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Calculus 2 : Calculus II

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 #442 : Parametric, Polar, And Vector

The graph of the vector function $r(t)=\left \langle t+6, \sin(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=\sin(x+6)$

$y=\sin(6-x)$

$y=\sin(x-6)$

$y=-\sin(x-6)$

$y=-\sin(x+6)$

Correct answer:

$y=\sin(x-6)$

Explanation:

We can find the graph of $r(t)=\left \langle t+6, \sin(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

x=t+6

t=x-6

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(x-6)$

Report an Error

Example Question #4 : Graphing Vectors

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

Possible Answers:

$y=\sin2\sqrt{x}$

$y=\sin\sqrt{\frac{x}2{}}$

$y=-\sin\sqrt{x}$

$y=\sin(\pm\sqrt{x})$

$y=-\sin2\sqrt{x}$

Correct answer:

$y=\sin(\pm\sqrt{x})$

Explanation:

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

$x=t^{2}$

$t=\pm\sqrt{x}$

We can now use this value to solve for :

$y=\sin(\pm\sqrt{x})$

Report an Error

Example Question #6 : Graphing Vectors

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

Possible Answers:

$y=\cos^{2}\left(\sqrt{\frac{x}{3}}\right)$

$y=-\cos^{2}\left(\sqrt{\frac{x}{3}}\right)$

$y=\cos^{2}(\pm\sqrt{3x})$

$y=-\cos^{2}(\sqrt{3x})$

None of the above

Correct answer:

$y=\cos^{2}(\pm\sqrt{3x})$

Explanation:

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

$x=\frac{t^{2}}{3}$

$t^{2}=3x$

$t=\pm\sqrt{3x}$

We can now use this value to solve for :

$y=\cos^{2}(\pm\sqrt{3x})$

Report an Error

Example Question #7 : Graphing Vectors

The graph of the vector function $r(t)=\left \langle 4-t,\sin(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=-\sin(4+x)$

$y=\sin(x-4)$

$y=\sin(4-x)$

$y=-\sin(4-x)$

$y=\sin(4+x)$

Correct answer:

$y=\sin(4-x)$

Explanation:

We can find the graph of $r(t)=\left \langle 4-t,\sin(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

x=4-t

t=4-x

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(4-x)$

Report an Error

Example Question #1 : Graphing Vectors

The graph of the vector function $r(t)=\left \langle 4t^{2},\cos(t)\right \rangle$ can also be represented by the graph of which of the following functions in rectangular form?

Possible Answers:

$y=-\cos(\sqrt{\frac{4}{x}})$

$y=\cos(\sqrt{\frac{4}{x}})$

$y=4\cos(\sqrt{x})$

$y=\cos(\sqrt{\frac{x}{4}})$

$y=-\cos(\sqrt{\frac{x}{4}})$

Correct answer:

$y=\cos(\sqrt{\frac{x}{4}})$

Explanation:

We can find the graph of $r(t)=\left \langle 4t^{2},\cos(t)\right \rangle$ in rectangular form by mapping the parametric coordinates to Cartesian coordinates (x,y) :

x=4t^2

$\frac{x}{4}=t^2$

$t=\sqrt{\frac{x}{4}}$

We can now use this value to solve for :

$y=\cos(t)$

$y=\cos(\sqrt{\frac{x}{4}})$

Report an Error

Example Question #1 : Graphing Vectors

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

Possible Answers:

$y=-2\sin(\sqrt{3x})$

$y=2\sin(\sqrt{3x})$

$y=2\sin(x\sqrt{3})$

$y=-2\sin(3\sqrt{x})$

$y=2\sin(3\sqrt{x})$

Correct answer:

$y=2\sin(\sqrt{3x})$

Explanation:

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

$x=\frac{t^2}{}3$

3x=t^2

$t=\sqrt{3x}$

We can now use this value to solve for :

$y=2\sin(t)$

$y=2\sin(\sqrt{3x})$

Report an Error

Example Question #101 : Vector

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

Possible Answers:

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

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

$y=\sin(x^{2})$

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

$y=\sin(\frac{1}{x})$

Correct answer:

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

Explanation:

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

$x=\frac{1}{\sqrt{t}}$

$\sqrt{t}=\frac{1}{x}$

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

We can now use this value to solve for :

$y=\sin(t)$

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

Report an Error

Example Question #1 : Vector Form

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

Possible Answers:

$y=-\sin(2x)$

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

$y=2\sin(x)$

$y=\sin(2x)$

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

Correct answer:

$y=\sin(2x)$

Explanation:

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

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

t=2x

We can now use this value to solve for :

$y=\sin(t)$

$y=\sin(2x)$

Report an Error

Example Question #1 : Vector Form

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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

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

We can now use this value to solve for :

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

Report an Error

Example Question #451 : Parametric, Polar, And Vector

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

x=5+2t

2t=x-5

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

We can now use this value to solve for :

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

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

Report an Error

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #442 : Parametric, Polar, And Vector

Example Question #4 : Graphing Vectors

Example Question #6 : Graphing Vectors

Example Question #7 : Graphing Vectors

Example Question #1 : Graphing Vectors

Example Question #1 : Graphing Vectors

Example Question #101 : Vector

Example Question #1 : Vector Form

Example Question #1 : Vector Form

Example Question #451 : Parametric, Polar, And Vector

All Calculus 2 Resources

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