Polar Form - AP Calculus BC

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Question

What is the polar form of $y=-x^{2}$ ?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=-x^{2}$ , then:

$r\sin\theta=-(r\cos \theta)^{2}$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=-r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=-r$

$\tan \theta\sec\theta=-r$

$r=-\tan \theta\sec\theta$

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Question

What is the polar form of $y=\frac{3}{10}x^{2}$ ?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=\frac{3}{10}x^{2}$ , then:

$r\sin\theta=\frac{3}{10}(r\cos \theta)^{2}$

$r\sin\theta=\frac{3}{10}r^{2}\cos^{2} \theta$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=\frac{3}{10}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{10}r$

$\tan \theta\sec\theta=\frac{3}{10}r$

$r=\frac{10}{3}\tan \theta\sec\theta$

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Question

What is the polar form of $y=4x^{2}$ ?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=4x^{2}$ , then:

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

Dividing both sides by $r\cos\theta$ , we get:

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

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

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

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

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Question

Rewrite the polar equation

$r ^{2} + 1 = \cos \theta$

in rectangular form.

Answer

$r ^{2} + 1 = \cos \theta$

$r\left ( r ^{2} + 1 \right )= r \cos \theta$

$\sqrt{x^{2} + y ^{2}} \left ( {x^{2} + y ^{2}} +1 \right ) = x$

$\sqrt{x^{2} + y ^{2}} =\frac{ x}{{x^{2} + y ^{2}} +1}$

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Question

Rewrite the polar equation

$r = \sin ^{2} \theta$

in rectangular form.

Answer

$r = \sin ^{2} \theta$

$r^{2} \cdot r = r^{2} \sin ^{2} \theta$

$r^{3} =\left ( r \sin \theta \right ) ^{2}$

$\left ( r^{2} \right )^{\frac{3}{2}} =\left ( r \sin \theta \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{\frac{3}{2}} =y ^{2}$

$\left [ \left (x^{2} +y^{2} \right )^{\frac{3}{2}} \right ] ^{2} =\left ( y ^{2} \right ) ^{2}$

$\left (x^{2} +y^{2} \right )^{3} = y ^{4}$

or $y ^{4} = \left (x^{2} +y^{2} \right )^{3}$

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Question

Give the polar form of the equation of the line with intercepts .

Answer

This line has slope $\frac{4-0}{0- (-6)} = \frac{2}{3}$ and -intercept , so its Cartesian equation is $y = \frac{2}{3}x + 4$ .

By substituting, we can rewrite this:

$r \sin \theta = \frac{2}{3}r \cos \theta + 4$

$r \sin \theta - \frac{2}{3}r \cos \theta = 4$

$3 r \sin \theta - 2 r \cos \theta = 12$

$r \left ( 3 \sin \theta - 2 \cos \theta \right )= 12$

$r = \frac{12}{3 \sin \theta - 2 \cos \theta }$

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Question

Rewrite in polar form:

$x^{2} = 3xy + 1$

Answer

$x^{2} = 3xy + 1$

$\left (r \cos \theta \right ) ^{2} = 3\left (r \cos \theta \right )\left (r \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta = 3 r^{2} \cos \theta; \sin \theta \right ) + 1$

$r ^{2} \cos ^{2} \theta - 3 r^{2} \cos \theta; \sin \theta \right ) = 1$

$r ^{2} \left ( \ \cos ^{2} \theta - 3 \cos \theta; \sin \theta \right ) = 1$

$r ^{2} = \frac{1}{\cos ^{2} \theta - 3 \cos \theta; \sin \theta }$

$r =\sqrt{ \frac{1}{\cos ^{2} \theta - 3 \cos \theta; \sin \theta }}$

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Question

Give the rectangular coordinates of the point with polar coordinates

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

Answer

$x = r \cos \theta = -4 \cos \frac{\pi}{3} = -4 \cdot \frac{1}{2} = -2$

$y = r \sin \theta = -4 \sin \frac{\pi}{3} = -4 \cdot \frac{ \sqrt{3}}{2} = -2 \sqrt{3}$

The point will have rectangular coordinates $\left (- 2, -2 \sqrt{3} \right )$ .

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Question

What would be the equation of the parabola $\small y=x^{2}$ in polar form?

Answer

We know $\small y=rsin\theta$ and $\small x=rcos\theta$ .

Subbing that in to the equation $\small y=x^{2}$ will give us $\small rsin\theta=r^{2}cos^{2}\theta$ .

Multiplying both sides by $\small 1/(rcos^{2}\theta)$ gives us

$\small \small \small \small \small r=sin\theta /cos^{2}\theta=tan\theta/cos\theta=tan\theta\cdot sec\theta$ .

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Question

A point in polar form is given as $\left(3,\frac{\pi}{3}\right)$ .

Find its corresponding coordinate.

Answer

To go from polar form to cartesion coordinates, use the following two relations.

$x=rcos(\theta)$

$y=rsin(\theta)$

In this case, our is and our $\theta$ is $\frac{\pi}{3}$ .

Plugging those into our relations we get

$x=\frac{3}{2}$

$y=\frac{3\sqrt3}{2}$ ,

which gives us our coordinate.

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Question

What is the magnitude and angle (in radians) of the following cartesian coordinate?

Give the answer in the format below.

$(r,\theta)$

Answer

Although not explicitly stated, the problem is asking for the polar coordinates of the point . To calculate the magnitude, , calculate the following:

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

To calculate $\theta$ , do the following:

$\theta=tan^{-1}\left(\frac{y}{x}\right)$ in radians. (The problem asks for radians)

$\theta = 0.927$

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Question

What is the following coordinate in polar form?

Provide the angle in degrees.

Answer

To calculate the polar coordinate, use

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

$r=\sqrt{29}$

$\theta=tan^{-1}\left(\frac{y}{x}\right)=68$

However, keep track of the angle here. 68 degree is the mathematical equivalent of the expression, but we know the point (-2,-5) is in the 3rd quadrant, so we have to add 180 to it to get 248.

Some calculators might already have provided you with the correct answer.

$\theta=248$ .

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Question

What is the equation $y=2x^{2}$ in polar form?

Answer

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=2x^{2}$ , then $r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ .

$r\sin \theta=2(r\cos \theta )^{2}=2r^{2}\cos^{2} \theta$ . Dividing both sides by $r\cos \theta$ ,

$\tan \theta=2r\cos \theta$

$\frac{\tan \theta}{\cos \theta}=2r$

$\tan \theta}{\sec \theta=2r$

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

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Question

What is the equation $y=\frac{6}{x}$ in polar form?

Answer

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{6}{x}$ , then $r\sin \theta=\frac{6}{r\cos \theta}$ . Multiplying both sides by $r\cos \theta$ ,

$r^{2}\sin \theta cos \theta=6$

$r^{2}=\frac{6}{\sin \theta cos \theta}$

$r=\sqrt\frac{6}{\sin \theta cos \theta}$

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Question

What is the equation $y=\frac{7}{3}x^{2}$ in polar form?

Answer

We can convert from rectangular form to polar form by using the following identities: $y=r\sin \theta$ and $x=r\cos \theta$ . Given $y=\frac{7}{3}x^{2}$ , then $r\sin \theta=\frac{7}{3}{r^{2}\cos^{2} \theta}$ . Simplifying accordingly,

$\tan \theta=\frac{7}{3}{r\cos\theta}$

$\tan \theta\sec \theta =\frac{7}{3}r$

$\frac{3}{7}\tan \theta\sec \theta =r$

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Question

Given and , what is in terms of (rectangular form)?

Answer

Knowing that and , we can isolate in both equations as follows:

$x=3t+15\rightarrow t=\frac{x-15}{3}$

$y=15t-6\rightarrow t=\frac{y+6}{15}$

Since both of these equations equal , we can set them equal to each other:

$\frac{y+6}{15}=\frac{x-15}{3}$

$y+6=15(\frac{x-15}{3})$

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Question

Convert the following function into polar form:

Answer

The following formulas were used to convert the function from polar to Cartestian coordinates:

$x=r\cos(\theta), y=r\sin(\theta), x^2+y^2=r^2$

Note that the last formula is a manipulation of a trignometric identity.

Simply replace these with x and y in the original function.

$f(x)=2r^2+5rcos(\theta)+2$

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Question

Convert from rectangular to polar form:

Answer

To convert from rectangular to polar form, we must use the following formulas:

$x=r\cos(\theta), y=r\sin(\theta)$

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

It is easier to find our angle $\theta$ first, which is done by plugging in our x and y into the second formula:

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

Find the angle by taking the inverse of the function:

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

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

Now find r by plugging in our angle and x and y into the first formula, and solving for r:

$2=r\cos\left(\frac{\pi}{4}\right)$

$r=2\sqrt{2}$

Our final answer is reported in polar coordinate form $(r, \theta)$ :

$\left(2\sqrt{2}, \frac{\pi}{4}\right)$

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Question

What is the equation $y=3x^{2}$ in polar form?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=3x^{2}$ , then:

$r\sin\theta=3(r\cos \theta)^{2}$

$r\sin\theta=3r^{2}\cos^{2} \theta$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=3r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=3r$

$\tan \theta\sec\theta=3r$

$r=\frac{1}{3}\tan \theta\sec\theta$

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Question

What is the equation $y=-\frac{1}{2}x^{2}$ in polar form?

Answer

We can convert from rectangular to polar form by using the following trigonometric identities: $y=r\sin\theta$ and $x=r\cos\theta$ . Given $y=-\frac{1}{2}x^{2}$ , then:

$r\sin\theta=-\frac{1}{2}(r\cos \theta)^{2}$

$r\sin\theta=-\frac{1}{2}r^{2}\cos^{2} \theta$

Dividing both sides by $r\cos\theta$ , we get:

$\tan \theta=-\frac{1}{2}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=-\frac{1}{2}r$

$\tan \theta\sec\theta=-\frac{1}{2}r$

$r=-2\tan \theta\sec\theta$

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