Question

.

Answer 1

The formula that we learnt that should be used can be found here:

http://tutorial.math.lamar.edu/Classes/CalcII/PolarTangents_files/eq0008MP.gif

Answer 2

Function: \(\color{#000000 }{ \displaystyle r(\theta)=\theta }\)
Point of tangency: \(\color{#000000 }{ \displaystyle \theta=\pi }\)

correct?

Answer 3

Yes!

Answer 4

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{\dfrac{dr}{d\theta }\sin \theta +r\cos \theta }{\dfrac{dr}{d\theta }\cos \theta -r\sin \theta} }\)

Answer 5

I tried that and ended up with 0, with I believe is incorrect

Answer 6

\(\color{#000000 }{ \displaystyle \frac{dr}{d\theta}=\frac{d}{d\theta}[r(\theta)]=\frac{d}{d\theta}[\theta]=1 }\)

Answer 7

And \(\theta=\pi\).

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{1\cdot \sin \pi +r\cos \pi }{1\cdot\cos \pi -r\sin \pi} }\)

Answer 8

And \(r(\theta)=\theta=\pi \)

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{1\cdot \sin \pi +\pi\cos \pi }{1\cdot\cos \pi -\pi\sin \pi} }\)

Answer 9

I'm following!

Answer 10

wouldn't it come out to -pi?

Answer 11

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{0-\pi}{-1 -0}=\pi }\)

Answer 12

Can that be the slope?

Answer 13

I know this may be frowned upon, but I usually check my answers with WolframAlpha, and it says the answer is 1 :(

Answer 14

can you link me?

Answer 15

http://www.wolframalpha.com/widgets/view.jsp?id=8cc76aa60d671138e037520930656242

Answer 16

But your answer seems more logical

Answer 17

I think they are putting in y=theta, as if y and theta are cartesians.

Answer 18

Okay makes sense! Could you help me with b) as well?

Answer 19

Please :)

Answer 20

r=sin(2θ)

dr/dθ = ?

Answer 21

2cos2theta

Answer 22

Yes, correct

Answer 23

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{\dfrac{dr}{d\theta }\sin \theta +r\cos \theta }{\dfrac{dr}{d\theta }\cos \theta -r\sin \theta} }\)

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{2\cos(2\theta)\sin \theta +\sin(2\theta)\cos \theta }{2\cos(2\theta)\cos \theta -\sin(2\theta)\sin \theta} }\)

Answer 24

that is the slope of your equation at any value of \(\theta\).

Answer 25

Now, you need to find the slope when \(\theta=0\).

Answer 26

And sin(0)=0, so my answer will be 0 correct?

Answer 27

you need the slope, not just r(0)

Answer 28

you are plugging \(\theta=0\) into

\(\color{#000000 }{ \displaystyle \frac{dy}{dx}=\frac{2\cos(2\theta)\sin \theta +\sin(2\theta)\cos \theta }{2\cos(2\theta)\cos \theta -\sin(2\theta)\sin \theta} }\)

Answer 29

oh, yes, yes .... =0
I see what you meant:)

Answer 30

Yes I think that's what I did with some mental math. Every sinθ in that equation will equal 0, so the top will be 0 and the bottom will be some number.

Answer 31

Oh okay thank you so much for the help!

Answer 32

yes, figured... good work

Answer 33

Thank you, you're awesome!

Answer 34

\(y{\tiny~}(\omega)\)