# 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 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).

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team