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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #91 : Vectors

What is the vector form of -i+2j+10k ?

Possible Answers:

$\left \langle 1,2,-10\right \rangle$

$\left \langle 1,-2,10\right \rangle$

$\left \langle 1,2,10\right \rangle$

$\left \langle -1,2,10\right \rangle$

$\left \langle 1,-2,-10\right \rangle$

Correct answer:

$\left \langle -1,2,10\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for -i+2j+10k , we can derive the vector form $\left \langle -1,2,10\right \rangle$ .

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Example Question #92 : Vector Form

What is the vector form of i+k ?

Possible Answers:

$\left \langle 1,1,0\right \rangle$

$\left \langle 0,1,1\right \rangle$

$\left \langle -1,0,1\right \rangle$

$\left \langle 1,0,-1\right \rangle$

$\left \langle 1,0,1\right \rangle$

Correct answer:

$\left \langle 1,0,1\right \rangle$

Explanation:

In order to derive the vector form, we must map the , , -coordinates to their corresponding , , and coefficients.

That is, given $a_{1}i+a_{2}j+a_{3}k$ , the vector form is $a=\left \langle a_{1},a_{2},a_{3}\right \rangle$ .

So for i+k , we can derive the vector form $\left \langle 1,0,1\right \rangle$ .

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Example Question #11 : Vectors & Spaces

What is the vector form of j+8k ?

Possible Answers:

$\left \langle 1,0,8\right \rangle$

None of the above

$\left \langle 1,8,0\right \rangle$

$\left \langle 0,8,1\right \rangle$

$\left \langle 0,1,8\right \rangle$

Correct answer:

$\left \langle 0,1,8\right \rangle$

Explanation:

Given j+8k , we need to map the , , and coefficients back to their corresponding , , and -coordinates.

Thus the vector form of j+8k is $\left \langle 0,1,8\right \rangle$ .

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Example Question #11 : Vectors

Express 4i-9j in vector form.

Possible Answers:

$\left \langle 4,-9,0\right \rangle$

$\left \langle 4,0,-9\right \rangle$

$\left \langle 0,4,-9\right \rangle$

$\left \langle 4,9,0\right \rangle$

$\left \langle 4,-9\right \rangle$

Correct answer:

$\left \langle 4,-9,0\right \rangle$

Explanation:

In order to express 4i-9j in vector form, we must use the coefficients of i, j, and to represent the -, -, and -coordinates of the vector.

Therefore, its vector form is

$\left \langle 4,-9,0\right \rangle$ .

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Example Question #441 : Parametric, Polar, And Vector

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:

$\sqrt{3} - \sqrt2$

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

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

$\sqrt{2} + 1$

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

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 #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 :

$y=\sin(t)$

$y=\sin(x-6)$

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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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Calculus 2 : Parametric, Polar, and Vector

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #91 : Vectors

Example Question #92 : Vector Form

Example Question #11 : Vectors & Spaces

Example Question #11 : Vectors

Example Question #441 : Parametric, Polar, And Vector

Example Question #2 : Graphing Vectors

Example Question #3 : Graphing Vectors

Example Question #442 : Parametric, Polar, And Vector

Example Question #4 : Graphing Vectors

Example Question #6 : Graphing Vectors

All Calculus 2 Resources

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