }}$

Report an Error

Example Question #1 : Polar

Rewrite the polar equation

$r ^{2} + 1 = \cos \theta$

in rectangular form.

Possible Answers:

$\sqrt{x^{2} + y ^{2}} =\frac{ y}{x^{2} + y ^{2}}$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} }$

$\sqrt{x^{2} + y ^{2}} =\frac{ y}{{x^{2} + y ^{2}} +1}$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

$\sqrt{x^{2} + y ^{2}} =\frac{x+ y}{{x^{2} + y ^{2}} +1}$

Correct answer:

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

Explanation:

$r ^{2} + 1 = \cos \theta$

$r\left ( r ^{2} + 1 \right )= r \cos \theta$

$\sqrt{x^{2} + y ^{2}} \left ( {x^{2} + y ^{2}} +1 \right ) = x$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

Report an Error

Example Question #161 : Parametric, Polar, And Vector

Give the polar form of the equation of the line with intercepts (0,4), (-6,0) .

Possible Answers:

$r = \frac{12}{ \tan \theta }$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

$r = \frac{12}{2 \sin \theta - 3 \cos \theta }$

$r = \frac{12}{3 \sin \theta + 2 \cos \theta }$

$r = 12\tan \theta$

Correct answer:

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

Explanation:

This line has slope $\frac{4-0}{0- (-6)} = \frac{2}{3}$ and -intercept (0.4) , so its Cartesian equation is $y = \frac{2}{3}x + 4$ .

By substituting, we can rewrite this:

$r \sin \theta = \frac{2}{3}r \cos \theta + 4$

$r \sin \theta - \frac{2}{3}r \cos \theta = 4$

$3 r \sin \theta - 2 r \cos \theta = 12$

$r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

Report an Error

Example Question #3 : Polar

Give the rectangular coordinates of the point with polar coordinates

$\left ( -4 , \frac{\pi }{3}\right )$ .

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

$x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$

$y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}$

The point will have rectangular coordinates $\left (- 2, -2 \sqrt{3} \right )$ .

Report an Error

Example Question #1 : Polar Form

What would be the equation of the parabola $\small y=x^{2}$ in polar form?

Possible Answers:

$\small r=cot\theta cos\theta$

$\small \small \small r=tan\theta sin\theta$

$\small r=tan\theta sec\theta$

$\small \small r=cot\theta csc\theta$

$\small r=\theta ^{2}$

Correct answer:

$\small r=tan\theta sec\theta$

Explanation:

We know $\small y=rsin\theta$ and $\small x=rcos\theta$ .

Subbing that in to the equation $\small y=x^{2}$ will give us $\small rsin\theta=r^{2}cos^{2}\theta$ .

Multiplying both sides by $\small 1/(rcos^{2}\theta)$ gives us

$\small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta$ .

Report an Error

Example Question #1 : Polar Form

A point in polar form is given as $\left(3,\frac{\pi}{3}\right)$ .

Find its corresponding (x,y) coordinate.

Possible Answers:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

(0,3)

$\left(\frac{3\sqrt3}{2},\frac{3}{2}\right)$

$\left(\frac{3}{2},\frac{3}{2}\right)$

$\left(3,\frac{3\sqrt3}{2}\right)$

Correct answer:

$\left(\frac{3}{2},\frac{3\sqrt3}{2}\right)$

Explanation:

To go from polar form to cartesion coordinates, use the following two relations.

$x=rcos(\theta)$

$y=rsin(\theta)$

In this case, our is and our $\theta$ is $\frac{\pi}{3}$ .

Plugging those into our relations we get

$x=\frac{3}{2}$

$y=\frac{3\sqrt3}{2}$ ,

which gives us our (x,y) coordinate.

Report an Error

View Calculus 2 Tutors

Justin
Certified Tutor

McMaster University, Bachelor of Engineering, Mechanical Engineering.

View Calculus 2 Tutors

Dylan
Certified Tutor

University of California-Berkeley, Bachelor of Science, Cognitive Science.

View Calculus 2 Tutors

Tyrus Julian
Certified Tutor

Purdue University-Main Campus, Bachelor of Science, Aerospace Engineering.

All Calculus 2 Resources

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

Popular Subjects

Algebra Tutors in Boston, Biology Tutors in San Diego, Reading Tutors in Los Angeles, LSAT Tutors in San Diego, Physics Tutors in Phoenix, Spanish Tutors in Los Angeles, Calculus Tutors in San Diego, French Tutors in Atlanta, Computer Science Tutors in San Diego, LSAT Tutors in Chicago

Popular Courses & Classes

GMAT Courses & Classes in Denver, MCAT Courses & Classes in Atlanta, GMAT Courses & Classes in San Diego, Spanish Courses & Classes in Seattle, Spanish Courses & Classes in San Francisco-Bay Area, LSAT Courses & Classes in Atlanta, Spanish Courses & Classes in Houston, GRE Courses & Classes in Washington DC, ISEE Courses & Classes in Philadelphia, SAT Courses & Classes in Washington DC

Popular Test Prep

MCAT Test Prep in Denver, LSAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Phoenix, MCAT Test Prep in Chicago, LSAT Test Prep in Washington DC, SAT Test Prep in Boston, GRE Test Prep in Houston, MCAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Atlanta, GRE 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, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #21 : Parametric Calculations

Example Question #21 : Parametric Calculations

Example Question #23 : Parametric Calculations

Example Question #1 : Polar

Example Question #1 : Polar Form

Example Question #1 : Polar

Example Question #161 : Parametric, Polar, And Vector

Example Question #3 : Polar

Example Question #1 : Polar Form

Example Question #1 : Polar Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: