Given that x²cos y-sin y=0 ,(0,π):

Question

a)verfiy that given point is on the curve.

b)use implicit differentiation to find the slope of the above curve at the given point.

c)find the equation for tangent and normal to the curve at that point.

Answer 1

(a) trés simples, non?
(b)
2x cosy - x^2 siny y' - cosy y' = 0
y' = (2x cosy)/(cosy + x^2 siny)

So, at (0,π), y'=0

(c) so, the tangent is a horizontal line, and the normal is a vertical line through (0,π).