Convert Rectangular Coordinates To Polar Coordinates and vice versa - Pre-Calculus

Card 0 of 96

Convert the rectangular coordinates to polar coordinates:

Tap to see back →

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

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

Using the rectangular coordinates given by the question,

$\tan \theta=\frac{6}{1}$$

The polar coordinates are $(6.08, 80.54^\circ)$

Convert the rectangular coordinates to polar form with an angle between $\small $0^{\circ}$$ and $\small $360^{\circ}$$ .

Tap to see back →

We must first recall that the polar coordinates of a point are expressed in the form $(r,\theta )$ , where is the radius (or the distance from the origin to the point) and $\theta$ is the angle formed between the positive x-axis to the radius.

The radius can be calculated using the distance formula.

$r=$\sqrt{(x_1-x_$2)^2$+(y_1-y_$2)^2$}$$

Our first point is the origin and our second point is the one in question

Therefore, substituting gives us

$\small \small $r=$\sqrt{(0-4)^2$$+(0-(-4))^2$$}$=$\sqrt{(-4)^2$$+(4)^2$}$=$\sqrt{16+16}$=$\sqrt{32}$=4$\sqrt{2}$$

Therefore, our radius is $\small 4$\sqrt{2}$$ .

We can find our angle $\small \theta$ using the formula

$\small tan(\theta )=\frac{y}{x}$$

Substituting the coordinates of our point $\small (4,-4)$ gives

$\small tan(\theta)=\frac{-4}{4}$=-1$

We can use our knowledge or a chart or calculator to determine that the angle that gives this tangent value is $\small $-45^{\circ}$$ or $\small $315^{\circ}$$ . Since we want a postive angle less than $\small $360^{\circ}$$ , we need to go with the latter option.

Therefore, the polar coordinates of our point are $\small $(4$\sqrt{2}$,315^{\circ}$)$

Convert the polar coordinates $\small (10, $150^{\circ}$)$ to rectangular form.

Tap to see back →

We begin by recalling that polar coordinates are expressed in the form $\small (r,\theta)$ , where $\small r$ is the radius (the distance from the origin to the point) and $\small \theta$ is the angle formed between the postive x-axis and the radius.

We can find our x-coordinate and y-coordinate in rectangular form quite easily by keeping in mind two equations.

$\small cos(\theta)=\frac{x}{r}$$ or $\small x=r(cos(\theta))$

$\small sin(\theta)=\frac{y}{r}$$ or $\small y=r(sin(\theta))$

Substituting in both of these gives respectively

$\small $x=(10)cos(150^{\circ}$)=10(-$\frac{\sqrt3}{2}$)=-5\sqrt3$

$\small y=10(sin(150^\circ))=10($\frac{1}{2}$)=5$

Therefore, the rectangular coordinates of our point are $\small (-5\sqrt3,5)$

Convert to polar coordinates.

Tap to see back →

Write the Cartesian to polar conversion formulas.

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

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

$\theta=tan^$-^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 .

Convert the following rectangular coordinates to polar coordinates:

Tap to see back →

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

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

Using the rectangular coordinates given by the question,

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

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

Convert the following rectangular coordinates to polar coordinates:

Tap to see back →

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

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

Using the rectangular coordinates given by the question,

$\tan \theta=\frac{14}{5}$$

The polar coordinates are $(14.87, 70.35^\circ)$

Convert the rectangular coordinates to polar coordinates:

$(\sqrt3, 5)$

Tap to see back →

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

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

Using the rectangular coordinates given by the question,

$\tan \theta=\frac{5}{\sqrt3}$$

The polar coordinates are $(5.29, 70.89^\circ)$

Convert the polar coordinates to rectangular coordinates:

Tap to see back →

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

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=5\cos60=2.5$

The rectangular coordinates are

Convert the polar coordinates to rectangular coordinates:

Tap to see back →

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

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=8\cos 45=5.66$

$y=8 \sin 45=5.66$

The rectangular coordinates are

Convert the polar coordinates to rectangular coordinates:

Tap to see back →

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

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=10\cos 35=8.19$

$y=10\sin 35=5.74$

The rectangular coordinates are

Convert the polar coordinates to rectangular coordinates:

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

Tap to see back →

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

Convert the polar coordinates to rectangular form:

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

Tap to see back →

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 $(-5, -5\sqrt3)$

Convert the polar coordinates to rectangular coordinates:

$(4, \pi)$

Tap to see back →

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

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=4\cos \pi=4(-1)=-4$

$y=4 \sin \pi=4(0)=0$

The rectangular coordinates are

Convert the polar coordinates to rectangular coordinates:

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

Tap to see back →

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}$, -4\right)$

How could you express $\small \small \left(-5, $\frac{\pi}{3}\right)$ in rectangular coordinates?

Tap to see back →

The polar coordinates given have an angle of $\small $\frac{\pi}{3}$$ but a negative radius, so our coordinates are located in quadrant III.

This means x and y are both negative. You can figure out these x and y coordinates using trigonometric ratios, or since the angle is $\small $\frac{\pi}{3}$$ , special right triangles.

The hypotenuse of this triangle is 5, but in the special right triangle it's 2, so we know we're multiplying each side by $\small $\frac{5}{2}$$ .

That makes the x-coordinate or adjacent side be

$\small $\frac{5}{2}$1 = $\frac{5}{2}$$

and the y-coordinate or opposite side be

$\small $\frac{5}{2}$$\sqrt{3}$ = $\frac{5\sqrt3}{2}$$ .

In this case, once again, both are negative, so our answer is

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

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

Round to the nearest hundredth.

Tap to see back →

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

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}$$

Convert $\left(7, $\frac{4\pi}{5}\right)$ into rectangular coordinates.

Tap to see back →

If the angle in the polar coordinates is $\frac{4\pi}{5}$ , that means it's in the second quadrant, and $\frac{\pi}{5}$ away from the x-axis.

This means that the x-coordinate will be negative, and the y-coordinate will be positive:

We can find the x-coordinate using cosine:

$cos\left($\frac{\pi}{5}\right) = $\frac{x}{7}$$

$0.80901 \approx $\frac{x}{7}$$ multiply both sides by 7

, however we know that the x-coordinate is negative, so we'll use -5.66.

We can find the y-coordinate using sine:

$sin\left($\frac{\pi}{5}\right) = $\frac{y}${}7$

$0.58779 \approx $\frac{y}{7}$$ multiply both sides by 7

the coordinates are .

How could you express in polar coordinates?

Tap to see back →

These rectangular coordinates form a right triangle whose side adjacent to the angle is 7.5, and whose opposite side is 4. This means we can find the angle using tangent:

$tan(\theta) = $\frac{4}{7.5}$$

$\theta = $tan^{-1}$($\frac{4}{7.5}$)$

$\theta \approx 0.49$

This would be the angle if these coordinates were in the first quadrant. Since both x and y are negative, this point is in the third. We can adjust the angle by adding $\pi$ , giving us $\theta = 3.63$ .

Now we just need to find the radius - this will be the hypotenuse of the triangle:

take the square root of both sides

So, our polar coordinates are

Which polar coordinates conicide with the rectangular point ?

Tap to see back →

Since the x and y coordinates indicate the same distance, we know that the triangle formed has two angles measuring $\frac{\pi}{4}$ .

The ratio of the legs to the hypotenuse is always $1:\sqrt2$ , so since the legs both have a distance of 6, the hypotenuse/ radius for our polar coordinates is $6\sqrt2$ .

Since the x-coordinate is negative but the y-coordinate is positive, this angle is located in the second quadrant.

$\pi - $\frac{\pi}{4}$ = $\frac{3\pi}{4}$$ , so our angle is $\frac{3\pi}{4}$ .

This makes our coordinates

$\left(6\sqrt2, $\frac{3\pi}{4}\right)$ .

Which coordinates would not describe a point at ?

Tap to see back →

Plotting the point listed gives this triangle:

Using Pythagorean Theorem or just knowing that this is a Pythagorean Triple, we get that the hypotenuse/ radius in polar coordinates is 5.

To find that angle, we can use tangent:

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

$\theta = $tan^{-1}$($\frac{3}{4}$) \approx 0.6435$

That's the angle in quadrant I. This point is in quadrant IV, so we can figure out the angle by subtracting from $2\pi$ :

$2\pi - 0.6435 \approx 5.6397$

We can also find the corresponding angle in quadrant II by subtracting from $\pi$ , and the corresponding angle in quadrant III by adding $\pi$ , giving us these angles:

The point originally converted to polar coordinates is , so we know that works.

If the radius is negative, we want the angle to be at 2.4981, so the point works.

Since our angle 5.6397 is exactly 0.6435 radians below the x-axis, the point will work.

Similarly, the negative version of 2.4981 would be -3.7851, so

works.

The one that does not work has a positive 5 radius and 0.6435 as the angle, which would be located in quadrant I.