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)

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)

anonymous · Answer

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

anonymous · Answer

solve for y' to find the slope

anonymous · Answer

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

anonymous · Answer

would the slope be 4pi/2