at the given point fine 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)

Question

at the given point fine 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)

anonymous · Answer

?

.sam. · Answer

"fine" or "find"? lol

anonymous · Answer

maybe 

at the given point find the slope of the line that is tangent to the curve and the line that is normal to the curve

anonymous · Answer

curve being 
$$2x^2y-\pi\cos(y)=3\pi$$?

hihi67 · Answer

@satellite73 
(a^2)^2/3b / 4a^5/(b^3)^2=

anonymous · Answer

something wrong with this problem, since if you replace x by 1 i don't think you get 
$$\pi$$

anonymous · Answer

Yea i meant find, my bad

anonymous · Answer

2x2y−πcos(y)=3π     This is the equation and there asking me to find the tangent line at (1, pi)