Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Polar

Study concepts, example questions & explanations for Calculus 2

Create An Account Create Tests & Flashcards

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 … 18 19 Next →

Example Question #1 :Polar

Rewrite the polar equation

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

in rectangular form.

Possible Answers:

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

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

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

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

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

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 #11 :Functions, Graphs, And Limits

Rewrite in polar form:

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

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 #1 :Polar

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

Possible Answers:

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

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

$r = 12\tan \theta$

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

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

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

例子问题# 681:Calculus Ii

Give the rectangular coordinates of the point with polar coordinates

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

Correct answer:

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

Explanation:

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

代替的,在方程 $\small y=x^{2}$ 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 #2 :Polar Form

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

Find its corresponding (x,y) coordinate.

Possible Answers:

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

(0,3)

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

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

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

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, ourisand 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

Example Question #3 :Polar Form

What is the magnitude and angle (in radians) of the following cartesian coordinate?

(3,4)

Give the answer in the format below.

$(r,\theta)$

Possible Answers:

(3,1.04)

(4,0.927)

(12,0.707)

(5,0.643)

(5,0.927)

Correct answer:

(5,0.927)

Explanation:

Although not explicitly stated, the problem is asking for the polar coordinates of the point (3,4) . To calculate the magnitude,, calculate the following:

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

r=5

To calculate $\theta$ , do the following:

$\theta=tan^{-1}\left(\frac{y}{x}\right)$ in radians. (The problem asks for radians)

$\theta = 0.927$

Report an Error

Example Question #1 :Polar Form

What is the following coordinate in polar form?

(-2,-5)

Provide the angle in degrees.

Possible Answers:

(16,112)

$(\sqrt{29},248)$

$(\sqrt{29},68)$

(10,248)

$(\sqrt{29},202)$

Correct answer:

$(\sqrt{29},248)$

Explanation:

To calculate the polar coordinate, use

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

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

Report an Error

Example Question #5 :Polar Form

What is the equation $y=2x^{2}$ in polar form?

Possible Answers:

$r=\frac{1}{2}\sin \theta}{\tan \theta$

$r=\frac{1}{2}\cos \theta}{\tan \theta$

$r=\frac{1}{2}\sin \theta}{\sec \theta$

$r=\frac{1}{2}\tan \theta}{\sec \theta$

$r=\frac{1}{2}\sin \theta}{\cos \theta$

Correct answer:

$r=\frac{1}{2}\tan \theta}{\sec \theta$

Explanation:

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=2x^{2}$ , then $r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ .

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

$\frac{\tan \theta}{\cos \theta}=2r$

$\tan \theta}{\sec \theta=2r$

$\frac{1}{2}\tan \theta}{\sec \theta=r$

Report an Error

← Previous 1 2 3 4 5 6 7 8 9 … 18 19 Next →

View Calculus 2 Tutors

Harminderjit Singh
Certified Tutor

University of Windsor, Bachelor of Science, Electrical Engineering.

View Calculus 2 Tutors

Pierre
Certified Tutor

Brigham Young University-Provo, Bachelor of Science, Electrical Engineering. Pennsylvania State University-Penn State Fayette...

View Calculus 2 Tutors

Mike
Certified Tutor

Concordia University-Ann Arbor, Master of Arts Teaching, Mathematics Teacher Education.

All Calculus 2 Resources

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

Popular Subjects

SAT Tutors in Miami,Physics Tutors in Los Angeles,MCAT Tutors in Denver,SAT Tutors in Phoenix,Biology Tutors in San Diego,Physics Tutors in Miami,MCAT Tutors in Seattle,Spanish Tutors in San Francisco-Bay Area,Algebra Tutors in Los Angeles,Statistics Tutors in Chicago

Popular Courses & Classes

行为Courses & Classes in Miami,行为Courses & Classes in Denver,行为Courses & Classes in Boston,GRE Courses & Classes in Atlanta,GMAT Courses & Classes in Houston,GRE Courses & Classes in Los Angeles,ISEE Courses & Classes in Seattle,LSAT Courses & Classes in Atlanta,ISEE Courses & Classes in Houston,LSAT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

GRE Test Prep in Miami,MCAT Test Prep in Boston,GRE Test Prep in New York City,行为Test Prep in Denver,SAT Test Prep in Los Angeles,行为Test Prep in Boston,SAT Test Prep in San Francisco-Bay Area,GRE Test Prep in Denver,行为Test Prep in Houston,LSAT Test Prep in Washington DC

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.

哟ur 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:

哟u 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.

把你的投诉我们指定的代理:

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.

我承认可能有不利的法律上的缺点equences 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 : Polar

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 :Polar

Example Question #11 :Functions, Graphs, And Limits

Example Question #1 :Polar

Example Question #1 :Polar

例子问题# 681:Calculus Ii

Example Question #1 :Polar Form

Example Question #2 :Polar Form

Example Question #3 :Polar Form

Example Question #1 :Polar Form

Example Question #5 :Polar Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
哟ur name:
Feedback: