# x=cos^2theta ,y=costheta Find all points (if any) of horizontal and vertical tangency to the curve. x=cos^2 theta

y=cos theta

First, take the derivative of x and y with respect to theta .

dx/(d theta) = 2costheta (-sin theta)

dx/(d theta)=-2sintheta cos theta

dy/(d theta) = -sin theta

Take note that the slope of a 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 is zero.

0= (dy/(d theta))/(dx/(d theta))

This implies that the graph of the parametric equation will have a horizontal tangent when dy/(d theta)=0 and dx/(d theta)!=0 .

So, set the derivative of y equal to zero.

dy/(d theta) = 0

-sin theta = 0

sin theta=0

theta= 0, pi

Then, plug-in these values of theta to dx/(d theta) to verify if the slope is zero, not indeterminate.

dx/(d theta) =-2sintheta cos theta

theta_1=0

dx/(d theta) =-2sin(0) cos (0) = 0

theta_2=pi

dx/(d theta) = -2sin(pi)cos(pi)=0

Since both dy/(d theta) and dx/(d theta) are zero at these values of theta, the slope is indeterminate.

m=dy/dx= (dy/(d theta))/(dx/(d theta))=0/0  (indeterminate)

Therefore, the parametric equation has no horizontal tangent.

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 happens when dx/(d theta)=0 , but dy/(d theta)!=0  .

So, set the derivative of x equal to zero.

dx/(d theta) = 0

-2sinthetacos theta=0

-2sin theta = 0

sin theta = 0

theta = 0, pi

cos theta = 0

theta = pi/2, (3pi)/2

Take note that at theta =0 and theta =pi , both dy/(d theta) and dx/(d theta) are zero.  So the slope at these two values of theta is indeterminate.

m=dy/dx= (dy/(d theta))/(dx/(d theta))=0/0  (inderterminate)

Plug-in theta =pi/2 and theta=(3pi)/2 to dy/(d theta) to verify that the slope is undefined at these values of theta.

dy/(d theta) = -sin theta

theta_1=pi/2

dy/(d theta) = -sin (pi/2)=-1

theta_2= (3pi)/2

dy/(d theta) = -sin ((3pi)/2)=1

Since  dy/(d theta) !=0 ,  the parametric equation has vertical tangent at  theta_1=pi/2 and theta=(3pi)/2 .

Then, plug-in these values to the parametric equation to get the points (x,y).

x=cos^2 theta

y=cos theta

theta_1= pi/2

x=cos^2(pi/2)=0

y= cos (pi/2)=0

theta_2= (3pi)/2

x=cos^((3pi)/2)=0

y=cos((3pi)/2)=0

Since theta_1 and theta_2 result to same x and y coordinates, there is only one point in which the curve has a vertical tangent.

Therefore, the graph of the parametric equation has vertical tangent at point (0,0).