How do you sketch the graph of the polar equation and find the tangents at the pole of #r=3(1-costheta)#?

1 Answer

The tangents are given by #theta = 0 rarr and larr theta = pi#. See the explanation, for this ticklish problem.

Explanation:

graph{x^2+y^2+3x-3sqrt(x^2+y^2)=0 [-10, 10, -5, 5]}

The cartesian form #x^2+y^2+3x-3sqrt(x^2+y^2=0# is used for this

Socratic graph.

#t = 3( 1 - cos theta )#, with period #2pi#.

#r'= 3 sin theta#

As the point #(r, theta)# is moved on the curve, the tangent vector

rotates. Here, from start at the pole to the finish ( at the return to the

pole), #theta# completes one period #[0, 2pi]#.

With respect to the pole, r = 0 but #theta = 0 and 2pi#, giving the

same direction..

Now, the formula for the slope of the tangent

at #(r, theta ) = ( 3(1-cos theta), theta)#

#slope = (r'sintheta + r cos theta)/(r'cos theta- r sin theta )#.

Here, this is

#(3 sin^2 theta+3(1-cos theta) cos theta)/#
#(3sin theta cos theta-3(1-cos theta)sin theta) #,

At the pole (0, 0), the slope ( in the form #0/0# ) has the limit #0_+#.

Likewise, for the tangent at the finish #(0, 2pi)#, the limit is #0_-#.

The tangents are given by #theta = 0 and theta = pi#.

Upon reading my answer, some eyebrows might be raised.

My approach is practical and real. I have followed the tangent

vector, from start to finish, in the tracing of the cardioid.