AP Calculus BC - Polar Form

Convert the following cartesian coordinates into polar form:

$(\sqrt{394},40.91)$

$(\sqrt{390},40.45)$

$(\sqrt{363},75.21)$

$(\sqrt{374},81.70)$

Explanation

Cartesian coordinates have and , represented as . 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:

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

$r=\sqrt{(15)^2 + (13)^2}$

$r=\sqrt{225+169}$

$r=\sqrt{394}$

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

$\theta=\arctan(\frac{13}{15})$

$\theta=40.91$

$(\sqrt{394},40.91)$

Calculate the polar form hypotenuse of the following cartesian equation:

$\sqrt{\frac{\tan(\theta)}{8\cos^2(\theta)}}$

$\sqrt{}\frac{\tan(\theta)}{8\cos(\theta)}}$

$\sqrt{\frac{\tan(\theta)}{\cos(\theta)}}$

$\frac{\tan(\theta)}{8\cos(\theta)}$

Explanation

In a cartesian form, the primary parameters are x and y. 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:

$y = 8x^3 = r\sin(\theta)$

$x= r\cos(\theta)$

$y = 8\cdot(r\cos(\theta))^3$

$y = 8\cdot r^3\cdot\cos^3(\theta) = r\sin(\theta)$

$r = \sqrt{\frac{\tan(\theta)}{8\cos^2(\theta)}}$

What is the polar form of $y=\frac{5}{7}x^{2}$ ?

$r=\frac{7}{5}\tan \theta\sec\theta$

$r=-\frac{7}{5}\tan \theta\sec\theta$

$r=\frac{5}{7}\tan \theta\sec\theta$

$r=-\frac{5}{7}\tan \theta\sec\theta$

$r=\tan \frac{5}{7} \theta\sec\theta$

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 $y=\frac{5}{7}x^{2}$ , then:

$r\sin\theta=\frac{5}{7}(r\cos \theta)^{2}$

$r\sin\theta=\frac{5}{7}r^{2}\cos^{2} \theta$

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

$\tan \theta=\frac{5}{7}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{5}{7}r$

$\tan \theta\sec\theta=\frac{5}{7}r$

$r=\frac{7}{5}\tan \theta\sec\theta$

What is the polar form of $y=-\frac{6}{5}x^{2}$ ?

$r=-\frac{5}{6}\tan \theta\sec\theta$

$r=\frac{5}{6}\tan \theta\sec\theta$

$r=-\frac{6}{5}\tan \theta\sec\theta$

$r=\frac{6}{5}\tan \theta\sec\theta$

$r=-\tan\frac{6}{5} \theta\sec\theta$

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 $y=-\frac{6}{5}x^{2}$ , then:

$r\sin\theta=-\frac{6}{5}(r\cos \theta)^{2}$

$r\sin\theta=-\frac{6}{5}r^{2}\cos^{2} \theta$

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

$\tan \theta=-\frac{6}{5}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=-\frac{6}{5}r$

$\tan \theta\sec\theta=-\frac{6}{5}r$

$r=-\frac{5}{6}\tan \theta\sec\theta$

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

$r=\frac{9}{2}\tan \theta\sec\theta$

$r=-\frac{9}{2}\tan \theta\sec\theta$

$r=\frac{2}{9}\tan \theta\sec\theta$

$r=-\frac{2}{9}\tan \theta\sec\theta$

$r=\tan \frac{2}{9}\theta\sec\theta$

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

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

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

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

$\tan \theta=\frac{2}{9}r\cos\theta$

$\frac{\tan \theta}{\cos\theta}=\frac{2}{9}r$

$\tan \theta\sec\theta=\frac{2}{9}r$

$r=\frac{9}{2}\tan \theta\sec\theta$

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

Explanation

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:

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

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

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

$r=-\frac{1}{9}\tan \theta\sec\theta$

$r=\frac{1}{9}\tan \theta\sec\theta$

$r=-9\tan \theta\sec\theta$

$r=9\tan \theta\sec\theta$

$r=-\tan 9\theta\sec\theta$

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 $y=-9x^{2}$ , then: