Question

Find slope at pi/6 for r=2sin(theta)

Answer 1

do you know what a derivative is?

Answer 2

yes

Answer 3

ok so then in this question, you will have to do the derivative of 2sin(theta)

\[(\delta/\delta x)(2\sin(\theta))=2(\delta/\delta x)(\sin(\theta)) = 2\cos(\theta)\]

Answer 4

nono, my bad, r=2sin(theta) is polar form

Answer 5

\[y=r sin(\theta) = 2 sin ^2(\theta)\]
\[x=r cos(\theta)=2 sin(\theta) cos(\theta)=sin(2 \theta)\]
\[{dy \over dx} = {{dy \over d\theta}\over{dx \over d\theta}}\].
Then substitute pi/6 in here.

Answer 6

then you just have to plug in pi/6 as theta and you should get your answer.

btw my previous answer it should be (d/d(theta))

Answer 7

slope is not d/d(theta).....also bloackcolder, mind finishing dy/dx? I kept getting ((sqrt(3)+3)/4) but the answer is sqrt(3)

Answer 8

Sure.
\[{dy \over dx}=\frac{4 sin(\theta)cos(\theta)}{2sin(2\theta)cos(2\theta)}=\frac{2sin(2\theta)}{2sin(2\theta)cos(2\theta)}=\frac{1}{cos(2\theta)}\]
At pi/6....
\[{1 \over cos(2\theta)}={1 \over cos({\pi \over 3})}=\sqrt3.\]

Answer 9

for ur dy/dx, just want to make sure, u did not use identity right?

Answer 10

I did.
\[2\sin\theta\cos\theta=\sin(2\theta)\]

Answer 11

ye, kk, so u got a suggestion for this? change to x, and y or do dr/d(theta)?
I used dr/d(theta) and i got a mess with +/-

Answer 12

Always find dy/dx for slopes. dr/d(theta) has a different meaning, I think.

Answer 13

alright, thanks alot man, gona do ur way instead