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Calculus 2 : Parametric, Polar, and Vector

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Example Questions

Example Question #321 : Parametric, Polar, And Vector

Given the Cartesian coordinate $\left ( -1,1\right )$ , what is $\theta$ in the polar form $\left ( r,\theta\right )$ ?

Possible Answers:

$\frac{7\pi}{6}$

$\frac{5\pi}{4}$

$\frac{3\pi}{4}$

$\frac{\pi}{4}$

$-\frac{\pi}{4}$

Correct answer:

$\frac{3\pi}{4}$

Explanation:

The formula to find theta in polar form is:

$\theta= tan^-^1\left(\frac{y}{x}\right)$

Plug in the Cartesian coordinates into the equation.

$\theta= tan^-^1\left(\frac{1}{-1}\right) = tan^-^1(-1) =-\frac{\pi}{4}$

However, this angle is located in the fourth quadrant and is not in the right quadrant. Add $\pi$ radians to get the correct angle since the Cartesian coordinate given is located in the second quadrant.

$-\frac{\pi}{4}+\pi = \frac{3\pi}{4}$

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Example Question #321 : Parametric, Polar, And Vector

Tom is scaling a mathematical mountain. The mountain's profile can be described by $\small \small \small -x^4+2x^3-x^2-x+3$ between $\small x=-1$ and $\small x\approx1.68$ . Tom climbs from $\small x=-1$ to the peak of the mountain. How far did he climb? You'll need an equation solver for certain parts of the problem. Round everything to the nearest hundredth.

Possible Answers:

$\small 3.15$

$\small 1.86$

$\small 7.18$

$\small 6.52$

$\small 3.33$

Correct answer:

$\small 3.33$

Explanation:

This is a two step problem. First step is to maximize the function to find the peak of the mountain. The next step is to use the arc length formula to find the distance he climbed.

To maximize, we'll take a derivative and set it equal to zero.

$\small \small \frac{\mathrm{d} }{\mathrm{d} x}(-x^4+2x^3-x^2-x+3)=-4x^3+6x^2-2x-1$ .

Setting this equal to zero, we get $\small \small x\approx-0.26$ .

The derivative is positive prior to this value and negative after, so it is a max. We now must take the arc length from $\small x=-1$ to $\small x\approx-0.26$ .

The formula for arc length is

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

For this case, the integral becomes

$\small \small \int_{-1}^{-.26}\sqrt{1+(-4x^3+6x^2-2x-1)^2}dx$ .

This will give us $\small 3.33$ . No units were given in the problem, so leaving the answer unitless is acceptable.

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Example Question #322 : Parametric, Polar, And Vector

Find the length of the polar valued function $r=cos(\theta)$ from r=2 to r=5 .

Possible Answers:

-sin(5)+sin(2)

Correct answer:

Explanation:

Recall the formula for length in polar coordinates is given by

$L=\int_{a}^{b}\sqrt{r^2+\left(\frac{dr}{d \theta}\right)^2} d \theta$ .

We were given the formula

$r=cos(\theta)$ .

In our case, this translates to

$\\L=\int_{2}^{5}\sqrt{cos^2(\theta )+((-sin(\theta ))^2}d \theta\\ \\= \int_{2}^{5}\sqrt{1}d \theta\\ \\=3$ .

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Example Question #1 : Polar Calculations

Convert $(-1, \frac{\pi}{4})$ to Cartesian coordinates.

Possible Answers:

$( -\sqrt2,-\sqrt2)$

$( \frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

$( -\frac{\sqrt2}{2},\frac{\sqrt2}{2})$

$( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

$( \frac{\sqrt2}{2},\frac{\sqrt2}{2})$

Correct answer:

$( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

Explanation:

Write the formulas to convert from polar to Cartesian.

$x=rcos(\theta)$

$y=rsin(\theta)$

The and $\theta$ values are known. Substitute both into each equation and solve for and .

$x=(-1)cos(\frac{\pi}{4}) = -\frac{\sqrt2}{2}$

$y=(-1)sin(\frac{\pi}{4}) = -\frac{\sqrt2}{2}$

The Cartesian coordinates are: $( -\frac{\sqrt2}{2}, -\frac{\sqrt2}{2})$

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Example Question #324 : Parametric, Polar, And Vector

Convert $r=-4cos(\theta)$ to Cartesian coordinates and find the coordinates of the center.

Possible Answers:

(-2,0)

$(-\frac{1}{2},0)$

(-4,0)

$(-\frac{1}{4},0)$

$(-\frac{1}{8},0)$

Correct answer:

(-2,0)

Explanation:

Write the conversion formulas.

x^2+y^2=r^2

$x=rcos(\theta)$

Notice the r^2 term. If we multiplied by on both sides of the $r=-4cos(\theta)$ equation, we will get:

$r^2=-4rcos(\theta)$

r^2=-4x

Substitute this back into the first equation.

x^2+y^2=-4x

Add on both sides.

x^2+4x+y^2=0

Complete the square with the terms.

(x^2+4x+4)+y^2=4

This would then become:

(x+2)^2+y^2 = 4

This is a circle centered at (-2,0) with a radius of 4.

The answer is: (-2,0)

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Example Question #9 : Polar Calculations

Convert $\Big (3, \frac{\pi}{6} \Big)$ to Cartesian coordinates.

Possible Answers:

$\Bigg( \frac{\sqrt3}{2}, \frac{1}{2} \Bigg)$

$\Bigg (\frac{3 \sqrt3}{2} , \frac{3}{2} \Bigg)$

$\big(\sqrt3, 3 \big)$

$\Bigg(\frac{3\sqrt2}{2}, \frac{3\sqrt2}{2}\Bigg)$

Correct answer:

$\Bigg (\frac{3 \sqrt3}{2} , \frac{3}{2} \Bigg)$

Explanation:

When converting from polar to Cartesian coordinates, we must use the formulas

$x = r\cos \theta \; ;\; y=r\sin \theta$

the values r = 3 and $\theta = \frac{\pi}{6}$ are given, so we can calculate that

$x = 3\cdot \cos(\frac{\pi}{6}) = \frac{3\sqrt3}{2}$

and $y = 3\cdot \sin(\frac{\pi}{6}) = \frac{3}{2}$

So the Cartesian coordinate form is

$\Bigg( \frac{3\sqrt3}{2} , \frac{3}{2} \Bigg)$

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Example Question #10 : Polar Calculations

Determine the equation in polar coordinates of y=2x

Possible Answers:

$\theta=tan^{-1}(2)$

$\theta=tan^{-1}(\frac{1}{2})$

$\theta=tan(2)$

$\theta=2$

Correct answer:

$\theta=tan^{-1}(2)$

Explanation:

y=2x can be immediately transformed into polar form by:

$rsin(\theta)=2rcos(\theta)$ .

Dividing by ,

$sin(\theta)=2cos(\theta)$

Dividing both sides by $cos(\theta)$

$tan(\theta)=2$

$\theta=tan^{-1}(2)$

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Example Question #841 : Calculus Ii

Possible Answers:

$r=\sqrt{}2$

$\Theta=\frac{4}{\Pi}$

r=2

$\Theta=\frac{5\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{\Pi}{4}$

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

r=2

$\Theta=\frac{\Pi}{4}$

Correct answer:

$r=\sqrt{2}$

$\Theta=\frac{5\Pi}{4}$

Explanation:

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Example Question #12 : Polar Calculations

Determine how many points of intersection exist for the curves

$r=cos(2\Theta)$

and

$r = \frac{1}{2}$ .

Possible Answers:

Correct answer:

Explanation:

Solving the equations $r=cos(2\Theta )$ and $r=\frac{1}{2}$ yields $cos(2\Theta)=\frac{1}{2}$ .

Hence,

$2\Theta=\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}.$

Therefore, the values of $\Theta$ between and $2\pi$ that satisfy both equations are:

$\Theta=\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}.$

From this, it can be deduced that there are four points of intersection between the given curves:

$(\frac{1}{2},\frac{\pi}{6}),(\frac{1}{2},\frac{5\pi}{6}),(\frac{1}{2},\frac{7\pi}{6}),(\frac{1}{2},\frac{11\pi}{6}).$

However, an identical graph to $r=\frac{1}{2}$ in polar coordinates is $r=-\frac{1}{2}$ , since these two equations describe the same circle with a radius $\frac{1}{2}$ units long. Therefore, the equations $r=cos(2\Theta)$ and $r=-\frac{1}{2}$ must also be solved to yield the remaining points of intersection:

$cos(2\Theta)=-\frac{1}{2}$ ,

$\Rightarrow 2\Theta=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3}, \frac{10\pi}{3}.$

$\Rightarrow \Theta=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}.$

From this, it can be deduced that there are four other points of intersection between the given curves:

$(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{2\pi}{3}),(-\frac{1}{2},\frac{4\pi}{3}),(-\frac{1}{2},\frac{5\pi}{3}).$

Hence, there are eight total points of intersection between the curves $r=cos(2\Theta)$ and $r=\frac{1}{2}$ .

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Example Question #13 : Polar Calculations

Convert $r=-3cos(\theta)$ to Cartesian coordinates

Possible Answers:

x+y=-3x

x^2+y^2=-3y

x+y=-3y

x^2+y^2=-3x

x^2+y^2=y

Correct answer:

x^2+y^2=-3x

Explanation:

we are given $r=-3cos(\theta)$ and we know that

$x=rcos(\theta )$

$y=rsin(\theta )$

r^2=x^2+y^2

we have r and 3cos(theta). Multiplying each side of the equation by r would give us

$r^2=-3rcos(\theta )$

substitute out the parts we know from the formulas above

x^2+y^2=-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 #321 : Parametric, Polar, And Vector

Example Question #321 : Parametric, Polar, And Vector

Example Question #322 : Parametric, Polar, And Vector

Example Question #1 : Polar Calculations

Example Question #324 : Parametric, Polar, And Vector

Example Question #9 : Polar Calculations

Example Question #10 : Polar Calculations

Example Question #841 : Calculus Ii

Example Question #12 : Polar Calculations

Example Question #13 : Polar Calculations

All Calculus 2 Resources

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