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Calculus 2 : Polar

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Example Questions

Example Question #202 : Ap Calculus Bc

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

Possible Answers:

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

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

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

None of the above

$r=\sqrt\frac{6}{\tan \theta}$

Correct answer:

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

Explanation:

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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Example Question #1 : Polar Form

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

Possible Answers:

None of the above

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

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

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

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

Correct answer:

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

Explanation:

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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Example Question #6 : Polar Form

Given x=3t+15 and y=15t-6 , what is in terms of (rectangular form)?

Possible Answers:

y=15x-81

y=5x-81

y=5x-3

y=15x-3

y=5x-15

Correct answer:

y=5x-81

Explanation:

Knowing that x=3t+15 and y=15t-6 , 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:

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

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

y+6=5({x-15})

y+6=5x-75

y=5x-81

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Example Question #1 : Polar Form

Convert the following function into polar form:

f(x)=2x^2+5x+2+2y^2

Possible Answers:

$r^2+1+5r\cos(\theta)$

$r^2+\sin(\theta)$

$2r^2+5r\sin(\theta)+2$

$2r^2+5r\cos(\theta )+2$

Correct answer:

$2r^2+5r\cos(\theta )+2$

Explanation:

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)=2x^2+5x+2+2y^2

f(x)=(x^2+y^2)2+5x+2

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

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Example Question #1 : Polar Form

Convert from rectangular to polar form:

(2, 2)

Possible Answers:

$(0, \pi)$

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

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

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

Correct answer:

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

Explanation:

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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Example Question #11 : Polar

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $y=3x^{2}$ , 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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Example Question #12 : Functions, Graphs, And Limits

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 $y=-\frac{1}{2}x^{2}$ , 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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Example Question #12 : Polar

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

Possible Answers:

$r=-\frac{2}{3}\tan \theta\sec\theta$

$r=\frac{3}{2}\tan \theta\sec\theta$

None of the above

$r=\frac{2}{3}\tan \theta\sec\theta$

$r=-\frac{3}{2}\tan \theta\sec\theta$

Correct answer:

$r=\frac{3}{2}\tan \theta\sec\theta$

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

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

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

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

$\tan \theta=\frac{2}{3}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{2}{3}r$

$3\tan \theta\sec\theta=2r$

$r=\frac{3}{2}\tan \theta\sec\theta$

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Example Question #181 : Parametric, Polar, And Vector

What is the polar form of $y=\frac{3}{5}x^{2}$ ?

Possible Answers:

None of the above

$r=-\frac{3}{5}\tan\theta\sec\theta$

$r=\frac{5}{3}\tan\theta\sec\theta$

$r=\frac{3}{5}\tan\theta\sec\theta$

$r=-\frac{5}{3}\tan\theta\sec\theta$

Correct answer:

$r=\frac{5}{3}\tan\theta\sec\theta$

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 $y=\frac{3}{5}x^{2}$ , then:

$r\sin\theta=\frac{3}{5}(r\cos\theta)^{2}$

$r\sin\theta=\frac{3}{5}r^{2}\cos^{2}\theta$

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

$\tan\theta=\frac{3}{5}r\cos\theta$

$\frac{\tan\theta}{\cos\theta}=\frac{3}{5}r$

$\tan\theta\sec\theta=\frac{3}{5}r$

$r=\frac{5}{3}\tan\theta\sec\theta$

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Example Question #3 : Polar Form

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

Possible Answers:

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

$r=\tan7\theta\sec\theta$

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

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

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

Correct answer:

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

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

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #202 : Ap Calculus Bc

Example Question #1 : Polar Form

Example Question #6 : Polar Form

Example Question #1 : Polar Form

Example Question #1 : Polar Form

Example Question #11 : Polar

Example Question #12 : Functions, Graphs, And Limits

Example Question #12 : Polar

Example Question #181 : Parametric, Polar, And Vector

Example Question #3 : Polar Form

All Calculus 2 Resources

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