Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Convert Polar Equations To Rectangular Form and vice versa

Study concepts, example questions & explanations for Precalculus

Create An Account

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #111 : Polar Coordinates And Complex Numbers

Convert the rectangular equation into polar form:

y=-4

Possible Answers:

$r=-4\sin\theta$

$r=-4\cos\theta$

$r=-4\sec\theta$

$r=-4\csc\theta$

Correct answer:

$r=-4\csc\theta$

Explanation:

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=-4$

Recall that,

$\frac{1}{\sin\theta}=\csc\theta$

$r=\frac{-4}{\sin\theta}=-4\csc\theta$

Report an Error

Example Question #1632 : Pre Calculus

How could you write the equation y=x^2 in polar coordinates?

Possible Answers:

$r = cos\theta tan \theta$

$r = tan^2 \theta$

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

$r = cot \theta$

$r = tan\theta$

Correct answer:

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

Explanation:

To convert from rectangular to polar, use the equivalent forms $x = r*cos\theta$ and $y = r*sin\theta$ . Substituting these in, we get:

$r*sin\theta = (r*cos\theta)^2$

$r*sin\theta = r^2*cos^2 \theta$ divide both sides by r

$sin \theta = r*cos^2 \theta$ divide both sides by $cos^2 \theta$ to get this equation in terms of r=

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

Note that we could simplify this a little bit if we wanted to $r = \frac{tan\theta}{cos\theta}$ but that wasn't one of the choices.

Report an Error

Example Question #1633 : Pre Calculus

How could you express the rectangular equation $y = \sqrt{x + \frac{1}{4}}$ in polar form?

Possible Answers:

$r = \frac{1}{4 \sin^2 \theta - \cos \theta}$

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

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

$r = \frac { \cos \theta + \frac {1}{4}}{ \sin ^2 \theta}$

$r = \frac{-\cos \theta \pm \sqrt { \cos 2 \theta }}{2 \sin ^2 \theta}$

Correct answer:

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

Explanation:

To convert from rectangular to polar, we can substitute in $y = r \sin \theta$ and $x = r \cos \theta$ . Our equation now becomes:

$r \sin \theta = \sqrt{ r \cos \theta + \frac {1}{4}}$ square both sides to remove the radical

$r^2 \sin^2 \theta = r \cos \theta + \frac{1}{4}$

Now we can see that in terms of r, this is a quadratic. We can solve using the quadratic formula if we subtract everything from the right side and get our equation equal to 0:

$r^2 \sin ^2 \theta - r \cos \theta - \frac{1}{4} = 0$

Put our coefficents a, b, and c into the quadratic formula:

$r = \frac{ \cos \theta \pm \sqrt{\cos ^2 \theta - 4* \sin ^2 \theta * -\frac{1}{4}}}{2 \sin ^2 \theta}$ multiplying $-4 * -\frac{1}{4}$ yields 1, so this now becomes:

$r = \frac{ \cos \theta \pm \sqrt{ \cos ^2 \theta + \sin ^2 \theta }}{2 \sin ^2 \theta}$ we can simplify this knowing the trigonometric identity that $\cos ^2 \theta + \sin ^2 \theta = 1$

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

Report an Error

Example Question #42 : Polar Coordinates

Which equation is the polar equivalent of the rectangular quadratic $y = -\frac{1}{4} x^2 + 1$ ?

Possible Answers:

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

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

$r = \frac {1}{ \sin \theta + \frac {1}{4} \cos \theta }$

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

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

Correct answer:

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

Explanation:

To convert from rectangular to polar, we can substitute $y = r \sin \theta$ and $x = r \cos \theta$ . That gives us:

$r \sin \theta = -\frac {1}{4} r^2 \cos ^2 \theta + 1$

We can see that this is a quadratic in terms of r, so to solve, just like any other quadratic, we want to subtract everything from the right side so that it is equal to 0.

$\frac { 1}{4} \cos ^2 \theta * r^2 + \sin \theta * r - 1 = 0$

Now we can use the quadratic formula to solve for r:

$r = \frac {- \sin \theta \pm \sqrt { \sin ^2 \theta - 4*\frac{1}{4}cos^2 \theta *(-1) }}{2*\frac{1}{4} \cos ^2 \theta}$

$r = \frac { - \sin \theta \pm \sqrt { \sin ^2 \theta + \cos ^2 \theta }}{ \frac {1}{2} \cos ^2 \theta}$

we can simplify using the trig identity $\sin ^2 \theta + \cos ^2 \theta = 1$

$r = \frac { - \sin \theta \pm 1 }{ \frac {1}{2} \cos ^ 2 \theta }$ to get rid of the fraction in the denominator, multiply top and bottom by 2

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

Report an Error

Example Question #43 : Polar Coordinates

How would you write the equation x^2 + y^2 = 9 as a polar equation?

Possible Answers:

r=9

$r = \theta$

$r^2 = 9\theta$

$r=3\theta$

r=3

Correct answer:

r=3

Explanation:

This simple rectangular equation represents a circle centered at the origin with radius 3,

x^2+y^2=r^2

since 3^2 = 9 .

The way to write that in polar form is just r = 3 .

Report an Error

Example Question #1641 : Pre Calculus

Write y = 2x + 3 in polar form.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To convert from rectangular to polar, substitute $y = r \sin \theta$ and $x = r \cos \theta$ :

$r \sin \theta = 2(r \cos \theta ) + 3$

$r \sin \theta - 2 r \cos \theta = 3$ factor out r

$r(\sin \theta - 2 \cos \theta ) = 3$ divide

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

Report an Error

Example Question #1641 : Pre Calculus

Write the equation y = 3x^2 in polar form.

Possible Answers:

$r = \frac{ \sin \theta }{ 3 \cos \theta }$

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

$r = 3 \tan \theta$

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

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

Correct answer:

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

Explanation:

To convert from rectangular to polar, substitute in $x = r \cos \theta$ and $y = r \sin \theta$ :

$r \sin \theta = 3 (r \cos \theta )^2$

$r \sin \theta = 3 r^2 \cos^2 \theta$

$0 = r^2 \cdot 3 \cos ^2 \theta - r \cdot \sin \theta$ factor out r

$0 = r [ r \cdot 3 \cos ^2 \theta - \sin \theta ]$ this gives us a trivial answer of r = 0, and a second answer found by setting the second [more interesting] answer equal to zero:

$0 = r\cdot 3 \cos^2 \theta - \sin \theta$

$\sin \theta = r \cdot 3 \cos ^2 \theta$

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

Report an Error

Example Question #113 : Polar Coordinates And Complex Numbers

Write $y = \frac{ 1}{2} x^2 + x$ in polar form.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To convert, substitute $x = r \cos \theta$ and $y = r \sin \theta$

$r \sin \theta = \frac{1}{2} (r \cos \theta ) ^ 2 + r \cos \theta$

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

$0 =r^2 \cdot \frac{1}{2} \cos ^2 \theta + r \cos \theta - r \sin \theta$ factor out r

$0 = r [ r \cdot \frac{1}{2 } \cos ^2 \theta + \cos \theta - \sin \theta ]$

This gives us the trivial answer r = 0, but also another answer from setting the second factor equal to zero:

$0 = r \cdot \frac{1}{2} \cos ^2 \theta + \cos \theta - \sin \theta$

$\sin \theta - \cos \theta = r \cdot \frac{1}{2} \cos ^2 \theta$ multiply by 2

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

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

Report an Error

Example Question #42 : Polar Coordinates

Write y = 4x^2 - 3x in polar form.

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

To convert, substitute $x = r \cos \theta$ and $y = r \sin \theta$

$r \sin \theta = 4(r \cos \theta ) ^2 - 3 r \cos \theta$

$r \sin \theta = 4 \cos ^2 \theta \cdot r^2 - 3 r \cos \theta$ divide both sides by r

$\sin \theta = 4 \cos ^2 \theta \cdot r - 3 \cos \theta$

$\sin \theta + 3 \cos \theta = 4 \cos ^2 \theta \cdot r$

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

The answer choice appears in a slightly different order,

$r = \frac{3 \cos \theta + \sin \theta }{ 4 \cos ^2 \theta }$ , but these are equivalent expressions.

Report an Error

Example Question #1642 : Pre Calculus

Convert (x^2 + 2x + y^2 )^2 = 4x^2 +4y^2 to polar form

Possible Answers:

$r = 2 \cos ^2 \theta + 2$

$r = 2 + \cos ^2 \theta$

$r = 2 \cos ^2 \theta + \sqrt{ \cos ^4 \theta + \sin ^2 \theta }$

$r = 2 - 2 \cos \theta$

$r = -2 \cos \theta + 2 \sqrt{ \cos ^4 \theta - \cos ^2 \theta + 1 }$

Correct answer:

$r = 2 - 2 \cos \theta$

Explanation:

Re-arrange the left side so that x^2 is next to y^2 , and factor the 4 out on the right side :

(x^2 + y^2 +2x)^2 = 4(x^2 +y^2)

Make the substitutions x^2 + y^2 = r^2 and $x = r \cos \theta$ :

$(r^2 + 2 r \cos \theta )^2 = 4r^2$ take the square root of both sides

$r^2 + 2 r \cos \theta = 2r$ divide both sides by r

$r + 2 \cos \theta = 2$ subtract $2 \cos \theta$ from both sides

$r = 2 - 2 \cos \theta$

Report an Error

View Pre-Calculus Tutors

Alejandro
Certified Tutor

Kennesaw State University, Bachelor of Science, Computer Software Engineering.

View Pre-Calculus Tutors

Dossevi
Certified Tutor

cole Ste Genevive Versailles, Bachelor of Science, Mathematics. Institut National Polytechnique de Grenoble, Master of Scienc...

View Pre-Calculus Tutors

Travis
Certified Tutor

Delaware State University, Bachelor of Science, Mathematics.

All Precalculus Resources

12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Phoenix, GRE Tutors in Washington DC, Chemistry Tutors in San Francisco-Bay Area, Chemistry Tutors in Denver, Reading Tutors in Washington DC, SSAT Tutors in Chicago, Physics Tutors in Dallas Fort Worth, French Tutors in Houston, Chemistry Tutors in Atlanta, Computer Science Tutors in Boston

Popular Courses & Classes

ACT Courses & Classes in New York City, ISEE Courses & Classes in Atlanta, ACT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in Denver, SSAT Courses & Classes in San Francisco-Bay Area, ISEE Courses & Classes in San Diego, Spanish Courses & Classes in Seattle, LSAT Courses & Classes in Denver, GMAT Courses & Classes in Philadelphia, LSAT Courses & Classes in San Diego

Popular Test Prep

SAT Test Prep in Washington DC, SAT Test Prep in Chicago, LSAT Test Prep in Denver, GMAT Test Prep in Houston, GMAT Test Prep in Miami, SAT Test Prep in San Diego, SAT Test Prep in Dallas Fort Worth, MCAT Test Prep in Philadelphia, SSAT Test Prep in Washington DC, GMAT 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

Precalculus : Convert Polar Equations To Rectangular Form and vice versa

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #111 : Polar Coordinates And Complex Numbers

Example Question #1632 : Pre Calculus

Example Question #1633 : Pre Calculus

Example Question #42 : Polar Coordinates

Example Question #43 : Polar Coordinates

Example Question #1641 : Pre Calculus

Example Question #1641 : Pre Calculus

Example Question #113 : Polar Coordinates And Complex Numbers

Example Question #42 : Polar Coordinates

Example Question #1642 : Pre Calculus

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: