Polar Coordinates

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Pre-Calculus › Polar Coordinates

Questions 1 - 10

1

Convert the following rectangular coordinates to polar coordinates:

$(5, 53.13^\circ)$

$(4, 45.12^\circ)$

$(5, 36.87^\circ)$

$(5, 49.57^\circ)$

Explanation

To convert from rectangular coordinates to polar coordinates $(r, \theta )$ :

$r=\sqrt{x^2+y^2}$

$\tan \theta=\frac{y}{x}$

Using the rectangular coordinates given by the question,

$r=\sqrt{3^2+4^2}=\sqrt{25}=5$

$\tan \theta=\frac{4}{3}$

$\theta=\tan^{-1}\frac{4}{3}=53.13^\circ$

The polar coordinates are $(5, 53.13^\circ)$

2

How could you express $\left(10, \frac{\pi}{10}\right)$ in rectangular coordinates?

Round to the nearest hundredth.

Explanation

In order to determine the rectangular coordinates, look at the triangle representing the polar coordinates:

Polar to rectangular a

We can see that both x and y are positive. We can figure out the x-coordinate by using the cosine:

$cos\left(\frac{\pi}{10}\right) = \frac{x}{10}$

$0.951 \approx \frac{x}{10}$ multiply both sides by 10.

We can figure out the y-coordinate by using the sine:

$sin\left(\frac{\pi}{10}\right) = \frac{y}{10}$

$0.309 \approx \frac{y}{10}$

3

Convert the polar coordinates to rectangular coordinates:

$\left(8, \frac{7\pi}{6}\right)$

$\left(-\frac{8\sqrt3}{2}, -4\right)$

$(\sqrt2, \sqrt2)$

$(4\sqrt2, 4\sqrt2)$

$(-4\sqrt2, 4\sqrt2)$

Explanation

To convert polar coordinates $(r, \theta)$ to rectangular coordinates ,

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=8\cos \frac{7\pi}{6}=8\left(-\frac{\sqrt3}{2}\right)=-\frac{8\sqrt3}{2}$

$y=8 \sin \frac{7\pi}{6}=8\left(-\frac{1}{2}\right)=-4$

The rectangular coordinates are $\left(-\frac{8\sqrt3}{2}, -4\right)$

4

Convert the polar coordinates to rectangular form:

$\left(10, \frac{4\pi}{3}\right)$

$(-5, -5\sqrt3)$

$(-5\sqrt3, -5)$

$(5\sqrt3, -5)$

$(5, 5\sqrt3)$

Explanation

To convert polar coordinates $(r, \theta)$ to rectangular coordinates ,

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=10\cos \frac{4\pi}{3}=10\left(-\frac{1}{2}\right)=-5$

$y=10 \sin \frac{4\pi}{3}=10\left(-\frac{\sqrt3}{2}\right)=-5\sqrt3$

The rectangular coordinates are $(-5, -5\sqrt3)$

5

How could you express $\left(10, \frac{\pi}{10}\right)$ in rectangular coordinates?

Round to the nearest hundredth.

Explanation

In order to determine the rectangular coordinates, look at the triangle representing the polar coordinates:

Polar to rectangular a

We can see that both x and y are positive. We can figure out the x-coordinate by using the cosine:

$cos\left(\frac{\pi}{10}\right) = \frac{x}{10}$

$0.951 \approx \frac{x}{10}$ multiply both sides by 10.

We can figure out the y-coordinate by using the sine:

$sin\left(\frac{\pi}{10}\right) = \frac{y}{10}$

$0.309 \approx \frac{y}{10}$

6

Convert to polar coordinates.

$\left(5,\frac{\pi}{2}\right)$

$(\sqrt5,0)$

$(\sqrt5,\pi)$

Explanation

Write the Cartesian to polar conversion formulas.

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

Substitute the coordinate point to the equations to solve for $(r,\theta)$ .

$r=\sqrt{5^2+0^2}=5$

$\theta=tan^-^1\left(\frac{y}{x}\right)=tan^-^1 \left(\frac{0}{5}\right)=0$

Ensuring that is located the first quadrant, the correct angle is zero.

Therefore, the answer is .

7

Convert the polar coordinates to rectangular coordinates:

$\left(5, \frac{2\pi}{3}\right)$

$\left(-\frac{5}{2}, \frac{5\sqrt3}{2}\right)$

$\left(\frac{2}{5}, \frac{\sqrt3}{2}\right)$

$\left(\frac{\sqrt2}{4}, \frac{5}{\sqrt3}\right)$

Explanation

To convert polar coordinates $(r, \theta)$ to rectangular coordinates ,

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=5\cos \frac{2\pi}{3}=5\left(-\frac{1}{2}\right)=-\frac{5}{2}$

$y=5 \sin \frac{2\pi}{3}=5\left(\frac{\sqrt3}{2}\right)=\frac{5\sqrt3}{2}$

The rectangular coordinates are $\left(-\frac{5}{2}, \frac{5\sqrt3}{2}\right)$

8

Convert to polar coordinates.

$\left(5,\frac{\pi}{2}\right)$

$(\sqrt5,0)$

$(\sqrt5,\pi)$

Explanation

Write the Cartesian to polar conversion formulas.

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

Substitute the coordinate point to the equations to solve for $(r,\theta)$ .

$r=\sqrt{5^2+0^2}=5$

$\theta=tan^-^1\left(\frac{y}{x}\right)=tan^-^1 \left(\frac{0}{5}\right)=0$

Ensuring that is located the first quadrant, the correct angle is zero.

Therefore, the answer is .

9

Write the equation in polar form

$r = 2+3 \sin \theta$

$r = 3 \sin \theta$

$r = 4 + 3 \sin \theta$

$r = 4 - 3 \sin \theta$

$r = 2 - 3 \sin \theta$

Explanation

First re-arrange the original equation so that the 4 is factored out on the right side, and put and next to each other:

Make the substitutions and $y = r \sin \theta$ :

$(r^2 - 3 r \sin \theta )^2 = 4r^2$ take the square root of both sides

$r^2 - 3 r \sin \theta = 2 r$ divide both sides by r

$r - 3 \cos \theta = 2$ add $3 \cos \theta$ to both sides

$r = 2 + 3 \cos \theta$

10

Convert from polar form to rectangular form:

$r=4\cos\theta$

Explanation

Start by multiplying both sides by .

$r=4\cos\theta$

$r^2=4r\cos\theta$

Keep in mind that

$x^2+y^2=4r\cos\theta$

Remember that $x=r\cos\theta$

So then,

Now, complete the square.

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