How do you find the polar equation to the tangent, at #theta = pi/12#, on #r = sin 3theta#?

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Answer 1 · 2017-01-24T10:19:40

So, #r =1/(4cos theta - 2sin theta )#. In cartesian form, this is #4x-2y-1=0#

The given equation represents a three-petal rose.

The equation to the tangent

at #P (f(alpha), alpha)#, on #r = f(theta)# is

#r(sin theta-tan psi cos theta) =f(alpha)(sin alpha-tan psi cos alpha)#,

where

#tan psi = m=#

# ((dr)/(d theta )sinalpha+ r cos alpha)##/((dr)/(d theta)cos theta- r sin theta)# , at P.

Briefly,

#r sin (theta-psi)= f(alpha)sin(alpha-psi)#

Here, # r = sin 3 theta and (dr)/(d theta)=3 cos 3 theta#.

P is# (1/sqrt2, pi/12)#,

m at P = #(3/2+1/2)/(3/2-1/2)= 2#.

Now, the equation to the tangent at P is

#r(sin theta- 2 cos theta) = 1/sqrt2(1/sqrt2-2/sqrt2)=-1/2#

So, #r =1//(4cos theta - 2sin theta )#

In cartesian form, this is 4x-2y-1=0#