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Calculus 2 : Graphing Vectors

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Example Questions

Example Question #2 : Vector Form

The graph of the vector function $r(t)=\left \langle \frac{1}{2}t,\sin(t)\right \rangle$ $\displaystyle r(t)=\left \langle \frac{1}{2}t,\sin(t)\right \rangle$can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

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

$y=-\sin(2x)$ $\displaystyle y=-\sin(2x)$

$y=2\sin(x)$ $\displaystyle y=2\sin(x)$

$y=\sin(2x)$ $\displaystyle y=\sin(2x)$

$y=\sin(\frac{x}2{})$ $\displaystyle y=\sin(\frac{x}2{})$

Correct answer:

$y=\sin(2x)$ $\displaystyle y=\sin(2x)$

Explanation:

We can find the graph of $r(t)=\left \langle \frac{1}{2}t,\cos(t)\right \rangle$ $\displaystyle r(t)=\left \langle \frac{1}{2}t,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

$x=\frac{1}{2}t$ $\displaystyle x=\frac{1}{2}t$

t=2x $\displaystyle t=2x$

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$ $\displaystyle y=\sin(2x)$

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Example Question #3 : Vector Form

The graph of the vector function $r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$ $\displaystyle r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=\cos\sqrt{\frac{3}{x-1}}$ $\displaystyle y=\cos\sqrt{\frac{3}{x-1}}$

$y=-\cos\sqrt{\frac{x-1}{3}}$ $\displaystyle y=-\cos\sqrt{\frac{x-1}{3}}$

$y=\cos\sqrt{\frac{x+1}{3}}$ $\displaystyle y=\cos\sqrt{\frac{x+1}{3}}$

$y=-\cos\sqrt{\frac{x+1}{3}}$ $\displaystyle y=-\cos\sqrt{\frac{x+1}{3}}$

$y=\cos\sqrt{\frac{x-1}{3}}$ $\displaystyle y=\cos\sqrt{\frac{x-1}{3}}$

Correct answer:

$y=\cos\sqrt{\frac{x+1}{3}}$ $\displaystyle y=\cos\sqrt{\frac{x+1}{3}}$

Explanation:

We can find the graph of $r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$ $\displaystyle r(t)=\left \langle 3t^{2}-1,\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

$x=3t^{2}-1$ $\displaystyle x=3t^{2}-1$

$x+1=3t^{2}$ $\displaystyle x+1=3t^{2}$

$\frac{x+1}{3}=t^{2}$ $\displaystyle \frac{x+1}{3}=t^{2}$

$t=\sqrt{\frac{x+1}{3}}$ $\displaystyle t=\sqrt{\frac{x+1}{3}}$

We can now use this value to solve for :

$y=\cos\sqrt{\frac{x+1}{3}}$ $\displaystyle y=\cos\sqrt{\frac{x+1}{3}}$

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Example Question #111 : Vector

The graph of the vector function $r(t)=\left \langle 5+2t,\sin^{2}(t)\right \rangle$ $\displaystyle r(t)=\left \langle 5+2t,\sin^{2}(t)\right \rangle$can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=-\sin^{2}(\frac{x+5}{2})$ $\displaystyle y=-\sin^{2}(\frac{x+5}{2})$

$y=\sin^{2}(\frac{x+5}{2})$ $\displaystyle y=\sin^{2}(\frac{x+5}{2})$

$y=\sin^{2}(\frac{x-5}{2})$ $\displaystyle y=\sin^{2}(\frac{x-5}{2})$

$y=\sin^{2}(\frac{2}{x-5})$ $\displaystyle y=\sin^{2}(\frac{2}{x-5})$

$y=-\sin^{2}(\frac{x-5}{2})$ $\displaystyle y=-\sin^{2}(\frac{x-5}{2})$

Correct answer:

$y=\sin^{2}(\frac{x-5}{2})$ $\displaystyle y=\sin^{2}(\frac{x-5}{2})$

Explanation:

We can find the graph of $r(t)=\left \langle 5+2t,\sin^{2}(t)\right \rangle$ $\displaystyle r(t)=\left \langle 5+2t,\sin^{2}(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

x=5+2t $\displaystyle x=5+2t$

2t=x-5 $\displaystyle 2t=x-5$

$t=\frac{x-5}{2}$ $\displaystyle t=\frac{x-5}{2}$

We can now use this value to solve for :

$y=\sin^{2}(t)$ $\displaystyle y=\sin^{2}(t)$

$y=\sin^{2}(\frac{x-5}{2})$ $\displaystyle y=\sin^{2}(\frac{x-5}{2})$

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Example Question #11 : Graphing Vectors

In which quadrant does the vector <8,13> $\displaystyle < 8,13>$ terminate from the origin?

Possible Answers:

$\displaystyle I$

III $\displaystyle III$

$\displaystyle II$

$\displaystyle IV$

Correct answer:

$\displaystyle I$

Explanation:

The vector

<8,13> $\displaystyle < 8,13>$

goes $\displaystyle 8$ units right from the origin and $\displaystyle 13$ units up, terminating in Quadrant I.

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Example Question #12 : Graphing Vectors

In which quadrant does the vector <7,-10> $\displaystyle < 7,-10>$ terminate from the origin?

Possible Answers:

$\displaystyle I$

III $\displaystyle III$

$\displaystyle II$

$\displaystyle IV$

Correct answer:

$\displaystyle IV$

Explanation:

The vector

<7,-10> $\displaystyle < 7,-10>$

goes $\displaystyle 7$ units right from the origin and $\displaystyle 10$ units down, terminating in Quadrant IV.

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Example Question #13 : Graphing Vectors

In which quadrant does the vector <-8,13> $\displaystyle < -8,13>$ terminate from the origin?

Possible Answers:

$\displaystyle II$

$\displaystyle IV$

III $\displaystyle III$

$\displaystyle I$

Correct answer:

$\displaystyle II$

Explanation:

The vector

<-8,13> $\displaystyle < -8,13>$

goes $\displaystyle 8$ units left from the origin and $\displaystyle 13$ units up, terminating in Quadrant II.

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Example Question #14 : Graphing Vectors

In which quadrant does the vector <-1,-1> $\displaystyle < -1,-1>$ terminate from the origin?

Possible Answers:

III $\displaystyle III$

$\displaystyle II$

$\displaystyle IV$

$\displaystyle I$

Correct answer:

III $\displaystyle III$

Explanation:

The vector

<-1,-1> $\displaystyle < -1,-1>$

goes $\displaystyle 1$ unit left from the origin and $\displaystyle 1$ units down, terminating in Quadrant III.

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Example Question #15 : Graphing Vectors

In which quadrant does the vector <8,13> $\displaystyle < 8,13>$ terminate from the origin?

Possible Answers:

Quadrant I

Quadrant IV

Quadrant III

Quadrant II

Correct answer:

Quadrant I

Explanation:

From the origin, the vector <8,13> $\displaystyle < 8,13>$ goes to the right $\displaystyle 8$ units and up $\displaystyle 13$ units, terminating in

Quadrant I.

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Example Question #16 : Graphing Vectors

In which quadrant does the vector <-5,-2> $\displaystyle < -5,-2>$ terminate from the origin?

Possible Answers:

Quadrant IV

Quadrant II

Quadrant I

Quadrant III

Correct answer:

Quadrant III

Explanation:

From the origin, the vector <-5,-2> $\displaystyle < -5,-2>$ goes to the left $\displaystyle 5$ units and down $\displaystyle 2$ units, terminating in

Quadrant III.

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Example Question #17 : Graphing Vectors

In which quadrant does the vector <57,-12> $\displaystyle < 57,-12>$ terminate from the origin?

Possible Answers:

Quadrant I

Quadrant IV

Quadrant II

Quadrant III

Correct answer:

Quadrant IV

Explanation:

From the origin, the vector <57,-12> $\displaystyle < 57,-12>$ goes to the right $\displaystyle 57$ units and down $\displaystyle 12$ units, terminating in

Quadrant IV.

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Calculus 2 : Graphing Vectors

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #2 : Vector Form

Example Question #3 : Vector Form

Example Question #111 : Vector

Example Question #11 : Graphing Vectors

Example Question #12 : Graphing Vectors

Example Question #13 : Graphing Vectors

Example Question #14 : Graphing Vectors

Example Question #15 : Graphing Vectors

Example Question #16 : Graphing Vectors

Example Question #17 : Graphing Vectors

All Calculus 2 Resources

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