Calculus of a Single Variable, Chapter 10, 10.4, Section 10.4, Problem 65

r=1-sin theta
To solve, express the polar equation in parametric form. To convert it to parametric equation, apply the formula
x = rcos theta
y=r sin theta
Plugging in r=1-sin theta , the formula becomes:
x=(1-sin theta)cos theta=cos theta -sin theta cos theta
y = (1-sin theta)sin theta=sin theta -sin^2 theta
So the equivalent parametric equation of r= 1-sin theta is:
x=cos theta -sin theta cos theta
y=sin theta -sin^2 theta
Then, take the derivative of x and y with respect to theta.
dx/(d theta) = -sintheta - (sintheta*(-sintheta) + costheta*costheta)
dx/(d theta)=-sintheta+sin^2theta-cos^2theta
dy/(d theta) = costheta - 2sinthetacostheta
Take note that the slope of the tangent is equal to dy/dx.
m= (dy)/(dx)
To get the dy/dx of a parametric equation, apply the formula:
dy/dx = (dy/(d theta))/(dx/(d theta))
When the tangent line is horizontal, the slope of the tangent is zero.
0 = (dy/(d theta)) / (dx/(d theta))
This implies that the polar curve will have a horizontal tangent when dy/(d theta)=0 and dx/(d theta) !=0.
Setting the derivative of y yields:
dy/(d theta) = 0
costheta - 2sinthetacostheta=0
costheta(1 - 2sintheta) =0
costheta = 0
theta=pi/2,(3pi)/2
1-2sintheta=0
-2sintheta=-1
sintheta=1/2
theta=pi/6,(5pi)/6
Take note that at theta=pi/2 , the value of dx/(d theta) is zero. Since both dy/(d theta) and dx/(d theta) are zero, the slope at this value of theta is indeterminate.
m=0/0 (indeterminate)
So the polar curve has horizontal tangents at:
theta_1 = pi/6 + 2pin
theta_2= (5pi)/6+2pin
theta_3= (3pi)/2+2pin
where n is any integer.
To determine the points (r, theta) , plug-in the values of theta to the polar equation.
r=1-sin theta
theta_1 = pi/6 + 2pin
r_1=1-sin(pi/6 + 2pin)=1-sin(pi/6) = 1-1/2=1/2
theta_2= (5pi)/6+2pin
r_2=1-sin((5pi)/6+2pin)=1-sin((5pi)/6)= 1 -1/2=1/2
theta_3= (3pi)/2+2pin
r_3=1-sin((3pi)/2+2pin)=1-sin((3pi)/2)=1-(-1)=2
Therefore, the polar curve has horizontal tangent at points
(1/2, pi/6+2pin) , (1/2, (5pi)/6+2pin) , and (2, (3pi)/2+2pin) .
Moreover, when the tangent line is vertical, the slope is undefined.
u n d e f i n e d =(dy/(d theta)) / (dx/(d theta))
This implies that the polar curve will have vertical tangent when dx/(d theta)=0 and dy/(d theta)!=0 .
Setting the derivative of x equal to zero yields:
dx/(d theta) = 0
-sintheta+sin^2theta-cos^2theta=0
-sin theta + sin^2 theta-(1-sin^2 theta) = 0
2sin^2 theta -sin theta -1=0
(2sin theta +1)(sin theta -1) = 0
2sin theta + 1=0
sin theta=-1/2
theta = (7pi)/6,(11pi)/6
sin theta -1=0
sin theta=1
theta=pi/2
Take note that at theta =pi/2 , both dy/(d theta ) and dx/(d theta) are zero. So the slope is indeterminate at this value of theta.
m=0/0 (indeterminate)
So the polar curve has vertical tangents at:
theta_1 =(7pi)/6+2pin
theta_2=(11pi)/6+2pin
where n is any integer.
To determine the points (r, theta) , plug-in the values of theta to the polar equation.
r=1-sin theta
theta_1=(7pi)/6+2pin
r_1=1-sin((7pi)/6+2pin)=1-sin((7pi)/6)=1-(-1/2)=3/2
theta_2=(11pi)/6+2pin
r_2=1-sin((11pi)/6+2pin)=1-sin((11pi)/6)=1-(-1/2)=3/2
Therefore, the polar curve has vertical tangent at points (3/2, (7pi)/6+2pin) and (3/2, (11pi)/6+2pin) .