Question

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Answer 1

http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents.aspx

Answer 2

I looked at that before but I still don't understand

Answer 3

I don't think I found the derivative correctly either?

Answer 4

first find dy/dx :

\[\dfrac{dy}{dx} = \dfrac{\frac{dr}{d\theta} \sin \theta + r\cos \theta
}{\frac{dr}{d\theta} \cos \theta - r\sin\theta
}\]

Answer 5

^^above gives u slope of tangent line

Answer 6

once u have the slope, u can use the given point to write out the equation in cartesian point-slope form

Answer 7

is dr/dtheta the derivative I found before to put into that formula?

Answer 8

Yep

Answer 9

How do I simplify it, though?

Answer 10

\[

\dfrac{dy}{dx} = \dfrac{\frac{dr}{d\theta} \sin \theta + r\cos \theta
}{\frac{dr}{d\theta} \cos \theta - r\sin\theta} = \dfrac{4\cos (4\theta) \sin \theta + r\cos \theta
}{4\cos (4\theta) \cos \theta - r\sin\theta}

\]

Answer 11

Oh that's what I got but I thought it could be be further simplified

Answer 12

ur point : (2,0)
change it to polar

Answer 13

\(r = \sqrt{2^2 + 0^2} = 2\)
\(\theta = \arctan (\frac{0}{2}) = 0\)

Answer 14

\[
\dfrac{dy}{dx} \Bigg|_{(2, 0)} = \dfrac{4\cos (4*0) \sin 0 + 2\cos 0
}{4\cos (4*0) \cos 0 - 2\sin 0}
\]

Answer 15

simplify

Answer 16

1/2 is then the slope?

Answer 17

yes

Answer 18

How do I get the full equation of the tangent line, though? Do I have to use the slope somehow now?

Answer 19

slop = 1/2
point = (2, 0)

Answer 20

point slope form : \(y - y_1 = m(x-x_1)\)

Answer 21

Oh, oh, yeah I know this

Answer 22

Thanks for that! Would you also happen to know how to find the area of this polar equation?

Answer 23

Area :

\[\iint \limits_R 1 ~r dr ~d\theta = \int \limits_{..}^{..} \frac{1}{2}r^2 d\theta \]

Answer 24

you just need to integrate from 0->2pi

Answer 25

\[ \int \limits_{0}^{2\pi} \frac{1}{2}r^2 d\theta \]

Answer 26

plugin the value of \(r\) and evaluate the integral

Answer 27

So I just need to plug in 2 into r and just solve that integral?

Answer 28

nope, area is not related to just a point.
you're done with finding the tangent at (2, 0) in the first part

Answer 29

plugin the equation for \(r\) above^

r=2+sin(4theta)

Answer 30

\[\int \limits_{0}^{2\pi} \frac{1}{2}r^2 d\theta = \int \limits_{0}^{2\pi} \frac{1}{2}(2+\sin(4\theta))^2 d\theta \]

Answer 31

And the answer to that integral will just be the area of the enclosed region of the polar equation, is that correct?

Answer 32

yup

Answer 33

Oh, sweet. Thanks so much!

Answer 34

you should get 9pi/2

http://www.wolframalpha.com/input/?i=+%5Cint+%5Climits_%7B0%7D%5E%7B2%5Cpi%7D++%5Cfrac%7B1%7D%7B2%7D%282%2B%5Csin%284%5Ctheta%29%29%5E2+++d%5Ctheta++

Answer 35

Okay, I'll do all that now, that seemed a lot more simpler than I thought. Thanks :)

Answer 36

u wlc :)