Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Parametric

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 #1 : Parametric

Rewrite as a Cartesian equation:

$x = t^{2} + 2t + 1, y = t^{2} - 2t + 1, t \in [-1, 1]$

Possible Answers:

y = x - 2

$y = x + 4 - 4\sqrt{x}$

$y = x + 4 + 4\sqrt{x}$

$y = x - 4 - 4\sqrt{x}$

y = x+2

Correct answer:

$y = x + 4 - 4\sqrt{x}$

Explanation:

$x = t^{2} + 2t + 1$

$x = \left ( t + 1\right )^{2}$

$\pm \sqrt{x} = t + 1$

$\sqrt{x} = t + 1$ or $-\sqrt{x} = t + 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t + 1 is nonnegative; we choose

$\sqrt{x} = t + 1$ .

Also,

$y = t^{2} - 2t + 1$

$y= \left ( t - 1\right )^{2}$

$\sqrt{y} = \pm \left ( t - 1\right )$

$\sqrt{y} = t- 1$ or $-\sqrt{y} = t- 1$

We are restricting to values on $\left [ -1, 1\right ]$ , so t - 1 is nonpositive; we choose

$-\sqrt{y} = t- 1$

or equivalently,

$\sqrt{y} = -t+ 1$

to make t - 1 nonpositive.

Then,

$\sqrt{x}+ \sqrt{y} = (t + 1) + (-t + 1)$

and

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

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

$y =\left ( 2 - \sqrt{x} \right )^2$

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

$y = x + 4 - 4\sqrt{x}$

Report an Error

Example Question #1 : Parametric, Polar, And Vector

Write in Cartesian form:

$x = \cos 2t, y = \sin t$

Possible Answers:

x + 2y = 1

$x = 2y^{2}$

x - 2y = 1

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

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

Correct answer:

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

Explanation:

Rewrite using the double-angle formula:

$x = \cos 2t =1 - 2\sin ^{2} t = 1 - 2y^{2}$

Then

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

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

which is the correct choice.

Report an Error

Example Question #1 : Parametric

Write in Cartesian form:

$x = 3t, y = 2t^{2}$

Possible Answers:

$y = \frac{2}{9} x^{2}$

$y = \frac{9}{2} x^{2}$

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

$x^{2}+ y^{2} = 36$

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

Correct answer:

$y = \frac{2}{9} x^{2}$

Explanation:

x = 3t , so

$x^{2} = (3t)^{2} = 9t^{2} = \frac{9}{2} \cdot 2t^{2} = \frac{9}{2} y$ .

$\frac{9}{2} y= x^{2}$ , so

$\frac{2}{9}\cdot \frac{9}{2} y = \frac{2}{9}x^{2}$

$y = \frac{2}{9}x^{2}$

Report an Error

Example Question #3 : Parametric

Write in Cartesian form:

x = 2t - 1

y = 4t+2

Possible Answers:

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

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

y = 2x

y = 2x - 4

y = 2x + 4

Correct answer:

y = 2x + 4

Explanation:

y = 4t+2 = 4t - 2 + 4 = 2 (2t - 1)+ 4 = 2x + 4 ,

so the Cartesian equation is

y = 2x + 4 .

Report an Error

Example Question #4 : Parametric

Write in Cartesian form:

$x = \ln (t+1)$

y = t^2 + 2t + 2

$t \in (-1, \infty )$

Possible Answers:

$y = e^{2x} + 1$

$y = e^{2x} + e$

$y = e^{2x} - e$

$y = e^{2x}+ 2$

$y = e^{2x} - 1$

Correct answer:

$y = e^{2x} + 1$

Explanation:

$x = \ln (t+1)$

$e ^{x} = \left [ \ln (t+1) \right ] ^{x}$

$e ^{x} = t + 1$

$y = t^2 + 2t + 2 = t^2 + 2t + 1 + 1 = (t + 1)^{2}+ 1 = \left (e^{x} \right )^{2} + 1 = e^{2x} + 1$

Therefore the Cartesian equation is $y = e^{2x} + 1$ .

Report an Error

Example Question #2 : Parametric, Polar, And Vector Functions

Rewrite as a Cartesian equation:

$x =10^ {t}, y = \sinh t, t \in (0,\infty )$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

$y = \sinh t = \frac{e ^{2t} -1}{2e ^{t}} = \frac{\left (e ^{ t} \right) ^{2} -1}{2e ^{t}}$

$e^{t} = 10 ^{\log e \cdot t} =\left ( 10 ^{ t} \right )^{\log e} = x ^{\log e}$ , so

$y = \frac{\left ( x ^{\log e} \right) ^{2} -1}{2 x ^{\log e}}= \frac{ x ^{2 \log e} -1}{2 x ^{\log e}}$

This makes the Cartesian equation

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

Report an Error

Example Question #7 : Parametric

$\small x=3t+4$ and $\small y=1/t$ . What is $\small y$ in terms of $\small x$ (rectangular form)?

Possible Answers:

$\small y=(x-3)/4$

$\small y=4/(x-3)$

$\small y=(x-4)/3$

$\small y=3/(x-4)$

Correct answer:

$\small y=3/(x-4)$

Explanation:

In order to solve this, we must isolate $\small t$ in both equations.

$\small \small x=3t+4\rightarrow t=(x-4)/3$ and

$\small y=1/t\rightarrow t=1/y$ .

Now we can set the right side of those two equations equal to each other since they both equal $\small t$ .

$\small (x-4)/3=1/y$ .

By multiplying both sides by $\small 3y/(x-4)$ , we get $\small y=3/(x-4)$ , which is our equation in rectangular form.

Report an Error

Example Question #1 : Functions, Graphs, And Limits

If x=3+t and y=4t+7 , what is in terms of (rectangular form)?

Possible Answers:

y=4x-5

y=4x-3

y=4x-12

y=4x-7

y=4x-2

Correct answer:

y=4x-5

Explanation:

Given x=3+t and y=4t+7 , we can find in terms of by isolating in both equations:

$x=3+t\rightarrow t=x-3$

$y=4t+7\rightarrow t=\frac{y-7}{4}$

Since both of these transformations equal , we can set them equal to each other:

$\frac{y-7}{4}=x-3$

y-7=4(x-3)

y-7=4x-12

y=4x-5

Report an Error

Example Question #1 : Parametric Form

Given x=7t-5 and y=2t+9 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{7}{2}({x+5})+9$

None of the above

$y=\frac{2}{7}({x+5})+9$

$y=\frac{5}{7}({x+2})+9$

$y=\frac{9}{7}({x+5})+2$

Correct answer:

$y=\frac{2}{7}({x+5})+9$

Explanation:

In order to find with respect to , we first isolate in both equations: $x=7t-5\rightarrow t=\frac{x+5}{7}$

$y=2t+9\rightarrow t=\frac{y-9}{2}$

Since both equations equal , we can then set them equal to each other and solve for :

$\frac{y-9}{2}=\frac{x+5}{7}$

${y-9}=2(\frac{x+5}{7})$

$y=\frac{2}{7}({x+5})+9$

Report an Error

Example Question #1 : Parametric Form

Given x=4t+9 and y=-t+2 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{4-x}{17}$

$y=\frac{x-17}{4}$

None of the above

$y=\frac{17-x}{4}$

$y=\frac{x-4}{17}$

Correct answer:

$y=\frac{17-x}{4}$

Explanation:

In order to find with respect to , we first isolate in both equations: $x=4t+9\rightarrow t=\frac{x-9}{4}$

$y=-t+2\rightarrow t=2-y$

Since both equations equal , we can then set them equal to each other and solve for :

$2-y=\frac{x-9}{4}$

$y=2-\frac{x-9}{4}$

$y=\frac{8}{4}-\frac{x-9}{4}$

$y=\frac{8-(x-9)}{4}$

$y=\frac{8-x+9}{4}$

$y=\frac{17-x}{4}$

Report an Error

View Calculus 2 Tutors

Edem
Certified Tutor

University of Ghana, Bachelor, Mathematics. African Institute of Mathematical Science, Master's/Graduate, Mathematics.

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

Samita
Certified Tutor

University of Houston, Bachelor of Science, Computer Science. University of Houston, Non Degree Bachelors, Biology, General.

All Calculus 2 Resources

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

Popular Subjects

MCAT Tutors in Dallas Fort Worth, LSAT Tutors in Boston, Math Tutors in Denver, MCAT Tutors in Phoenix, Algebra Tutors in Los Angeles, ISEE Tutors in Miami, Math Tutors in Houston, SSAT Tutors in Atlanta, Algebra Tutors in Houston, Math Tutors in Phoenix

Popular Courses & Classes

SSAT Courses & Classes in Houston, ISEE Courses & Classes in Chicago, Spanish Courses & Classes in Los Angeles, Spanish Courses & Classes in Seattle, MCAT Courses & Classes in San Diego, ACT Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in New York City, MCAT Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in Los Angeles

Popular Test Prep

MCAT Test Prep in Los Angeles, LSAT Test Prep in Chicago, SAT Test Prep in Denver, SAT Test Prep in Houston, SAT Test Prep in Philadelphia, ISEE Test Prep in Phoenix, ACT Test Prep in Dallas Fort Worth, MCAT Test Prep in San Diego, MCAT Test Prep in Dallas Fort Worth, ACT Test Prep in San Francisco-Bay Area

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

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1 : Parametric

Example Question #1 : Parametric, Polar, And Vector

Example Question #1 : Parametric

Example Question #3 : Parametric

Example Question #4 : Parametric

Example Question #2 : Parametric, Polar, And Vector Functions

Example Question #7 : Parametric

Example Question #1 : Functions, Graphs, And Limits

Example Question #1 : Parametric Form

Example Question #1 : Parametric Form

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: