We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Polar Form

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 10 Next →

Example Question #772 :Calculus Ii

Calculate the polar form hypotenuse of the following cartesian equation: y^2=100x^8

Possible Answers:

$\frac{\tan(\theta)}{10\cos^3(\theta)}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{\cos(\theta)}}$

Correct answer:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Explanation:

In a cartesian form, the primary parameters areand. In polar form, they areand $\theta$

is the hypotenuse, and $\theta$ is the angle created by.

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

y^2=100x^8

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

$r^3 = \frac{\tan(\theta)}{10\cos^3(\theta)}$

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Report an Error

Example Question #773 :Calculus Ii

What is the polar form of $y=\frac{4}{3}x^{2}$

Possible Answers:

$r=\tan\frac{3}{4} \theta\sec\theta$

$r=\frac{4}{3}\tan \theta\sec\theta$

$r=-\frac{4}{3}\tan \theta\sec\theta$

$r=\frac{3}{4}\tan \theta\sec\theta$

$r=-\frac{3}{4}\tan \theta\sec\theta$

Correct answer:

$r=\frac{3}{4}\tan \theta\sec\theta$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=\frac{4}{3}x^{2}$ , then:

$r\sin\theta=\frac{4}{3}(r\cos \theta)^{2}$

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

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=\frac{4}{3}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{4}{3}r$

$\tan \theta\sec\theta=\frac{4}{3}r$

$r=\frac{3}{4}\tan \theta\sec\theta$

Report an Error

Example Question #774 :Calculus Ii

What is the polar form of $y=-x^{2}$ ?

Possible Answers:

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

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

$r=-\tan \theta\sec\theta$

$r=-\frac{\tan \theta}{\sec\theta}$

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

Correct answer:

$r=-\tan \theta\sec\theta$

Explanation:

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

$r\sin\theta=-(r\cos \theta)^{2}$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=-r\cos\theta$

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

$\tan \theta\sec\theta=-r$

$r=-\tan \theta\sec\theta$

Report an Error

Example Question #775 :Calculus Ii

What is the polar form of y=6+14x ?

Possible Answers:

$r=-\frac{6}{\sin\theta-14\cos \theta}$

$r=\frac{\sin\theta-14\cos \theta}{6}$

$r=-\frac{6}{\sin\theta+14\cos \theta}$

$r=\frac{6}{\sin\theta-14\cos \theta}$

$r=\frac{6}{\sin\theta+14\cos \theta}$

Correct answer:

$r=\frac{6}{\sin\theta-14\cos \theta}$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given y=6+14x , then:

$r\sin\theta=6+14(r\cos \theta)$

$r\sin\theta-14(r\cos \theta)=6$

$r(\sin\theta-14\cos \theta)=6$

$r=\frac{6}{\sin\theta-14\cos \theta}$

Report an Error

Example Question #776 :Calculus Ii

What is the polar form of y=2-9x ?

Possible Answers:

$r=-\frac{2}{\sin\theta+9\cos \theta}$

$r=\frac{2}{\sin\theta-9\cos \theta}$

$r=\frac{\sin\theta-9\cos \theta}{2}$

$r=-\frac{2}{\sin\theta-9\cos \theta}$

$r=\frac{2}{\sin\theta+9\cos \theta}$

Correct answer:

$r=\frac{2}{\sin\theta+9\cos \theta}$

Explanation:

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given y=2-9x , then:

$r\sin\theta=2-9(r\cos \theta)$

$r\sin\theta+9(r\cos \theta)=2$

$r(\sin\theta+9\cos \theta)=2$

$r=\frac{2}{\sin\theta+9\cos \theta}$

Report an Error

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

View Calculus 2 Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View Calculus 2 Tutors

Mohammed
Certified Tutor

South Carolina State University, Bachelor of Science, Mathematics. South Carolina State University, Masters in Education, Mat...

View Calculus 2 Tutors

Francisco
Certified Tutor

University of Texas at Brownsville, Bachelors, Mathematics. University of Texas at Brownsville, Masters, Mathematics.

All Calculus 2 Resources

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

Popular Subjects

Physics Tutors in Los Angeles,GMAT Tutors in Atlanta,Reading Tutors in Los Angeles,GRE Tutors in Miami,GRE Tutors in Atlanta,Statistics Tutors in Boston,ACT Tutors in Houston,Computer Science Tutors in Houston,Algebra Tutors in Denver,物理教师在芝加哥

Popular Courses & Classes

MCAT Courses & Classes in Seattle,Spanish Courses & Classes in New York City,ISEE Courses & Classes in Dallas Fort Worth,SSAT Courses & Classes in Dallas Fort Worth,LSAT Courses & Classes in Houston,LSAT Courses & Classes in Phoenix,GMAT Courses & Classes in Dallas Fort Worth,SAT Courses & Classes in Los Angeles,SAT Courses & Classes in Phoenix,ACT Courses & Classes in Boston

Popular Test Prep

LSAT Test Prep in Los Angeles,LSAT Test Prep in Philadelphia,ISEE Test Prep in Atlanta,LSAT Test Prep in Miami,GRE Test Prep in San Francisco-Bay Area,GMAT Test Prep in Dallas Fort Worth,SAT Test Prep in Denver,SSAT Test Prep in San Diego,MCAT Test Prep in Boston,SSAT 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.

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:

copyrigh的物理或电子签名t 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 : Polar Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #772 :Calculus Ii

Example Question #773 :Calculus Ii

Example Question #774 :Calculus Ii

Example Question #775 :Calculus Ii

Example Question #776 :Calculus Ii

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: