`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...
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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).
