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 #91 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #92 : Polar Form

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

Possible Answers:

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

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

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

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

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

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 #93 : Polar Form

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

Possible Answers:

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

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

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

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

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

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

Example Question #1 : Graphing Polar Form

Which of the following substitutions will help solve the following integral?

$\int\frac{\sqrt{25x^2-4}}{x}dx$

Possible Answers:

$x = \frac{2}{5}sec\theta$

u = 25x^2 - 4

$u = \frac{1}{x}$

$x = \frac{2}{5} tan\theta$

$x = \frac{2}{5}sin\theta$

Correct answer:

$x = \frac{2}{5}sec\theta$

Explanation:

As we can see in this integral, there is no reverse chain-rule u-substitution possible. The logical step is to use a trigonometric substitution. If one recalls that trig substitutions of the type x^2-a^2 could be solved with the substitution $\frac{2}{5}sec\theta$ , then the answer is easily seen. However, we can also use a right triangle: Screen_shot_2015-04-18_at_6.33.31_pm

And thus we have:

$sec \theta = \frac{5x}{2}$

or:

$x = \frac{2}{5}sec\theta$

Report an Error

Example Question #1 : Graphing Polar Form

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

Faker_cosx

R_cosx

R_cosx_1

R_sinx

R_cos2x

Correct answer:

R_cosx

Explanation:

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

Report an Error

Example Question #5 : Parametric Form

Draw the graph of $r=sin\theta$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R_cosx

R_sinx_1

Faker_cosx

R_sin2x

R_sinx

Correct answer:

R_sinx

Explanation:

Between and $\frac{\pi }{2}$ , the radius approaches from .

From $\frac{\pi }{2}$ to $\pi$ the radius goes from to .

Between $\pi$ and $\frac{3\pi }{2}$ , the curve is redrawn in the opposite quadrant, the first quadrant as the radius approaches .

From $\frac{3\pi }{2}$ and $2\pi$ , the curve is redrawn in the second quadrant as the radius approaches from .

Report an Error

Example Question #11 : Polar Form

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R_cosx_1

R_sin2x

R2_cos2x

R_sin2x

R_cos2x

Correct answer:

R_cos2x

Explanation:

Because this function has a period of $\pi$ , the x-intercepts of the graph y=cos(2x) happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

Report an Error

Example Question #1 : Parametric Form

Draw the graph of $r=sin(2\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

R_cos2x

Faker_cosx

R_sinx_1

R_sinx

R_sin2x

Correct answer:

R_sin2x

Explanation:

Because this function has a period of $\pi$ , the amplitude of the graph y=sin(2x) appear at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches 1 from 0.

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches 0 from 1.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches -1 from 0 and is drawn in the opposite quadrant, the fourth quadrant because it has a negative radius.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches 0 from -1, and is also drawn in the fourth quadrant.

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches 1 from 0. Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches 0 from 1.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches -1 from 0. Because it is a negative radius, it is drawn in the opposite quadrant, the second quadrant. Likewise, as the radius approaches 0 from -1. Between $\frac{7\pi }{4}$ and $2\pi$ , the curve is drawn in the second quadrant.

Report an Error

Example Question #261 : Parametric, Polar, And Vector

Graph $r^2=cos(2\theta )$ where $0\leq \theta \leq2\pi$ .

Possible Answers:

R_sin2x

R_cos2x

R2_sin2x

R_cosx

R2_cos2x

Correct answer:

R2_cos2x

Explanation:

Taking the graph of y=cos(2x) , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{4}$ , $\frac{3\pi }{4}$ to $\frac{5\pi }{4}$ , and $\frac{7\pi }{4}$ to $2\pi$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 1 at and traces to 0 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at -1 and traces to zero in the opposite quadrant, the third quadrant.

From $\frac{3\pi }{4}$ to $\pi$ , the curves are traced from 0 to 1 and 0 to -1 in the fourth quadrant. Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

Report an Error

Example Question #1 : Graphing Polar Form

Draw the curve of $r^2=sin(2\theta )$ from $0\leq \theta \leq2\pi$ .

Possible Answers:

R2_cos2x

R_sin2x

R2_sin2x

R_sinx

R_sinx_1

Correct answer:

R2_sin2x

Explanation:

Taking the graph of y=sin(2x) , we only want the areas in the positive first quadrant because the radius is squared and cannot be negative.

This leaves us with the areas from to $\frac{\pi }{2}$ and $\pi$ to $\frac{3\pi }{2}$ .

Then, when we take the square root of the radius, we get both a positive and negative answer with a maximum and minimum radius of $\pm 1$ .

To draw the graph, the radius is 0 at and traces to 1 at $\frac{\pi }{4}$ . As well, the negative part of the radius starts at 0 and traces to-1 in the opposite quadrant, the third quadrant.

From $\frac{\pi }4{}$ to $\frac{\pi }{2}$ , the curves are traced from 1 to 0 and -1 to 0 in the third quadrant.

Following this pattern, the graph is redrawn again from the areas included in $\pi$ to $2\pi$ .

Report an Error

View Calculus 2 Tutors

Jeremy
Certified Tutor

University of Illinois at Chicago, Bachelor of Science, Physics.

View Calculus 2 Tutors

Magdy
Certified Tutor

Cairo University, Egypt, Bachelor of Science, Electrical Engineering. New Mexico State University-Main Campus, Doctor of Phil...

View Calculus 2 Tutors

Anabel
Certified Tutor

UDELAR, Diploma, Chemistry. IPA, Bachelor of Education, Mathematics.

All Calculus 2 Resources

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

Popular Subjects

Math Tutors in Los Angeles, Computer Science Tutors in Chicago, Algebra Tutors in Miami, Reading Tutors in Miami, English Tutors in San Francisco-Bay Area, Statistics Tutors in Philadelphia, Spanish Tutors in Houston, Statistics Tutors in Los Angeles, Biology Tutors in Miami, French Tutors in Seattle

Popular Courses & Classes

ACT Courses & Classes in Dallas Fort Worth, SAT Courses & Classes in Philadelphia, SSAT Courses & Classes in Atlanta, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in Phoenix, ISEE Courses & Classes in Miami, SSAT Courses & Classes in Los Angeles, ACT Courses & Classes in Chicago, SAT Courses & Classes in San Diego, ISEE Courses & Classes in San Diego

Popular Test Prep

ISEE Test Prep in Phoenix, MCAT Test Prep in Los Angeles, MCAT Test Prep in Atlanta, LSAT Test Prep in Phoenix, ISEE Test Prep in San Diego, SAT Test Prep in Denver, ACT Test Prep in Chicago, GRE Test Prep in Boston, SSAT Test Prep in San Diego, 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

Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : Polar Form

Example Question #92 : Polar Form

Example Question #93 : Polar Form

Example Question #1 : Graphing Polar Form

Example Question #1 : Graphing Polar Form

Example Question #5 : Parametric Form

Example Question #11 : Polar Form

Example Question #1 : Parametric Form

Example Question #261 : Parametric, Polar, And Vector

Example Question #1 : Graphing Polar Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: