Calculus of a Single Variable, Chapter 10, 10.3, Section 10.3, Problem 35

x=5+3cos theta
y= -2+sin theta
First, take the derivative of x and y with respect to theta.
dx/(d theta) = -3sin theta
dy/(d theta) = cos theta
Take note that the slope of a tangent is equal to dy/dx.
m= dy/dx
To determine 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 is zero.
0= (dy/(d theta))/(dx/(d theta))
This occurs when dy/(d theta)=0 and dx/(d theta) !=0 . So setting the derivative of y equal to zero yields:
dy/(d theta) = 0
cos theta = 0
theta_1= pi/2+2pin
theta_2=(3pi)/2 + 2pin
(where n is any integer)
So the graph of the parametric equation has horizontal tangent at these values of theta.
To determine the points (x,y), plug-in the values of theta to the given parametric equation.
theta_1 =pi/2+2pin
x=5+3cos(pi/2+2pin)=5+3cos(pi/2)=5+3*0=5
y=-2+sin(pi/2+2pin)=-2+sin(pi/2)=-2+1=-1
theta_2 = (3pi)/2+2pin
x=5+3cos((3pi)/2+2pin)=5+3cos((3pi)/2)=5+3*0=5
y=-2+sin((3pi)/2+2pin)=-2+sin((3pi)/2)=-2+(-1)=-3
Therefore, the graph of the parametric equation has horizontal tangent at points (5,-1) and (5,-3).
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 occurs when dx/(d theta)=0 and dy/(d theta)!=0 . So, setting the derivative of x equal to zero yields:
dx/(d theta) = 0
-3sin theta = 0
sin theta = 0
theta_1 = 2pin
theta_2= pi+2pin
(where n is any integer)
So the graph of the parametric equation has vertical tangent at these values of theta.
To determine the points (x,y), plug-in the values of theta to the given parametric equation.
theta_1=2pin
x=5+3cos(2pin)=5+3cos(2pi) =5+3*1=8
y=-2+sin(2pin)=-2+2sin(2pi)=-2+0=-2
theta_2=pi+2pin
x=5+3cos(pi+2pin)=5+3cos(pi)=5+3(-1)=2
y=-2+sin(pi+2pin)=-2+sin(pi)=-2+0=-2
Therefore, the graph of the parametric equation has vertical tangent at points (8,-2) and (2,-2).