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Calculus 2 : Parametric Form

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Example Questions

Example Question #2 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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Example Question #2 : Parametric Form

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

Possible Answers:

y=4x-12 $\displaystyle y=4x-12$

y=4x-5 $\displaystyle y=4x-5$

y=4x-7 $\displaystyle y=4x-7$

y=4x-2 $\displaystyle y=4x-2$

y=4x-3 $\displaystyle y=4x-3$

Correct answer:

y=4x-5 $\displaystyle y=4x-5$

Explanation:

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

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

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

Since both of these transformations equal $\displaystyle t$, we can set them equal to each other:

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

$\displaystyle y-7=4(x-3)$

$\displaystyle y-7=4x-12$

y=4x-5 $\displaystyle y=4x-5$

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Example Question #1 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

None of the above

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

Correct answer:

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

Explanation:

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

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

Since both equations equal $\displaystyle t$, we can then set them equal to each other and solve for $\displaystyle y$:

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

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

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

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Example Question #1 : Parametric

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

Possible Answers:

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

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

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

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

None of the above

Correct answer:

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

Explanation:

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

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

Since both equations equal $\displaystyle t$, we can then set them equal to each other and solve for $\displaystyle y$:

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

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

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

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

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

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

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Example Question #1 : Parametric Form

Given x=10t-3 $\displaystyle x=10t-3$ and y=7t+8 $\displaystyle y=7t+8$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

$y=\frac{3}{10}({x+7)+8$ $\displaystyle y=\frac{3}{10}({x+7)+8$

None of the above

$y=\frac{8}{10}({x+3)+7$ $\displaystyle y=\frac{8}{10}({x+3)+7$

$y=\frac{7}{10}({x+3)+8$ $\displaystyle y=\frac{7}{10}({x+3)+8$

$y=\frac{8}{10}({x+7)+3$ $\displaystyle y=\frac{8}{10}({x+7)+3$

Correct answer:

$y=\frac{7}{10}({x+3)+8$ $\displaystyle y=\frac{7}{10}({x+3)+8$

Explanation:

In order to find $\displaystyle y$ with respect to $\displaystyle x$, we first isolate $\displaystyle t$ in both equations:

$x=10t-3\rightarrow t=\frac{x+3}{10}$ $\displaystyle x=10t-3\rightarrow t=\frac{x+3}{10}$

$y=7t+8\rightarrow t=\frac{y-8}{7}$ $\displaystyle y=7t+8\rightarrow t=\frac{y-8}{7}$

Since both equations equal $\displaystyle t$, we can then set them equal to each other and solve for $\displaystyle y$:

$\frac{y-8}{7}=\frac{x+3}{10}$ $\displaystyle \frac{y-8}{7}=\frac{x+3}{10}$

$y-8=7(\frac{x+3}{10})$ $\displaystyle y-8=7(\frac{x+3}{10})$

$y=\frac{7}{10}({x+3)+8$ $\displaystyle y=\frac{7}{10}({x+3)+8$

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Example Question #1 : Parametric Form

If x=6t+5 $\displaystyle x=6t+5$ and y=5t+6 $\displaystyle y=5t+6$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

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

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

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

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

None of the above

Correct answer:

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

Explanation:

Given x=6t+5 $\displaystyle x=6t+5$ and y=5t+6 $\displaystyle y=5t+6$, we can find $\displaystyle y$ in terms of $\displaystyle x$ by isolating $\displaystyle t$ in both equations:

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

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

Since both of these transformations equal $\displaystyle t$, we can set them equal to each other:

$\frac{y-6}{5}=\frac{x-5}{6}$ $\displaystyle \frac{y-6}{5}=\frac{x-5}{6}$

${y-6}=5\left(\frac{x-5}{6}\right)$ $\displaystyle {y-6}=5\left(\frac{x-5}{6}\right)$

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

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Example Question #3 : Parametric Form

If x=4t-3 $\displaystyle x=4t-3$ and y=2t+7 $\displaystyle y=2t+7$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

$y=\frac{3}{x+2}+7$ $\displaystyle y=\frac{3}{x+2}+7$

$y=\frac{2}{x+3}+7$ $\displaystyle y=\frac{2}{x+3}+7$

$y=\frac{x+3}{2}+7$ $\displaystyle y=\frac{x+3}{2}+7$

$y=\frac{x+7}{3}+2$ $\displaystyle y=\frac{x+7}{3}+2$

$y=\frac{x+2}{7}+3$ $\displaystyle y=\frac{x+2}{7}+3$

Correct answer:

$y=\frac{x+3}{2}+7$ $\displaystyle y=\frac{x+3}{2}+7$

Explanation:

Given x=4t-3 $\displaystyle x=4t-3$ and y=2t+7 $\displaystyle y=2t+7$, we can find $\displaystyle y$ in terms of $\displaystyle x$ by isolating $\displaystyle t$ in both equations:

$x=4t-3\rightarrow t=\frac{x+3}{4}$ $\displaystyle x=4t-3\rightarrow t=\frac{x+3}{4}$

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

Since both of these transformations equal $\displaystyle t$, we can set them equal to each other:

$\frac{y-7}{2}=\frac{x+3}{4}$ $\displaystyle \frac{y-7}{2}=\frac{x+3}{4}$

$y-7=2\left(\frac{x+3}{4}\right)$ $\displaystyle y-7=2\left(\frac{x+3}{4}\right)$

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

$y=\frac{x+3}{2}+7$ $\displaystyle y=\frac{x+3}{2}+7$

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Example Question #4 : Parametric Form

Given x=9t-2 $\displaystyle x=9t-2$ and y=5t+1 $\displaystyle y=5t+1$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

$y=\frac{1}{9}({x+2}})+5$ $\displaystyle y=\frac{1}{9}({x+2}})+5$

$y=\frac{2}{9}({x+5}})+1$ $\displaystyle y=\frac{2}{9}({x+5}})+1$

$y=\frac{5}{9}({x+2}})+1$ $\displaystyle y=\frac{5}{9}({x+2}})+1$

$y=\frac{9}{2}({x+5}})+1$ $\displaystyle y=\frac{9}{2}({x+5}})+1$

$y=\frac{9}{5}({x+2}})+1$ $\displaystyle y=\frac{9}{5}({x+2}})+1$

Correct answer:

$y=\frac{5}{9}({x+2}})+1$ $\displaystyle y=\frac{5}{9}({x+2}})+1$

Explanation:

Knowing that x=9t-2 $\displaystyle x=9t-2$ and y=5t+1 $\displaystyle y=5t+1$, we can isolate $\displaystyle t$ in both equations as follows:

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

$y=5t+1\rightarrow t=\frac{y-1}{5}$ $\displaystyle y=5t+1\rightarrow t=\frac{y-1}{5}$

Since both of these equations equal $\displaystyle t$, we can set them equal to each other:

$\frac{y-1}{5}=\frac{x+2}{9}$ $\displaystyle \frac{y-1}{5}=\frac{x+2}{9}$

${y-1}=5(\frac{x+2}{9})$ $\displaystyle {y-1}=5(\frac{x+2}{9})$

$y=\frac{5}{9}({x+2}})+1$ $\displaystyle y=\frac{5}{9}({x+2}})+1$

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Example Question #5 : Parametric Form

Given x=t-16 $\displaystyle x=t-16$ and y=8t+3 $\displaystyle y=8t+3$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

$\displaystyle y=8x+128$

y=8x+15 $\displaystyle y=8x+15$

$\displaystyle y=8x+131$

y=8x+13 $\displaystyle y=8x+13$

$\displaystyle y=8x+113$

Correct answer:

$\displaystyle y=8x+131$

Explanation:

Knowing that x=t-16 $\displaystyle x=t-16$ and y=8t+3 $\displaystyle y=8t+3$, we can isolate $\displaystyle t$ in both equations as follows:

$x=t-16\rightarrow t=x+16$ $\displaystyle x=t-16\rightarrow t=x+16$

$y=8t+3\rightarrow t=\frac{y-3}{8}$ $\displaystyle y=8t+3\rightarrow t=\frac{y-3}{8}$

Since both of these equations equal $\displaystyle t$, we can set them equal to each other:

$\frac{y-3}{8}=x+16$ $\displaystyle \frac{y-3}{8}=x+16$

$\displaystyle y-3=8(x+16)$

$\displaystyle y-3=8x+128$

$\displaystyle y=8x+131$

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Example Question #6 : Parametric Form

Given x=14-t $\displaystyle x=14-t$ and y=7t+6 $\displaystyle y=7t+6$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

None of the above

$\displaystyle y=7x-104$

$\displaystyle y=7-104x$

$\displaystyle y=104x-7$

$\displaystyle y=104-7x$

Correct answer:

$\displaystyle y=104-7x$

Explanation:

Since we know x=14-t $\displaystyle x=14-t$ and y=7t+6 $\displaystyle y=7t+6$, we can solve each equation for $\displaystyle t$:

$x=14-t\rightarrow t=14-x$ $\displaystyle x=14-t\rightarrow t=14-x$

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

Since both equations equal $\displaystyle t$, we can set them equal to each other and solve for $\displaystyle y$:

$\frac{y-6}{7}=14-x$ $\displaystyle \frac{y-6}{7}=14-x$

$\displaystyle y-6=7(14-x)$

$\displaystyle y-6=98-7x$

$\displaystyle y=104-7x$

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Calculus 2 : Parametric Form

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Parametric, Polar, And Vector

Example Question #2 : Parametric Form

Example Question #1 : Parametric, Polar, And Vector

Example Question #1 : Parametric

Example Question #1 : Parametric Form

Example Question #1 : Parametric Form

Example Question #3 : Parametric Form

Example Question #4 : Parametric Form

Example Question #5 : Parametric Form

Example Question #6 : Parametric Form

All Calculus 2 Resources

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