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

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Question

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

Answer

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 $\small (4\sqrt{2},315^{\circ})$

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Question

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

Answer

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 $\small (-5\sqrt3,5)$

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Question

Convert to polar coordinates.

Answer

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 .

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Question

Convert the following rectangular coordinates to polar coordinates:

Answer

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 $(5, 53.13^\circ)$

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Question

Convert the following rectangular coordinates to polar coordinates:

Answer

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,

$r=\sqrt{5^2+14^2}=\sqrt{221}=14.87$

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

$\theta=\tan^{-1}\frac{14}{5}=70.35^\circ$

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

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Question

Convert the rectangular coordinates to polar coordinates:

Answer

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,

$r=\sqrt{1^2+6^2}=\sqrt{37}=6.08$

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

$\theta=\tan^{-1}\frac{6}{1}=80.54^\circ$

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

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Question

Convert the rectangular coordinates to polar coordinates:

$(\sqrt3, 5)$

Answer

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,

$r=\sqrt{(\sqrt3)^2+5^2}=\sqrt{28}=5.29$

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

$\theta=\tan^{-1}\frac{5}{\sqrt3}=70.89^\circ$

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

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Question

Convert the polar coordinates to rectangular coordinates:

Answer

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

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Question

Convert the polar coordinates to rectangular coordinates:

Answer

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

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Question

Convert the polar coordinates to rectangular coordinates:

Answer

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

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Question

Convert the polar coordinates to rectangular coordinates:

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

Answer

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

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Question

Convert the polar coordinates to rectangular form:

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

Answer

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 $(-5, -5\sqrt3)$

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Question

Convert the polar coordinates to rectangular coordinates:

$(4, \pi)$

Answer

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

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Question

Convert the polar coordinates to rectangular coordinates:

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

Answer

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

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Question

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

Answer

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

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

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Question

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

Round to the nearest hundredth.

Answer

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:

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

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

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Question

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

Answer

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 .

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Question

How could you express in polar coordinates?

Answer

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

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Question

Which polar coordinates conicide with the rectangular point ?

Answer

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

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

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Question

Which coordinates would not describe a point at ?

Answer

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.

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