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Calculus 2 : Parametric Form

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Example Questions

Example Question #11 : Parametric Form

Given x=2t-7 and y=4t+6 , what is in terms of (rectangular form)?

Possible Answers:

y=2x+20

y=8x+2

y=6x+8

y=6x+2

y=2x+6

Correct answer:

y=2x+20

Explanation:

Since we know x=2t-7 and y=4t+6 , we can solve each equation for :

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

$y=4t+6\rightarrow t=\frac{y-6}{4}$

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

$\frac{y-6}{4}=\frac{x+7}{2}$

$y-6=4\left(\frac{x+7}{2}\right)$

y-6=2(x+7)

y-6=2x+14

y=2x+20

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Example Question #12 : Parametric Form

Given x=16t-15 and y=8t+1 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{x+15}{2}-1$

$y=\frac{x+2}{15}+1$

$y=\frac{x+15}{2}+1$

$y=\frac{x+2}{15}-1$

None of the above

Correct answer:

$y=\frac{x+15}{2}+1$

Explanation:

Since we know x=16t-15 and y=8t+1 , we can solve each equation for :

$x=16t-15\rightarrow t=\frac{x+15}{16}$

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

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

$\frac{y-1}{8}=\frac{x+15}{16}$

$y-1=8\left(\frac{x+15}{16} \right)$

$y-1=\frac{x+15}{2}$

$y=\frac{x+15}{2}+1$

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Example Question #13 : Parametric Form

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

Possible Answers:

y=2x+7

y=2x-7

y=7x-2

y=7x+2

None of the above

Correct answer:

y=2x+7

Explanation:

We know that x=3t-2 and y=6t+3 , so we can solve both equations for :

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

$y=6t+3\rightarrow t=\frac{y-3}{6}$

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

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

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

y-3=2(x+2)

y=2(x+2)+3

y=2x+4+3

y=2x+7

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Example Question #14 : Parametric Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

We know that x=4t-1 and y=t+4 , so we can solve both equations for :

$x=4t-1\rightarrow t=\frac{x+1}{4}$

$y=t+4\rightarrow t=y-4$

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

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

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

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Example Question #15 : Parametric Form

Given x=10t+7 and y=6t-9 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{3}{5}(x-7)+9$

$y=\frac{3}{5}(x+7)-9$

$y=\frac{3}{5}(x-7)-9$

None of the above

$y=\frac{3}{5}(x+7)+9$

Correct answer:

$y=\frac{3}{5}(x-7)-9$

Explanation:

We know that x=10t+7 and y=6t-9 , so we can solve both equations for :

$x=10t+7\rightarrow t=\frac{x-7}{10}$

$y=6t-9\rightarrow t=\frac{y+9}{6}$

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

$\frac{y+9}{6}=\frac{x-7}{10}$

$y+9=6(\frac{x-7}{10})$

$y=6(\frac{x-7}{10})-9$

$y=\frac{3}{5}(x-7)-9$

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Example Question #21 : Parametric, Polar, And Vector

Given x=6t+13 and y=10t+16 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{3}{5}x-\frac{17}{3}$

$y=\frac{3}{5}x+\frac{17}{3}$

$y=\frac{5}{3}x+\frac{17}{3}$

None of the above.

$y=\frac{5}{3}x-\frac{17}{3}$

Correct answer:

$y=\frac{5}{3}x-\frac{17}{3}$

Explanation:

We know x=6t+13 and y=10t+16 , so we can solve both equations for :

$x=6t+13\rightarrow t=\frac{x-13}{6}$

$y=10t+16\rightarrow t=\frac{y-16}{10}$

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

$\frac{y-16}{10}=\frac{x-13}{6}$

$y-16=10\left(\frac{x-13}{6}\right)$

$y-16=5\left(\frac{x-13}{3}\right)$

$y=5\left(\frac{x-13}{3}\right)+16$

$y=\frac{5}{3}x-\frac{65}{3}+16$

$y=\frac{5}{3}x-\frac{65}{3}+\frac{48}{3}$

$y=\frac{5}{3}x-\frac{17}{3}$

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Example Question #23 : Parametric

Given x=7t-6 and y=5t+6 , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{5}{7}x+\frac{72}{7}$

$y=\frac{5}{7}x+\frac{7}{72}$

$y=\frac{7}{5}x+\frac{7}{72}$

$y=\frac{7}{5}x+\frac{72}{7}$

None of the above.

Correct answer:

$y=\frac{5}{7}x+\frac{72}{7}$

Explanation:

We know x=7t-6 and y=5t+6 , so we can solve both equations for :

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

$y=5t+6\rightarrow t=\frac{y-6}{5}$

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

$\frac{y-6}{5}=\frac{x+6}{7}$

7({y-6})=5(x+6)

$y-6=\frac{5}{7}(x+6)$

$y=\frac{5}{7}(x+6)+6$

$y=\frac{5}{7}x+\frac{30}{7}+6$

$y=\frac{5}{7}x+\frac{30}{7}+\frac{42}{7}$

$y=\frac{5}{7}x+\frac{72}{7}$

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Example Question #22 : Parametric, Polar, And Vector

Given $x=\frac{2}{t}+6$ and $y=\frac{3}{t}-9$ , what is in terms of (rectangular form)?

Possible Answers:

$y=\frac{3}2x-18$

$y=\frac{2}3x-18$

None of the above.

$y=\frac{3}2x+18$

$y=\frac{2}3x+18$

Correct answer:

$y=\frac{3}2x-18$

Explanation:

We know $x=\frac{2}{t}+6$ and $y=\frac{3}{t}-9$ , so we can solve both equations for :

$x=\frac{2}{t}+6\rightarrow t=\frac{2}{x-6}$

$y=\frac{3}{t}-9\rightarrow t=\frac{3}{y+9}$

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

$\frac{3}{y+9}=\frac{2}{x-6}$

3(x-6)={2}(y+9)

$\frac{3}2{}(x-6)=y+9$

$\frac{3}2{}(x-6)-9=y$

$y=\frac{3}2x-\frac{18}{2}-9$

$y=\frac{3}2x-9-9$

$y=\frac{3}2x-18$

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Example Question #23 : Parametric, Polar, And Vector

Convert the following parametric function into rectangular coordinates:

x=t^2+3, y=2t

Possible Answers:

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

$y=\sqrt{x-3}$

$y=\pm 2\sqrt{x-3}$

$y=\sqrt{4x-3}$

Correct answer:

$y=\pm 2\sqrt{x-3}$

Explanation:

To eliminate the parameter, we can solve for t in terms of y easiest:

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

Next, substitute all of the t's in the equation for x with what we defined above:

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

To finish, subtract 3, multiply by 4 and take the square root of both sides. We need plus or minus because both positive and negative squared give a positive result.

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Example Question #21 : Parametric, Polar, And Vector

If x=4t-6 and y=12t+5 , what is in terms of (rectangular form)?

Possible Answers:

None of the above

y=23x+3

y=3x+23

y=23x-3

y=3x-23

Correct answer:

y=3x+23

Explanation:

Given x=4t-6 and y=12t+5 , we can find the rectangular form by solving both equations for :

$x=4t-6\rightarrow t=\frac{x+6}{4}$

$y=12t+5\rightarrow t=\frac{y-5}{12}$

Since both equations equal , we can set them equal to each other:

$\frac{y-5}{12}=\frac{x+6}{4}$

${y-5}=12(\frac{x+6}{4})$

y-5=3(x+6)

y=3(x+6)+5

y=3x+18+5

y=3x+23

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Calculus 2 : Parametric Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #11 : Parametric Form

Example Question #12 : Parametric Form

Example Question #13 : Parametric Form

Example Question #14 : Parametric Form

Example Question #15 : Parametric Form

Example Question #21 : Parametric, Polar, And Vector

Example Question #23 : Parametric

Example Question #22 : Parametric, Polar, And Vector

Example Question #23 : Parametric, Polar, And Vector

Example Question #21 : Parametric, Polar, And Vector

All Calculus 2 Resources

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