Find the equation of the tangent plane to the surface z=4x^3*y^2+2y at point (1, -2, 12)

Question

amistre64 · Answer

the gradient defines the normal of the plane, and you are given the anchor point for it

anonymous · Answer

You could find the tangent line at that point and it would help a bit more

anonymous · Answer

Would the equation be 48x + 14y + z = 64?

amistre64 · Answer

lets see :)$$z=4x^3y^2+2y$$
$$F(x,y,z)=4x^3y^2+2y-z$$
$$F_x=12x^2y^2$$$$F_y=8x^3y+2$$$$F_z=-1$$
the normal to the plane at any point is:$$<F_x,F_y,F_z>$$

anonymous · Answer

Oh lol do it that way...it's a lot faster that way.....I mean the way amistre64 did

amistre64 · Answer

(1, -2, 12)
$$Fx = 12(1^2)(-2)^2=48$$$$Fx = 8(1^3)(-2)+2=-14$$$$Fz=-1$$

$$P_{tan}=48(x-1)-14(y+2)-(z-12)=0$$

anonymous · Answer

Oh, thanks so much! Just did a careless mistake simplifying at the end. Way easier than it seemed :D

amistre64 · Answer

:)  good luck

amistre64 · Answer

lol, the trouble with calculus is all that algebra getting in the way

anonymous · Answer

Indeed :p