# Graph of the polar equation at the pole

• May 14th 2009, 03:07 PM
terrytriangle
Graph of the polar equation at the pole
Find the tangents at the pole for r=2(1-sin theta)

I've been having alot of trouble with this concept. So far Ive tried setting r= to 0 and solving for theta but i dont know if thats right...
• May 14th 2009, 03:44 PM
galactus
There's a little more to it than that.

$\displaystyle \frac{dy}{dx}=\frac{rcos(t)+sin(t)\cdot\frac{dr}{d t}}{-rsin(t)+cos(t)\cdot\frac{dr}{dt}}$

Find the derivative of $\displaystyle r=2(1-sin(t))$ and plug it all in.

Then, set to 0 and solve for theta.

I just used t for less typing.

Upon doing all that, you should get $\displaystyle \frac{sin(2t)-cos(t)}{cos(2t)+sin(t)}$

Just set $\displaystyle sin(2t)-cos(t)=0$ and solve for t.

Spoiler:
$\displaystyle cos(t)(2sin(t)-1)=0$

$\displaystyle cos(t)=0 \;\ or \;\ 2sin(t)-1=0$
• May 14th 2009, 06:41 PM
terrytriangle
thanks. so for my next problem its r=3sin2t
so i would solve for dy/dx, set the numerator equal to zero, and solve for t?