Question

at the given point find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve.

(2x^2)y -pi(cos)y = 3pi tangent at (1,pi)

(2x^2)y - pi cos y =3pi, tangent at (1,pi)

Answer 1

\[4xy+2x^2y'+\pi \sin(y)y'=0\] is a start

Answer 2

solve for y' to find the slope

Answer 3

\[y'=-\frac{4xy}{2x^2+\pi\sin(y)}\] replace x by 1, y by pi to get the slope

Answer 4

would the slope be 4pi/2