`x=3costheta ,y=3sintheta` Find all points (if any) of horizontal and vertical tangency to the curve.

`x=3cos theta`

`y=3sin theta`

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

`dx/(d theta) = -3sin theta`

`dy/(d theta) = 3cos 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...

`x=3cos theta`

`y=3sin theta`

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

`dx/(d theta) = -3sin theta`

`dy/(d theta) = 3cos 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`

`3cos theta =0`

`cos theta=0`

`theta = pi/2, (3pi)/2`    

These are the values of theta in which the graph of parametric equation will have horizontal tangents.

Then, substitute these values to the parametric equation to get the points (x,y).

`theta = pi/2`

`x=3cos theta=3cos(pi/2)=3*0=0`

`y=3sin theta=3sin (pi/2) = 3*1=3`

`theta=(3pi)/2`

`x=3cos theta=3cos(3pi)/2=3*0=0`

`y=3sin theta=3sin(3pi)/2=3*(-1)=-3`

Therefore, the parametric equation has horizontal tangent at points (0,3) and (0,-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 happens when  `dx/(d theta)=0` , but `dy/(d theta)!=0` .

So, set the derivative of x equal to zero.

`dx/(d theta) = 0`

`-3sin theta = 0`

`sin theta = 0`

`theta = 0, pi`

These are the values of theta in which the graph of parametric equation will have vertical tangents.

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

`theta= 0`

`x=3cos theta = 3cos0=3*1=3`

`y=3sin theta=3sin0=3*0=0`

`theta =pi`

`x=3cos theta = 3cospi=3*(-1)=-3`

`y=3sin theta=3sinpi=3*0=0`

Therefore, the parametric equation has vertical tangent at points (3,0) and (-3,0).