Polar Coordinates and Complex Numbers - Pre-Calculus

Card 0 of 20

Question

Convert the polar coordinates to rectangular coordinates:

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

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,

$x=5\cos \frac{3\pi}{2}=5(0)=0$

$y=5 \sin \frac{3\pi}{2}=5(-1)=-5$

The rectangular coordinates are

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Question

Convert to polar coordinates.

Answer

Write the Cartesian to polar conversion formulas.

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

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

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

$\theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{2}{0}\right)=\frac{\pi}{2}$

Since is located in between the first and second quadrant, this is the correct angle.

Therefore, the answer is $\left(2,\frac{\pi}{2}\right)$ .

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Question

Convert to polar form.

Answer

Write the Cartesian to polar conversion formulas.

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

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

$r=\sqrt{(-3)^2+(-3)^2}=\sqrt{18}=3\sqrt2$

$\theta=tan^-^1 \left(\frac{y}{x}\right)=tan^-^1 \left(\frac{-3}{-3}\right)=\frac{\pi}{4}$

Since is not located in between the first quadrant, this is not the correct angle. The correct location of this coordinate is in the third quadrant. Add $\pi$ radians to get the correct angle.

$\theta=\frac{\pi}{4}+\pi=\frac{5\pi}{4}$

Therefore, the answer is $\left(3\sqrt2,\frac{5\pi}{4}\right)$ .

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Question

Convert from polar form to rectangular form:

$r=4\cos\theta$

Answer

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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Question

Convert the polar equation to rectangular form:

$r=8\sin\theta$

Answer

Start by multiplying both sides by .

$r=8\sin\theta$

$r^2=8r\sin\theta$

Remember that

$x^2+y^2=8r\sin\theta$

Keep in mind that $y=r\sin\theta$

So then,

Now, complete the square.

This is a graph of a circle with a radius of and a center at

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Question

Convert the polar equation into rectangular form:

$r=8\sec\theta$

Answer

Remember that $\sec\theta=\frac{1}{\cos\theta}$

So then $r=8\sec\theta$ becomes

$r=\frac{8}{\cos\theta}$

Now, multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=8$

Since $x=r\cos\theta,$ the rectangular form of this equation is

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Question

Convert the polar equation into rectangular form:

$r=16\csc\theta$

Answer

Recall that $\csc\theta=\frac{1}{\sin\theta}$

Now, substitute in that value into the given equation.

$r=\frac{16}{\sin\theta}$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$r\sin\theta=16$

Remember that $y=r\sin\theta$

The rectangular form of this equation is then

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Question

Convert the polar equation into rectangular form:

$r^2=4\sin(2\theta)$

Answer

Start by using the double angle formula for $\sin$ .

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

Substitute that into the equation gives the following:

$r^2=4(2\sin\theta\cos\theta)$

$r^2=8\sin\theta\cos\theta$

Because we need $r\cos\theta$ and $r\sin\theta$ to get and respectively, multiply both sides by .

$(r^2)(r^2)=8(r\sin\theta)(r\cos\theta)$

Now, recall that .

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Question

Convert the polar equation to rectangular form:

$\theta=\frac{5\pi}{6}$

Answer

Start by taking the tangent.

$\tan\theta=\tan\frac{5\pi}{6}$

$\tan\theta=-\frac{\sqrt3}{3}$

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

$\frac{y}{x}=-\frac{\sqrt3}{3}$

$y=-\frac{\sqrt3}{3}x$

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Question

Convert the polar equation to rectangular form:

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

Answer

Start by taking the tangent.

$\tan\theta=\tan\frac{5\pi}{4}$

$\tan\theta=1$

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

$\frac{y}{x}=1$

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Question

Convert the polar equation to rectangular form:

$r=4\sec\theta\tan\theta$

Answer

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Substituting this into the given equation gives

$r=4\frac{1}{\cos\theta}\tan\theta$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=4\tan\theta$

Recall that $x=r\cos\theta$ and that $\tan\theta=\frac{y}{x}$

$x=4\frac{y}{x}$

$y=\frac{1}{4}x^2$

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Question

Convert the polar equation into rectangular form:

$r=2\csc\theta\cot\theta$

Answer

Recall that $\csc\theta=\frac{1}{sin\theta}$ and $\cot\theta=\frac{1}{tan\theta}$

$r=2\frac{1}{\sin\theta}\frac{1}{\tan\theta}$

Multiply both sides by $\sin\theta$

$r\sin\theta=2(\frac{1}{\tan\theta})$

Recall that $\tan\theta=\frac{y}{x}$ and $r\sin\theta=y$

$y=2(\frac{1}{\frac{y}{x}})$

$y=2(\frac{x}{y})$

$y=\sqrt{2x}$

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Question

Convert the polar equation into rectangular form:

$r=-3\sec\theta$

Answer

Recall that $\sec\theta=\frac{1}{\cos\theta}$

Plugging this into the equation gives us

$r=-3\frac{1}{\cos\theta}$

Multiply both sides by $\cos\theta$ to get rid of the fraction.

$r\cos\theta=-3$

Recall that $r\cos\theta=x$

So then the rectangular form of the equation is

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Question

Convert the polar equation into rectangular form.

$6r=\csc\theta$

Answer

Recall that $\csc\theta=\frac{1}{\sin\theta}$

$6r=\frac{1}{\sin\theta}$

Multiply both sides by $\sin\theta$ to get rid of the fraction.

$6r\sin\theta=1$

Recall that $y=r\sin\theta$ .

$y=\frac{1}{6}$

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Question

Convert the polar equation into rectangular form.

$r=2+\cos\theta$

Answer

$r=2+\cos\theta$

Start by multiplying both sides by .

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

Now, isolate the to one side.

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

Square both sides.

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

Recall that and that $x=r\cos\theta$ .

$x^2+y^2=\frac{(x^2+y^2-x)^2}{4}$

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Question

Convert the rectangular equation to polar form:

Answer

Because $x=r\cos\theta$ and $y=r\sin\theta$ , substitute those values in the rectangular form.

$(r\cos\theta+4)^2+(r\sin\theta)^2=16$

Now, expand the equation.

$r^2\cos^2\theta+8r\cos\theta+16+r^2\sin^2\theta=16$

Subtract from both sides.

$r^2\cos^2\theta+8r\cos\theta+r^2\sin^2\theta=0$

$r^2(\cos^2\theta+\sin^2\theta)+8r\cos\theta=0$

Recall the trigonometric identity $\sin^2\theta+\cos^2\theta=1$

$r^2+8r\cos\theta=0$

Factor the equation.

$r(r+8\cos\theta)=0$

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Question

Convert the rectangular equation to polar form:

Answer

Recall that $x=r\cos\theta$

Substitute that into the equation.

$r\cos\theta=9$

Recall that,

$\frac{1}{\cos\theta}=\sec\theta$

$r=\frac{9}{\cos\theta}=9\sec\theta$

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Question

Convert the rectangular equation to polar form:

Answer

Recall that $x=r\cos\theta$ .

Substitute that into the equation.

$r\sin\theta=-5$

Now, isolate the on one side.

$r=\frac{-5}{\sin\theta}$

Recall that,

$\frac{1}{\cos\theta}=\sec\theta$

$r=-5\sec\theta$

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Question

Convert the rectangular equation into polar form.

Answer

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=2$

Recall that,

$\frac{1}{\sin\theta}=\csc\theta$

$r=\frac{2}{\sin\theta}=2\csc\theta$

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Question

Convert the rectangular equation into polar form:

Answer

Recall that $y=r\sin\theta$

Substitute that into the equation.

$r\sin\theta=-4$

Recall that,

$\frac{1}{\sin\theta}=\csc\theta$

$r=\frac{-4}{\sin\theta}=-4\csc\theta$

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