Find an equation (in terms of x and y) of the tangent line to the curve r=3sin(5theta) at theta=pi/3

Answer 1

r = 3sin5Į

since x = rcosĮ and y = rsinĮ

dy/dx = dy/dĮ / dx/dĮ

= (r'sinĮ + rcosĮ)/(r'cosĮ - rsinĮ)
= (15cos5ĮsinĮ + 3sin5ĮcosĮ)/(15cos5ĮcosĮ - 3sin5ĮsinĮ)
= (15 * 1/2 * �ã3/2 + 3 * (-�ã3/2) * 1/2)
--------------------------------
(15 * 1/2 * 1/2 - 3 * -�ã3/2 * �ã3/2)

= (15�ã3/4 - 3�ã3/4)/(15/4 + 9/4)
= 12�ã3 / 24
= �ã3/2

So, now you have a point (pi/3,-3�ã3/2) and a slope.

(y+�ã3/2) = �ã3/2 (x-pi/3)

Answer 2

Interpret

Į as theta
�ã as sqrt

sorry.