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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #55 : Parametric Form

Given $x=\frac{1}{2}t-2$ and $y=\frac{1}{3}t-5$ , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{2}{3}(x-2)+5$

$y=\frac{2}{3}(x+2)+5$

$y=\frac{2}{3}(x-2)-5$

$y=-\frac{2}{3}(x+2)-5$

$y=\frac{2}{3}(x+2)-5$

Correct answer:

$y=\frac{2}{3}(x+2)-5$

Explanation:

Given $x=\frac{1}{2}t-2$ and $y=\frac{1}{3}t-5$ , let's solve both equations for :

$x=\frac{1}{2}t-2\rightarrow t=2(x+2)$

$y=\frac{1}{3}t-5\rightarrow t=3(y+5)$

Since both equations equal , let's set them equal to each other and solve for :

3(y+5)=2(x+2)

$y+5=\frac{2}{3}(x+2)$

$y=\frac{2}{3}(x+2)-5$

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Example Question #56 : Parametric Form

Given x=6-t and y=3t-2 , what is in terms of (rectangular form)?

Possible Answers:

None of the above

y=16x-3

y=16-3x

y=16x+3

y=16+3x

Correct answer:

y=16-3x

Explanation:

Given x=6-t and y=3t-2 , let's solve both equations for :

$x=6-t\rightarrow t=6-x$

$y=3t-2\rightarrow t=\frac{y+2}{3}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y+2}{3}=6-x$

y+2=3(6-x)

y=3(6-x)-2

y=18-3x-2

y=16-3x

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Example Question #51 : Parametric Form

Convert the following equation from parametric to rectangular form:

$x=t^2, y=\sec(t)+\tan(t)$

Possible Answers:

y=0

y=x

$y=\sec(\pm \sqrt{x})+\tan(\pm \sqrt{x})$

$y=\sec(-\sqrt{x})+\tan(-\sqrt{x})$

Correct answer:

$y=\sec(\pm \sqrt{x})+\tan(\pm \sqrt{x})$

Explanation:

To convert from parametric to rectangular form, we must eliminate the parameter from one of the equations:

$t=\pm\sqrt{x}$

Now, replace all of the t's in the equation for y with the above term:

$y=\sec(\pm \sqrt{x})+\tan(\pm \sqrt{x})$

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Example Question #52 : Parametric Form

Convert the following parametric equation to rectangular form:

x=t-1, y=8t

Possible Answers:

y=8x

y=8(x-1)

y=8(x+1)

y=x

Correct answer:

y=8(x+1)

Explanation:

To convert the equation into parametric form, we must eliminate the parameter by finding t in terms of either x or y:

t=x+1

Now, replace all of the t's in the equation for y with the above term:

y=8(x+1)

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Example Question #53 : Parametric Form

Convert the following parametric equation into rectangular form:

$x=2t, y=\cos(t)+2$

Possible Answers:

$y=\cos(x)+2$

$y=\cos\left(\frac{x}{2}\right)+2$

y=x

$y=\cos(x)-2$

Correct answer:

$y=\cos\left(\frac{x}{2}\right)+2$

Explanation:

To convert from parametric form to rectangular form, we must eliminate the parameter:

To do this we will first get t in terms of x.

$t=\frac{x}{2}$

Now, replace all of the t's in the equation for y:

$y=\cos\left(\frac{x}{2}\right)+2$

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Example Question #60 : Parametric Form

Given x=9-t and y=10-2t , what is in terms of (rectangular form)?

Possible Answers:

y=8x-2

None of the above

y=8x+2

y=2x-8

y=2x+8

Correct answer:

y=2x-8

Explanation:

Given x=9-t and y=10-2t , let's solve both equations for :

$x=9-t\rightarrowt=9-x$

$y=10-2t\rightarrow t=-\frac{y-10}{2}$

Since both equations equal , let's set them equal to each other and solve for :

$-\frac{y-10}{2}=9-x$

y-10=-2(9-x)

y=-2(9-x)+10

y=-18+2x+10

y=2x-8

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Example Question #61 : Parametric

Given $x=-\frac{1}{2}t$ and y=3t , what is in terms of (rectangular form)?

Possible Answers:

None of the above

y=6x

$y=-\frac{1}{6}x$

$y=\frac{1}{6}x$

y=-6x

Correct answer:

y=-6x

Explanation:

Given $x=-\frac{1}{2}t$ and y=3t , let's solve both equations for :

$x=-\frac{1}{2}t\rightarrow t=-2x$

$y=3t\rightarrow t=\frac{y}{3}$

Since both equations equal , let's set them equal to each other and solve for :

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

y=-6x

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Example Question #62 : Parametric

Given x=6t-7 and y=8t+9 , what is in terms of (rectangular form)?

Possible Answers:

None of the above

$y=\frac{8}{6}(x-7)-9$

$y=\frac{8}{6}(x-7)+9$

$y=\frac{8}{6}(x+7)-9$

$y=\frac{4}{3}(x+7)+9$

Correct answer:

$y=\frac{4}{3}(x+7)+9$

Explanation:

Given x=6t-7 and y=8t+9 , let's solve both equations for :

$x=6t-7\rightarrow t=\frac{x+7}{6}$

$y=8t+9\rightarrow t=\frac{y-9}{8}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y-9}{8}=\frac{x+7}{6}$

$y-9=8\left(\frac{x+7}{6}\right)$

$y=\frac{4}{3}(x+7)+9$

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Example Question #63 : Parametric

When $x=\frac{3}{4}t-5$ and y=7t+4 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{28}{3}(x+5)+4$

None of the above

$y=\frac{28}{3}(x-5)-4$

$y=\frac{28}{3}(x+5)-4$

$y=\frac{28}{3}(x-5)+4$

Correct answer:

$y=\frac{28}{3}(x+5)+4$

Explanation:

Given $x=\frac{3}{4}t-5$ and y=7t+4 , wet's solve both equations for :

$x=\frac{3}{4}t-5\rightarrow t=\frac{4}{3}(x+5)$

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

Since both equations equal , let's set them equal to each other and solve for :

$y-4=(7)(\frac{4}{3})(x+5)$

$y=(7)(\frac{4}{3})(x+5)+4$

$y=\frac{28}{3}(x+5)+4$

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Example Question #64 : Parametric

When $x=-t+\frac{1}{2}$ and y=3t-2 , what is in terms of (rectangular form)?

Possible Answers:

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

$y=3x+\frac{1}{2}$

None of the above

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

$y=-3x+\frac{1}{2}$

Correct answer:

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

Explanation:

Given $x=-t+\frac{1}{2}$ and y=3t-2 , wet's solve both equations for :

$x=-t+\frac{1}{2}\rightarrow t=\frac{1}{2}-x$

$y=3t-2\rightarrow t=\frac{y+2}{3}$

Since both equations equal , let's set them equal to each other and solve for :

$\frac{y+2}{3}=\frac{1}{2}-x$

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

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

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

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

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

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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #55 : Parametric Form

Example Question #56 : Parametric Form

Example Question #51 : Parametric Form

Example Question #52 : Parametric Form

Example Question #53 : Parametric Form

Example Question #60 : Parametric Form

Example Question #61 : Parametric

Example Question #62 : Parametric

Example Question #63 : Parametric

Example Question #64 : Parametric

All Calculus 2 Resources

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