Polar Form - AP Calculus BC

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Question

Rewrite in polar form:

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

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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

Graph the equation $r=cos(\theta )$ where $0\leq \theta \leq2\pi$ .

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Answer

At angle the graph as a radius of . As it approaches $\frac{\pi }{2}$ , the radius approaches .

As the graph approaches $\pi$ , the radius approaches .

Because this is a negative radius, the curve is drawn in the opposite quadrant between $\frac{3\pi }{2}$ and $2\pi$ .

Between $\pi$ and $\frac{3\pi }{2}$ , the radius approaches from and redraws the curve in the first quadrant.

Between $\frac{3\pi }{2}$ and $2\pi$ , the graph redraws the curve in the fourth quadrant as the radius approaches from .

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Question

Draw the graph of $r=cos(2\theta )$ from $0\leq \theta \leq2\pi$ .

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Answer

Because this function has a period of $\pi$ , the x-intercepts of the graph happen at a reference angle of $\frac{\pi }{4}$ (angles halfway between the angles of the axes).

Between and $\frac{\pi }{4}$ the radius approaches from .

Between $\frac{\pi }{4}$ and $\frac{\pi }{2}$ , the radius approaches from and is drawn in the opposite quadrant, the third quadrant because it has a negative radius.

From $\frac{\pi }{2}$ to $\frac{3\pi }{4}$ the radius approaches from , and is drawn in the fourth quadrant, the opposite quadrant.

Between $\frac{3\pi }{4}$ and $\pi$ , the radius approaches from .

From $\pi$ and $\frac{5\pi }{4}$ , the radius approaches from .

Between $\frac{5\pi }{4}$ and $\frac{3\pi }{2}$ , the radius approaches from . Because it is a negative radius, it is drawn in the opposite quadrant, the first quadrant.

Then between $\frac{3\pi }{2}$ and $\frac{7\pi }{4}$ the radius approaches from and is draw in the second quadrant.

Finally between $\frac{7\pi }{4}$ and $2\pi$ , the radius approaches from .

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Question

What is the following coordinate in polar form?

Provide the angle in degrees.

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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?

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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?

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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

Convert the following function into polar form:

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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

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

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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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Question

What is the polar form of $y=-7x^{2}$ ?

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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 $y=-7x^{2}$ , then:

$r\sin\theta=-7(r\cos\theta)^{2}$

$r\sin\theta=-7r^{2}\cos^{2}\theta$

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

$\tan\theta=-7r\cos\theta$

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

$\tan\theta\sec\theta=-7r$

$r=-\frac{1}{7}\tan\theta\sec\theta$

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Question

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

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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 ?

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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 , then:

$r\sin\theta=-10(r\cos\theta)+5$

$r\sin\theta+10(r\cos\theta)=5$

$r(\sin\theta+10\cos\theta)=5$

$r=\frac{5}{\sin\theta+10\cos\theta}$

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Question

What is the polar form of $y=\frac{2}{3}x+9$ ?

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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 $y=\frac{2}{3}x+9$ , then:

$r\sin\theta= \frac{2}{3}(r\cos\theta)+9$

$r\sin\theta-\frac{2}{3}(r\cos\theta)=9$

$r(\sin\theta-\frac{2}{3}\cos\theta)=9$

$r=\frac{9}{\sin\theta-\frac{2}{3}\cos\theta}$

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Question

What is the polar form of ?

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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 , then:

$r\sin\theta=3-4(r\cos\theta)$

$r\sin\theta+4(r\cos\theta)=3$

$r(\sin\theta+4\cos\theta)=3$

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

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Question

Find the derivative of the following function:

$r=3\theta+e^\theta$

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Answer

The derivative of a polar function is given by the following:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin(\theta)+r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta)-r\sin(\theta)}$

First, we must find $\frac{dr}{d\theta}=3+e^\theta$

We found the derivative using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

Finally, we plug in the above derivative and the original function into the above formula:

$\frac{(3+e^\theta)\sin(\theta)+(3\theta+e^\theta)\cos(\theta)}{(3+e^\theta)\cos(\theta)-(3\theta+e^\theta)\sin(\theta)}$

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Question

What is the polar form of ?

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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 , then:

$r\sin\theta=2(r\cos \theta)-7$

$r\sin\theta-2(r\cos \theta)=-7$

$r(\sin\theta-2\cos \theta)=-7$

$r=-\frac{7}{\sin\theta-2\cos \theta}$

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Question

What is the derivative of $r=4sin\theta+2$ ?

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Answer

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=4sin\theta+2$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=4cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta}{\frac{dr}{d\theta}cos\theta-rsin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+(4sin\theta+2)cos\theta}{4cos\theta\cos\theta-(4sin\theta+2)sin\theta}$

$\frac{dy}{dx}=\frac{4cos\theta\sin\theta+4sin\theta\cos\theta+2\cos\theta}{4cos\theta\cos\theta-4sin\theta\sin\theta-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(cos^{2}\theta-sin^{2}\theta)-2sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4((1-\sin^{2}\theta)-sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4(1-2\sin^{2}\theta)-2sin\theta}$

$\frac{dy}{dx}=\frac{8cos\theta\sin\theta+2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

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Question

What is the polar form of $y=\frac{3}{9}x^{2}$ ?

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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 $y=\frac{3}{9}x^{2}$ , then:

$r\sin\theta=\frac{3}9{}(r\cos \theta)^{2}$

$r\sin\theta=\frac{3}{9}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{3}9{}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{3}{9}r$

$\tan \theta\sec\theta=\frac{3}{9}r$

$r=\frac{9}{3}\tan \theta\sec\theta$

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

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Question

Convert the following cartesian coordinates into polar form:

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Answer

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have $(r,\theta)$

is the hypotenuse, and $\theta$ is the angle.

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

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

Solution:

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

$r = \sqrt{(-3)^2+11^2 }= \sqrt{(9+121)}$

$r = \sqrt{130}$

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

$\theta = \arctan(-\frac{11}{3}) = -74.74$

$(\sqrt{130},-74.74)$

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Question

Convert the following cartesian coordinates into polar form:

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Answer

Cartesian coordinates have x and y, represented as (x,y). Polar coordinates have $(r,\theta)$

is the hypotenuse, and $\theta$ is the angle.

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

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

Solution:

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

$r=\sqrt{(6)^2 + (-14)^2}$

$r=\sqrt{36+196}$

$r=\sqrt{232}$

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

$\theta=\arctan(\frac{-14}{6})$

$\theta=-66.8$

$(\sqrt{232},-66.8)$

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Question

What is the derivative of $r=2+5\sin\theta$ ?

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Answer

In order to find the derivative $\frac{dy}{dx}$ of a polar equation $r=2+5\sin\theta$ , we must first find the derivative of with respect to $\theta$ as follows:

$\frac{dr}{d\theta}=5\cos\theta$

We can then swap the given values of $\frac{dr}{d\theta}$ and into the equation of the derivative of an expression into polar form:

$\frac{dy}{dx}=\frac{\frac{dr}{d\theta}\sin\theta+r\cos\theta}{\frac{dr}{d\theta}\cos\theta-r\sin\theta}$

$\frac{dy}{dx}=\frac{(5\cos\theta)\sin\theta+(2+5\sin\theta)\cos\theta}{(5\cos\theta)\cos\theta-(2+5\sin\theta)\sin\theta}$

$\frac{dy}{dx}=\frac{5\cos\theta\sin\theta+2\cos\theta+5\sin\theta\cos\theta}{5\cos^{2}\theta-2\sin\theta-5\sin^{2}\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(\cos^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

Using the trigonometric identity $\sin^{2}\theta+\cos^{2}\theta=1$ , we can deduce that $\cos^{2}\theta=1-\sin^{2}\theta$ . Swapping this into the denominator, we get:

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-\sin^{2}\theta-\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5(1-2\sin^{2}\theta)-2\sin\theta}$

$\frac{dy}{dx}=\frac{10\cos\theta\sin\theta+2\cos\theta}{5-10\sin^{2}\theta-2\sin\theta}$

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