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Calculus 2 : Graphing Vectors

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Example Questions

Example Question #951 : Calculus Ii

What is the arclength, from x = 0 to $x=\frac{\pi}{3}$ , of the curve:

$y = ln \left | cos x\right |$

Hint:

$\int (sec (x)) dx= ln \left | sec (x) +tan(x)\right | + C$

Possible Answers:

$ln (\sqrt{2} + \sqrt3)$

$\sqrt{3} - \sqrt2$

$\sqrt{2} + 1$

$ln (\sqrt{3} + {2})$

$ln \left(\sqrt{2} + \frac{1}{2}\right)$

Correct answer:

$ln (\sqrt{3} + {2})$

Explanation:

Arclength is given by the formula:

$Arclength = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}}\right)^2 dx$

We should find dy/dx first, which we find to be:

$\frac{\mathrm{d} }{\mathrm{d} x}(ln(cos(x)) = \frac{1}{cos(x)}\cdot (-sin(x)) = -tan(x)$

Now let's proceed with the integral:

$\int_{0}^{\frac{\pi}{3}} \sqrt{1 + (-tan x)^2}dx = \int_{0}^{\frac{\pi}{3}} \sqrt{(sec^2x)}dx = \int_{0}^{\frac{\pi}{3}} sec(x)dx$

(here we apply the integration described in the hint)

$= \left[ln \left | tan(x) + sec(x)\right |\bigg]_{0}^{\frac{\pi}{3}} = ln (\sqrt{3} + {2})$

which is obtained by evaluating at both boundaries.

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Example Question #2 : Graphing Vectors

Find the area of the polar equation:

$r = \sqrt{cos(\theta )}$

Possible Answers:

$\frac{1}{2}$

$\pi$

Correct answer:

Explanation:

When you plot the graph of $r = \sqrt{cos(\theta )}$ , the bounds are between $\theta = -\frac{\pi }{2}$ and $\theta = \frac{\pi }{2}$ .

Use the area formula for polar equations:

$Area = \int_{\theta_{1} }^{\theta_{2}} \frac{1}{2}r^2 d\theta$

And so we find the area to be:

$\frac{1}{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} (\sqrt{cos( \theta)})^2d\theta = \frac{1}{2}\int_{-\frac{\pi }{2}}^{\frac{\pi }{2}} cos (\theta)d\theta = \frac{1}{2}[sin \theta ]_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$= \frac{1}{2}(1+1) = 1$

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Example Question #3 : Graphing Vectors

The graph of the vector valued function

$\vec{r(t)}=<t^3, sin(t^2)>$

looks most like which of the following scalar functions graphed on a Cartesian coordinate system?

Possible Answers:

$y=sin^3({x})$

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

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

$y=x^{3sin(x^2)}$

Correct answer:

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

Explanation:

To determine this, we must convert our parametric form back into Cartesian form

To do this, we will solve for and reverse substitute.

x=t^3

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

Since we know the parameterization for y, we substitute in for

$y=sin(t^2)=sin((x^\frac{1}{3})^2)= sin((x^\frac{2}{3}))$

Our answer therefore is

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

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Example Question #961 : Calculus Ii

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:

$y=\sin(x-6)$

$y=\sin(x+6)$

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

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

$y=\sin(6-x)$

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 :

$y=\sin(t)$

$y=\sin(x-6)$

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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 :

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

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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 :

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

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Example Question #1 : 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 :

$y=\sin(t)$

$y=\sin(4-x)$

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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:

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

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

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

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

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

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 :

$y=\cos(t)$

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

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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 :

$y=2\sin(t)$

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

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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 :

$y=\sin(t)$

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

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Calculus 2 : Graphing Vectors

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #951 : Calculus Ii

Example Question #2 : Graphing Vectors

Example Question #3 : Graphing Vectors

Example Question #961 : Calculus Ii

Example Question #4 : Graphing Vectors

Example Question #6 : Graphing Vectors

Example Question #1 : Graphing Vectors

Example Question #1 : Graphing Vectors

Example Question #1 : Graphing Vectors

Example Question #101 : Vector

All Calculus 2 Resources

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