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AP Calculus BC : Polar Form

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #24 : Polar

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

Possible Answers:

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

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

None of the above

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

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

Correct answer:

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

Explanation:

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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Example Question #21 : Functions, Graphs, And Limits

What is the polar form of y=2x-7 ?

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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=2x-7 , 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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Example Question #22 : Functions, Graphs, And Limits

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #23 : Functions, Graphs, And Limits

Convert the following cartesian coordinates into polar form: (-3,11)

Possible Answers:

$(\sqrt{130},74.74)$

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

(8,74.74)

(8,-74.74)

Correct answer:

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

Explanation:

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:

x=-3

y=11

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

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

$r = \sqrt{130}$

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

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

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

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Example Question #24 : Functions, Graphs, And Limits

Convert the following cartesian coordinates into polar form: (6,-14)

Possible Answers:

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

$(\sqrt{210},-42.8)$

$(\sqrt{225},54.8)$

$(\sqrt{230},62.8)$

Correct answer:

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

Explanation:

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:

x=6

y=-14

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

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

$r=\sqrt{36+196}$

$r=\sqrt{232}$

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

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

$\theta=-66.8$

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

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Example Question #11 : Polar Form

Calculate the polar form hypotenuse of the following cartesian equation: y^2=100x^8

Possible Answers:

$\sqrt[3]{\frac{\tan(\theta)}{\cos(\theta)}}$

$\frac{\tan(\theta)}{10\cos^3(\theta)}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos(\theta)}}$

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Correct answer:

$\sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

Explanation:

In a cartesian form, the primary parameters are and . In polar form, they are and $\theta$

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

2 things to know when converting from Cartesian to polar.

$x= r\cos(\theta)$

$y= r\sin(\theta)$

You want to calculate the hypotenuse,

Solution:

y^2=100x^8

$y=10x^4=r\sin(\theta)$

$y=10(r\cos(\theta))^4$

$y=10r^4\cos^4(\theta) = r\sin(\theta)$

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

$r = \sqrt[3]{\frac{\tan(\theta)}{10\cos^3(\theta)}}$

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Example Question #1 : Graphing Polar Form

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

Possible Answers:

R_cosx_1

R_cos2x

R_sinx

R_cosx

Faker_cosx

Correct answer:

R_cosx

Explanation:

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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Example Question #2 : Graphing Polar Form

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

Possible Answers:

R_cosx_1

R_cos2x

R_sin2x

R2_cos2x

R_sin2x

Correct answer:

R_cos2x

Explanation:

Because this function has a period of $\pi$ , the x-intercepts of the graph y=cos(2x) 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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Example Question #1 : Derivatives Of Polar Form

Find the derivative of the following function:

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

Possible Answers:

$3+e^\theta$

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

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

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

Correct answer:

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

Explanation:

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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Example Question #12 : Polar Form

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

Possible Answers:

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

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

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

$\frac{8cos\theta\sin\theta-2\cos\theta}{4-8\sin^{2}\theta-2sin\theta}$

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

Correct answer:

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

Explanation:

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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AP Calculus BC : Polar Form

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #24 : Polar

Example Question #21 : Functions, Graphs, And Limits

Example Question #22 : Functions, Graphs, And Limits

Example Question #23 : Functions, Graphs, And Limits

Example Question #24 : Functions, Graphs, And Limits

Example Question #11 : Polar Form

Example Question #1 : Graphing Polar Form

Example Question #2 : Graphing Polar Form

Example Question #1 : Derivatives Of Polar Form

Example Question #12 : Polar Form

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