Find the slope dy dx of the polar curve r = 1 + 2 sin θ at θ = pi over 2 . ( 10 points)

Question

A) Undefined
B) 1
C) -1
D) 0

Answer 1

Why are you switching names, you are making the same typo error as April
when you type "dy dx" instead of dy/dx

let's convert r = 1 + 2 sin θ to rectangular.
sin θ = y/r and x^2 + y^2 = r^2 from the standard triangle
configuration, so
r = 1 + 2y/r
r^2 = r + 2y
x^2 + y^2 = √(x^2 + y^2) + 2y
2x + 2y y' = (1/2)(x^2 + y^2)^(-1/2) * (2x + 2y y')

When θ = π/2
r = 1 + 2sin π/2 = 1+2(1) = 3
the polar coordinate point (3,π/2) converts to (0,3) in rectangular form
so for dy/dx or y'
2x + 2y y' = (1/2)(x^2 + y^2)^(-1/2) * (2x + 2y y')
0 + 6y' = (1/2)(0 + 9)^(-1/2) * (0 + 6y')
6y' = (1/2)(1/3)*(6y')
only true if y' = 0 , the slope is zero

www.wolframalpha.com/input/?i=r+%3D+1+%2B+2+sin+θ+