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

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`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.


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`


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


`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`


`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`


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

`theta_2= (3pi)/2`



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).

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