Hello, Holly!
Your answers are correct.
But there is another vertical tangent at the cusp: .
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
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
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