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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

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

Example Questions

Example Question #11 : Parametric Form

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

Possible Answers:

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

None of the above

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

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

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

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

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Example Question #21 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

None of the above.

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

Correct answer:

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

Explanation:

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

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

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

Since both equations equal $\displaystyle t$, let's set both equations equal to each other and solve for $\displaystyle y$:

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

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

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

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

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

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

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

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Example Question #21 : Parametric, Polar, And Vector

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

Possible Answers:

None of the above.

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

Since both equations equal $\displaystyle t$, let's set both equations equal to each other and solve for $\displaystyle y$:

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

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

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

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

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

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

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

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Example Question #21 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

None of the above.

Correct answer:

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

Explanation:

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

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

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

Since both equations equal $\displaystyle t$, let's set both equations equal to each other and solve for $\displaystyle y$:

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

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

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

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

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

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

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

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Example Question #23 : Parametric, Polar, And Vector

Convert the following parametric function into rectangular coordinates:

$\displaystyle x=t^2+3, y=2t$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

Possible Answers:

y=3x+23 $\displaystyle y=3x+23$

y=3x-23 $\displaystyle y=3x-23$

y=23x-3 $\displaystyle y=23x-3$

None of the above

y=23x+3 $\displaystyle y=23x+3$

Correct answer:

y=3x+23 $\displaystyle y=3x+23$

Explanation:

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

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

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

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

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

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

$\displaystyle y-5=3(x+6)$

$\displaystyle y=3(x+6)+5$

$\displaystyle y=3x+18+5$

y=3x+23 $\displaystyle y=3x+23$

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Example Question #22 : Parametric, Polar, And Vector

If $x=\frac{1}{t}+7$ $\displaystyle x=\frac{1}{t}+7$ and y=14t-3 $\displaystyle y=14t-3$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

$y=\frac{14}{x-7}-3$ $\displaystyle y=\frac{14}{x-7}-3$

$y=\frac{x-7}{14}+3$ $\displaystyle y=\frac{x-7}{14}+3$

$y=\frac{14}{x-7}+3$ $\displaystyle y=\frac{14}{x-7}+3$

None of the above

$y=\frac{x-7}{14}-3$ $\displaystyle y=\frac{x-7}{14}-3$

Correct answer:

$y=\frac{14}{x-7}-3$ $\displaystyle y=\frac{14}{x-7}-3$

Explanation:

Given $x=\frac{1}{t}+7$ $\displaystyle x=\frac{1}{t}+7$ and y=14t-3 $\displaystyle y=14t-3$, we can find the rectangular form by solving both equations for $\displaystyle t$:

$x=\frac{1}{t}+7\rightarrow t=\frac{1}{x-7}$ $\displaystyle x=\frac{1}{t}+7\rightarrow t=\frac{1}{x-7}$

$y=14t-3\rightarrow t=\frac{y+3}{14}$ $\displaystyle y=14t-3\rightarrow t=\frac{y+3}{14}$

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

$\frac{y+3}{14}=\frac{1}{x-7}$ $\displaystyle \frac{y+3}{14}=\frac{1}{x-7}$

$y+3=\frac{14}{x-7}$ $\displaystyle y+3=\frac{14}{x-7}$

$y=\frac{14}{x-7}-3$ $\displaystyle y=\frac{14}{x-7}-3$

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Example Question #23 : Parametric, Polar, And Vector

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

Possible Answers:

y=8x+5 $\displaystyle y=8x+5$

y=5x+8 $\displaystyle y=5x+8$

y=8x-5 $\displaystyle y=8x-5$

y=5x-8 $\displaystyle y=5x-8$

None of the above

Correct answer:

y=5x-8 $\displaystyle y=5x-8$

Explanation:

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

$x=t+4\rightarrow t=x-4$ $\displaystyle x=t+4\rightarrow t=x-4$

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

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

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

$\displaystyle y-12=5(x-4)$

$\displaystyle y=5(x-4)+12$

$\displaystyle y=5x-20+12$

y=5x-8 $\displaystyle y=5x-8$

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Example Question #26 : Parametric, Polar, And Vector

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

Possible Answers:

None of the above

$y=12\left(\frac{x+12}{5}\right)-5$ $\displaystyle y=12\left(\frac{x+12}{5}\right)-5$

$y=12\left(\frac{x-12}{5}\right)+5$ $\displaystyle y=12\left(\frac{x-12}{5}\right)+5$

$y=12\left(\frac{x+12}{5}\right)+5$ $\displaystyle y=12\left(\frac{x+12}{5}\right)+5$

$y=12\left(\frac{x-12}{5}\right)-5$ $\displaystyle y=12\left(\frac{x-12}{5}\right)-5$

Correct answer:

$y=12\left(\frac{x-12}{5}\right)+5$ $\displaystyle y=12\left(\frac{x-12}{5}\right)+5$

Explanation:

Since we have x=5t+12 $\displaystyle x=5t+12$ and y=12t+5 $\displaystyle y=12t+5$, let's solve each equation for $\displaystyle t$:

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

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

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

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

$y-5=12\left(\frac{x-12}{5}\right)$ $\displaystyle y-5=12\left(\frac{x-12}{5}\right)$

$y=12\left(\frac{x-12}{5}\right)+5$ $\displaystyle y=12\left(\frac{x-12}{5}\right)+5$

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Example Question #24 : Parametric, Polar, And Vector

Given x=3t-11 $\displaystyle x=3t-11$ and y=6t-2 $\displaystyle y=6t-2$, what is $\displaystyle y$ in terms of $\displaystyle x$ (rectangular form)?

Possible Answers:

y=2x-20 $\displaystyle y=2x-20$

y=20x-2 $\displaystyle y=20x-2$

y=2x+20 $\displaystyle y=2x+20$

None of the above

y=20x+2 $\displaystyle y=20x+2$

Correct answer:

y=2x+20 $\displaystyle y=2x+20$

Explanation:

Since we have x=3t-11 $\displaystyle x=3t-11$ and y=6t-2 $\displaystyle y=6t-2$, let's solve each equation for $\displaystyle t$:

$x=3t-11\rightarrow t=\frac{x+11}{3}$ $\displaystyle x=3t-11\rightarrow t=\frac{x+11}{3}$

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

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

$\frac{y+2}{6}=\frac{x+11}{3}$ $\displaystyle \frac{y+2}{6}=\frac{x+11}{3}$

$y+2=6\left(\frac{x+11}{3}\right)$ $\displaystyle y+2=6\left(\frac{x+11}{3}\right)$

$\displaystyle y+2=2(x+11)$

$\displaystyle y+2=2x+22$

y=2x+20 $\displaystyle y=2x+20$

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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 #11 : Parametric Form

Example Question #21 : Parametric, Polar, And Vector

Example Question #21 : Parametric, Polar, And Vector

Example Question #21 : Parametric, Polar, And Vector

Example Question #23 : Parametric, Polar, And Vector

Example Question #21 : Parametric, Polar, And Vector

Example Question #22 : Parametric, Polar, And Vector

Example Question #23 : Parametric, Polar, And Vector

Example Question #26 : Parametric, Polar, And Vector

Example Question #24 : Parametric, Polar, And Vector

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