1. ## polar equations

find the points of horizontal and vertical tangency (if any) to the polar curve.
r=1-sin theta

i got horizontal tangent at (1/2, pi/6) (2, 3pi/2) (1/2, 5pi/6)
and vertical tangent at (3/2, -pi/6) (3/2, 7pi/6)

can anyone check my work and see if those are correct? i am not sure about the vertical tangent points...
thanks a bunch

2. Hello, Holly!

But there is another vertical tangent at the cusp: .$\displaystyle \left(0,\tfrac{\pi}{2}\right)$

3. Your results are correct , however you'd probably list

(3/2,-pi/6) as (3/2, 11pi/6)

Don't know the uncertainty you have on the VTs

the demominator of dy/dx = 2sin^2(t)-sin(t)- 1 = 0

(2sin(t)+1)(sin(t)-1) = 0

sin(t) = -1/2 which yields 7pi/6 , 11pi/6

or sin(t) = 1 but this is the cusp at pi/2

4. Thanks, I had the point at pi/2 for the horizontal tangent but I threw it out because I got pi/2 for the vertical tangent too and I thought you had to get rid of it for some reason??

5. A technical and subtle but important point. because you have 0/0 at pi/2

for dy/dx the lim as theta -> pi/2+ = -inf

lim as theta ->pi/2- = inf

So it is a cusp and not a vertical tangent

6. thanks! and its not a horizontal tangent either? just a cusp?

7. Yes -- just a cusp

Think about the cardioid 1 + cos(theta)

at pi you wouldn't have a horizontal tangent but a cusp.

Typically with polar graphs 0/0 yields a cusp

Another good example is y = x^(2/3) where you have a cusp at x = 0

since y ' = 2/3/{x^(1/3)} as you approach 0+ you get inf but as you approach 0- you get - inf

But for y = x^ (1/3) you have a vertical tangent since

y ' = 1/3/{x^(2/3)} you get infinity as x approaches 0 either from the left or right

8. thanks! that was a good example easy to understand