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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #51 : Parametric, Polar, And Vector

Given x=10-t and y=5t+3 , what is in terms of (rectangular form)?

Possible Answers:

y=5-53x

y=5+53x

y=53+5x

None of the above

y=53-5x

Correct answer:

y=53-5x

Explanation:

Given x=10-t and y=5t+3 , let's solve both equations for :

$x=10-t\rightarrow t=10-x$

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

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

$\frac{y-3}{5}=10-x$

y-3=5(10-x)

y=5(10-x)+3

y=50-5x+3

y=53-5x

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Example Question #52 : Parametric, Polar, And Vector

Given x=8-3t and y=7-4t , what is in terms of (rectangular form)?

Possible Answers:

None of the above

$y=\frac{4}{3}(x-8)-7$

$y=\frac{4}{3}(x+8)-7$

$y=\frac{4}{3}(x-8)+7$

$y=\frac{4}{3}(x+8)+7$

Correct answer:

$y=\frac{4}{3}(x-8)+7$

Explanation:

Given x=8-3t and y=7-4t , let's solve both equations for :

$x=8-3t\rightarrow t=-\frac{x-8}{3}$

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

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

$-\frac{y-7}{4}=-\frac{x-8}{3}$

$-(y-7)=-4\left(\frac{x-8}{3}\right)$

$y-7=\frac{4}{3}(x-8)$

$y=\frac{4}{3}(x-8)+7$

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Example Question #53 : Parametric, Polar, And Vector

Given x=7t-4 and y=2t+9 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{2}{7}(x-4})+9$

$y=\frac{2}{7}(x-4})-9$

$y=\frac{2}{7}(x+4})-9$

$y=\frac{2}{7}(x+4})+9$

None of the above

Correct answer:

$y=\frac{2}{7}(x+4})+9$

Explanation:

Given x=7t-4 and y=2t+9 , let's solve both equations for :

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

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

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

$\frac{y-9}{2}=\frac{x+4}{7}$

$y-9=2\left(\frac{x+4}{7}\right)$

$y=\frac{2}{7}(x+4})+9$

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Example Question #54 : Parametric, Polar, And Vector

Given x=2t-11 and y=-t+9 , what is in terms of (rectangular form)?

Possible Answers:

None of the above

$y=9+\frac{x+11}{2}$

$y=9-\frac{x-11}{2}$

$y=9-\frac{x+11}{2}$

$y=9+\frac{x-11}{2}$

Correct answer:

$y=9-\frac{x+11}{2}$

Explanation:

Given x=2t-11 and y=-t+9 , let's solve both equations for :

$x=2t-11\rightarrow t=\frac{x+11}{2}$

$y=-t+9\rightarrow t=9-y$

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

$9-y=\frac{x+11}{2}$

$y=9-\frac{x+11}{2}$

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Example Question #55 : Parametric, Polar, And Vector

Find $\frac{dy}{dx}$ when $x=t^{2}-5$ and $y=t^{3}-5t^{2}-10t-50$ .

Possible Answers:

$\frac{dy}{dx}=3t^{2}-10t-10$

$\frac{dy}{dx}=\frac{3t^{2}-10t-10}{2t}$

$\frac{dy}{dx}=\frac{3t^{2}-10t-10}{t^{2}-5}$

$\frac{dy}{dx}=t^{3}-5t^{2}-10t-50$

$\frac{dy}{dx}=2t$

Correct answer:

$\frac{dy}{dx}=\frac{3t^{2}-10t-10}{2t}$

Explanation:

If x=x(t) and y=y(t) , then we can use the chain rule to define $\frac{dy}{dx}$ as

$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}$ .

We use the power rules

$\frac{d}{dt}(t^{c})=ct^{c-1}$ and $\frac{d}{dt}(t)=1$ where is a constant, the constant rule

$\frac{d}{dt}(c)=0$ where is a constant, and the additive property of derviatives

$\frac{d}{dt}(a+b)=\frac{d}{dt}(a)+\frac{d}{dt}(b)$ .

In this case

$\frac{dy}{dt}=\frac{d}{dt}(t^{3}-5t^{2}-10t-50)=3t^{2}-10t-10$

and

$\frac{dx}{dt}=\frac{d}{dt}(t^{2}-5)=2t$ ,

therefore

$\frac{dy}{dx}=\frac{dy}{dt}/\frac{dx}{dt}=\frac{3t^{2}-10t-10}{2t}$

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Example Question #56 : Parametric, Polar, And Vector

Given x=12-t and y=2t+5 , what is in terms of (rectangular form)?

Possible Answers:

None of the above

y=29+2x

y=29x-2

y=29-2x

y=29x+2

Correct answer:

y=29-2x

Explanation:

Given x=12-t and y=2t+5 , let's solve both equations for :

$x=12-t\rightarrow t=12-x$

$y=2t+5\rightarrow t=\frac{y-5}{2}$

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

$\frac{y-5}{2}=12-x$

y-5=2(12-x)

y=2(12-x)+5

y=24-2x+5

y=29-2x

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Example Question #51 : Parametric Form

Given x=2-3t and y=7-5t , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{5}{3}(x+2)-7$

$y=\frac{5}{3}(x-2)-7$

$y=\frac{5}{3}(x+2)+7$

$y=\frac{5}{3}(x-2)+7$

None of the above

Correct answer:

$y=\frac{5}{3}(x-2)+7$

Explanation:

Given x=2-3t and y=7-5t , let's solve both equations for :

$x=2-3t\rightarrow t=-\frac{x-2}{3}$

$y=7-5t\rightarrow t=-\frac{y-7}{5}$

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

$-\frac{y-7}{5}=-\frac{x-2}{3}$

$y-7=(-5)(-1)\frac{x-2}{3}$

$y-7=\frac{5}{3}(x-2)$

$y=\frac{5}{3}(x-2)+7$

Report an Error

Example Question #51 : Parametric Form

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

Possible Answers:

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

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

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

None of the answers provided

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

Correct answer:

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

Explanation:

Given x=5t-9 and y=2t+7 , let's solve both equations for :

$x=5t-9\rightarrow t=\frac{x+9}{5}$

$y=2t+7\rightarrow t=\frac{y-7}{2}$

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

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

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

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

Report an Error

Example Question #53 : Parametric Form

Given x=8t+1 and y=-4t , what is in terms of (rectangular form)?

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

Explanation:

Given x=8t+1 and y=-4t , let's solve both equations for :

$x=8t+1\rightarrow t=\frac{x-1}{8}$

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

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

$-\frac{y}{4}=\frac{x-1}{8}$

$y=-4\left(\frac{x-1}{8}\right)$

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

Report an Error

Example Question #54 : Parametric Form

Given x=5t+11 and y=-2t-7 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{2}{5}(x-11)+7$

$y=-\frac{2}{5}(x-11)+7$

$y=\frac{2}{5}(x-11)-7$

None of the above

$y=-\frac{2}{5}(x-11)-7$

Correct answer:

$y=-\frac{2}{5}(x-11)-7$

Explanation:

Given x=5t+11 and y=-2t-7 , let's solve both equations for :

$x=5t+11\rightarrow t=\frac{x-11}{5}$

$y=-2t-7\rightarrow t=-\frac{y+7}{2}$

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

$-\frac{y+7}{2}=\frac{x-11}{5}$

$y+7=-2\left(\frac{x-11}{5}\right)$

$y+7=-\frac{2}{5}(x-11)$

$y=-\frac{2}{5}(x-11)-7$

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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 #51 : Parametric, Polar, And Vector

Example Question #52 : Parametric, Polar, And Vector

Example Question #53 : Parametric, Polar, And Vector

Example Question #54 : Parametric, Polar, And Vector

Example Question #55 : Parametric, Polar, And Vector

Example Question #56 : Parametric, Polar, And Vector

Example Question #51 : Parametric Form

Example Question #51 : Parametric Form

Example Question #53 : Parametric Form

Example Question #54 : Parametric Form

All Calculus 2 Resources

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