Question

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

Answer 1

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

Explanation:

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#