Call Now to Set Up Tutoring:

(888) 888-0446

Precalculus : Polar Coordinates and Complex Numbers

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 #11 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form.

$r=2+\cos\theta$

Possible Answers:

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

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

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

x^2+y^2=4

Correct answer:

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

Explanation:

$r=2+\cos\theta$

Start by multiplying both sides by .

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

Now, isolate the to one side.

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

Square both sides.

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

Recall that r^2=x^2+y^2 and that $x=r\cos\theta$ .

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

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

Report an Error

Example Question #12 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert the polar equation into rectangular form:

$r=16\csc\theta$

Possible Answers:

x^2+y^2=16

y=16

x^2+(y+8)^2=16

x=16

Correct answer:

y=16

Explanation:

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

Now, substitute in that value into the given equation.

$r=\frac{16}{\sin\theta}$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$r\sin\theta=16$

Remember that $y=r\sin\theta$

The rectangular form of this equation is then y=16

Report an Error

Example Question #13 : Convert Polar Equations To Rectangular Form And Vice Versa

What would be the rectangular equation form for the polar equation $r = \cos^2 \theta$ ?

Possible Answers:

$y=\sqrt{x^{\frac{4}{3}}- x^2}$

$y=\sqrt[6]{x^4-x^6}$

$y=\sqrt{x^{\frac{2}{3}}-x^2}$

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

$y=\sqrt[3]{x^2}$

Correct answer:

$y=\sqrt{x^{\frac{4}{3}}- x^2}$

Explanation:

To convert from polar coordinates to rectangular coordinates, know that r is the hypotenuse of a right triangle with legs x and y, so $r = \sqrt{x^2 + y^2}$ .

The cosine of theta is this triangle's adjacent side over the hypotenuse r, so $\cos \theta = \frac{x}{\sqrt{x^2 + y^2}}$ . Making these substitutions into $r = \cos ^2 \theta$ we get:

$\sqrt{x^2 + y^2 } = (\frac{x}{\sqrt{x^2+y^2}})^2$ square the right side to simplify

$\sqrt{x^2 + y^2 } = \frac{x^2}{x^2 + y^2}$ square both sides to remove the radical

$x^2 + y^2 = \frac{x^4}{(x^2+y^2)^2}$ multiply both sides by the right denominator

(x^2 + y^2 )^3 = x^4 take both sides to the $\frac{1}{3}$ power

$x^2 + y^2 = x^{\frac{4}{3}}$ subtract x^2 from both sides

$y^2 = x^{\frac{4}{3}} - x^2$ take the square root

$y = \sqrt{x^{\frac{4}{3}} - x^2}$

Report an Error

Example Question #14 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is equivalent to $r = \frac{1}{2} \cos \theta \sin \theta$ in rectangular form?

Possible Answers:

$x^2 + y^2 = \frac{xy}{2(x^2 + y^2)}$

$\sqrt[3]{4} (x^2 + y^2 ) = \sqrt{x^2y^2}$

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

4(x^2 + y^2 )^3 = x^2 y^2

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

Correct answer:

4(x^2 + y^2 )^3 = x^2 y^2

Explanation:

To convert from polar form to rectangular form, substitute in $r = \sqrt{x^2 + y^2 }$ , $y = r \sin \theta$ , and $x = r\cos\theta$ . Equivalently, $\sin \theta = \frac{y }{r} = \frac{ y }{ \sqrt{x^2 + y^2 }}$ and $\cos \theta = \frac{ x }{ r } = \frac{ x }{\sqrt{x^2 + y^2}}$ :

Substituting these into the original polar equation, we get:

$\sqrt{x^2 +y^2} = \frac {1}{2} \cdot \frac{y}{ \sqrt{x^2 + y^2}} \cdot \frac{x}{\sqrt{x^2 + y^2}}$ multiply the second two fractions

$\sqrt{x^2 + y^2 } = \frac{1}{2} \cdot \frac{xy}{x^2 + y^2 }$ now multiply these fractions

$\sqrt{x^2 + y^2 } = \frac{xy}{2(x^2 + y^2 )}$ square both sides

$x^2 + y^2 = \frac{ x^2 y^2 }{4 (x^2 + y^2 )^2}$ multiply both sides by the denominator

4(x^2 + y^2 )^3 = x^2 y^2

Report an Error

Example Question #15 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation $r= 2 \cos \theta$ in rectangular form

Possible Answers:

y=x(2-x)

y = 2x^2

$y= \sqrt{2x -x^2 }$

$y = \sqrt{3x^2}$

y^2= x^2 - 2x

Correct answer:

$y= \sqrt{2x -x^2 }$

Explanation:

To convert to rectangular form, it is easiest to first multiply both sides by r:

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

Now we can make the substitutions r^2 = x^2 + y^2 and $r \cos \theta = x$ :

x^2 + y^2 = 2 x

We want to solve for y, so subtract x squared from both sides:

y^2 = 2x - x^2 now take the square root of both sides

$y = \sqrt{2x - x^2 }$

Report an Error

Example Question #16 : Convert Polar Equations To Rectangular Form And Vice Versa

Convert $r=\frac{\sin \theta + 1 }{2 \cos ^2 \theta }$ to rectangular form

Possible Answers:

$y = x^2 - \frac{1}{4}$

y = 2x^2 -1

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

$y = 1 - \frac{x^2 }{4}$

$y = x^2 + \frac{1}{4}$

Correct answer:

$y = x^2 - \frac{1}{4}$

Explanation:

First, multiply both sides by the denominator:

$r \cdot 2 \cos ^2 \theta = \sin \theta + 1$ multiply both sides by r

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

Now we can make the substitutions $r = \sqrt{x^2 + y^2 }$ $x=r \cos \theta$ and $y=r \sin \theta$ :

$2x^2 = y + \sqrt{x^2 + y^2 }$ subtract y from both sides

$2x^2 - y = \sqrt{x^2 + y^2}$ square both sides

4x^4 - 4 x^2 y + y^2 = x^2 + y^2 subtract y squared from both sides

4x^4 - 4x^2 y = x^2 we are trying to get this in the form of y=, so subtract 4x^2 from both sides

-4x^2 y = x^2 - 4x^4 divide both sides by -4x^2

$y = \frac{x^2} {-4x^2 } - \frac{4 x^4 }{-4x^2 }$ simplify

$y = \frac{-1}{4} + x^2$ or $y=x^2 - \frac{1}{4}$

Report an Error

Example Question #11 : Polar Coordinates

Convert the equation $r = \frac{\sin \theta - 1 }{ \cos ^2 \theta}$ to rectangular form

Possible Answers:

$y = \frac{1}{2} (x^2 - 1 )$

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

y = - x^2 + 1

$y = x^2 - \frac{1}{2}$

y=x^2 - 1

Correct answer:

$y = \frac{1}{2} (x^2 - 1 )$

Explanation:

First, multiply both sides by the denominator:

$r \cos ^2 \theta = \sin \theta - 1$ multiply both sides by r

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

To convert, make the substitutions $r = \sqrt{x^2 +y^2}$ , $y=r \sin \theta$ , and $x = r \cos \theta$

$x^2 = y - \sqrt{x^2 + y^2 }$ subtract y from both sides

$x^2 - y = -\sqrt{x^2 + y^2 }$ square both sides

x^4 - 2x^2 y + y^2 = x^2 + y^2 subtract y squared from both sides

x^4 - 2x^2 y = x^2 we want to get y by itself, so subtract x^4 from both sides

-2x^2 y = x^2 - x^4

divide both sides by -2x^2

$y = \frac{ x^2}{-2x^2} - \frac{x^4}{-2x^2} = -\frac{1}{2} + \frac{1}{2}x^2 = \frac{1}{2} (x^2 -1)$

Report an Error

Example Question #18 : Convert Polar Equations To Rectangular Form And Vice Versa

Which rectangular equation is equivalent to $r = \frac{\cos \theta }{\sin ^2 \theta + 2 }$ ?

Possible Answers:

$9y^2 + 2(x - \frac{1}{2})^2 = 1$

$3y^2 - 2(x-\frac{1}{2})^2 = 1$

3y^2 -2(x-1)^2 = 1

$9 y^2 = -2(x + \frac{1}{4} ) ^2$

$24y^2 +16(x-\frac{1}{4})^2 = 1$

Correct answer:

$24y^2 +16(x-\frac{1}{4})^2 = 1$

Explanation:

First multiply both sides by the denominator:

$r (\sin ^2 \theta + 2 ) = \cos \theta$ multiply both sides by r

$r^2 (\sin ^2 \theta + 2 ) = r \cos \theta$ distribute

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

Now we can make the substitutions $x^2 + y^2 = r^2, x = r \cos \theta$ , and $y = r \sin \theta$ :

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

y^2 +2x^2 + 2y^2 = x combine like terms

3y^2 + 2x^2 = x subtract 2 x squared from both sides

3y^2 = -2x^2 + x We want to complete the square on the right, so factor our the -2:

$3y^2 = -2( x^2 -\frac{1}{2}x \enspace \enspace )$ to complete the square, add $(\frac{1}{2}\cdot \frac{1}{2})^2 = \frac{1}{16}$ inside the parentheses. This multiplied by the -2 outside the parentheses is $-\frac{1}{8}$ , so this means we're actually subtracting $\frac{1}{8}$ from both sides:

$3y^3 - \frac{1}{8} = -2(x - \frac{1}{4}) ^2$ add $\frac{1}{8}$ and $2(x-\frac{1}{4})^2$ to both sides:

$3y^3 + 2(x - \frac{1}{4} ) ^2 = \frac{1}{8 }$ multiply both sides by 8

$24y^2 + 16(x-\frac{1}{4})^2 = 1$

Report an Error

Example Question #19 : Convert Polar Equations To Rectangular Form And Vice Versa

Which is the rectangular form for $r = \frac{\sin \theta } { 2 \cos ^2 \theta + 1 }$ ?

Possible Answers:

$y = \frac{-1 + \sqrt{1-12x^2}}{2}$

$y = \frac{1 + \sqrt{1 + 12x^2}}{2}$

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

$y = \frac{1}{4} (x^2 - 1 )$

$y = x^2 - \frac{1}{4}$

Correct answer:

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

Explanation:

First multiply both sides by the right denominator:

$r( 2 \cos ^2 \theta + 1 ) = \sin \theta$ multiply both sides by r

$r^2 (2\cos ^2 \theta + 1) = r \sin \theta$ distribute

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

Now we can start to convet to rectangular by making the substitutions $y = r \sin \theta$ , $x = r \cos \theta$ , and r^2 = x^2 + y^2 :

2x^2 + x^2 + y^2 = y combine like terms:

3x^2 + y^2 = y subtract y from both sides, and re-order this in decending order of powers of y:

y^2 - y + 3x^2 = 0 this is a quadratic, so we can use the quadratic equation to get y by itself:

$y = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(3x^2)}}{2(1)} = \frac{1 \pm \sqrt{1 - 12x^2 }}{2 }$

The answer choice that works is

$y = \frac{1 + \sqrt{1 - 12x^2}}{2}$

Report an Error

Example Question #20 : Convert Polar Equations To Rectangular Form And Vice Versa

Write the equation for $r = \frac{\cos \theta - \frac{1}{2}}{ \cos ^2 \theta }$ in rectangular form

Possible Answers:

$y = \sqrt{\frac{3}{4}x^2 - x^4}$

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

$y = \sqrt{2x^4 - 4x^3 + 1.5x^2 }$

$y = \sqrt{x^4 - 2x^3 + \frac{3}{4} x }$

$y = \sqrt{4x^4 - 8 x^3 + 3x^2 }$

Correct answer:

$y = \sqrt{4x^4 - 8 x^3 + 3x^2 }$

Explanation:

Multiply both sides by the right denominator:

$r \cos ^2 \theta = \cos \theta - \frac{1}{2}$ multiply both sides by r

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

Now we can substitute in $r = \sqrt{x^2 + y^2 }$ and $x = r \cos \theta$ to start converting to rectangular form:

$x^2 = x - \frac{1}{2} \sqrt{x^2 + y^2 }$ subtract x from both sides

$x^2 - x = -\frac{ 1}{2} \sqrt{x^2 + y^2 }$ square both sides

$x^4 -2x^3 + x^2 = \frac{1}{4} (x^ 2 + y^2 )$ multiply both sides by 4

4x^4 - 8 x ^2 + 4 x ^2 = x^2 + y^2 subtract x squared from both sides

4x^4 - 8x^3 + 3x^2 = y^2 take the square root of both sides

$y = \sqrt{4x^4 - 8x^3 + 3x^2 }$

Report an Error

View Pre-Calculus Tutors

Nate
Certified Tutor

University of Wisconsin-Oshkosh, non degree, Prepharmacy. University of Wisconsin-Madison, Pharmaceutical Chemist, Pharmacy.

View Pre-Calculus Tutors

Robert
Certified Tutor

Rensselaer Polytechnic Institute, Bachelors, Electrical Engineering. Brooklyn Polytechnic, Masters, Computer Science.

View Pre-Calculus Tutors

MrG
Certified Tutor

Long Island University-Brooklyn Campus, Master of Science, Applied Mathematics.

All Precalculus Resources

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

Popular Subjects

French Tutors in Dallas Fort Worth, Spanish Tutors in New York City, English Tutors in San Francisco-Bay Area, SAT Tutors in San Francisco-Bay Area, MCAT Tutors in Washington DC, Computer Science Tutors in Houston, Reading Tutors in Philadelphia, LSAT Tutors in Dallas Fort Worth, French Tutors in Houston, Computer Science Tutors in Los Angeles

Popular Courses & Classes

GRE Courses & Classes in Denver, GRE Courses & Classes in Boston, LSAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Denver, GRE Courses & Classes in New York City, SAT Courses & Classes in Atlanta, Spanish Courses & Classes in Houston, SAT Courses & Classes in Miami, ISEE Courses & Classes in Atlanta, MCAT Courses & Classes in Miami

Popular Test Prep

ISEE Test Prep in Chicago, SSAT Test Prep in Dallas Fort Worth, LSAT Test Prep in Chicago, MCAT Test Prep in San Diego, ISEE Test Prep in Houston, LSAT Test Prep in Phoenix, SAT Test Prep in Houston, LSAT Test Prep in Boston, ACT Test Prep in Denver, SSAT Test Prep in Houston

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 : Polar Coordinates and Complex Numbers

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #11 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #12 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #13 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #14 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #15 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #16 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #11 : Polar Coordinates

Example Question #18 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #19 : Convert Polar Equations To Rectangular Form And Vice Versa

Example Question #20 : Convert Polar Equations To Rectangular Form And Vice Versa

All Precalculus Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: