Question

HELP!! multivariable calculus--> find the tangent plane to z=f(x,y) at the point P: f(x,y)=x^2*y-2x^3+y+3 with P=(-1,2,9)

Answer 1

the partials will be needed for tangent planes

Answer 2

can you solve for Fx and Fy?

Answer 3

Fz is -1 if we allow: 0 = f(x,y) - z

Answer 4

I feel lost on how to start this

Answer 5

you need to know what partial derivatives are ....

Answer 6

partials assume that anything other than the "wrt" variable is a constant.

f(x,y)=x^2*y-2x^3+y+3

Fx, assume all other letters are constants

Fx=2x*y -6x^2+0+0
---------------------------

Fy, assume all other letters are constants

Fy=x^2 -0 +1+0

Answer 7

this creates expressions for a gradient of f: (Fx,Fy,Fz), if memory serves, this also defines the normal of the tangent plane

Answer 8

hence, given a point (a,b,c)

Fx(x-a) + Fy(x-b) + Fz(z-c) = 0

Answer 9

http://tutorial.math.lamar.edu/Classes/CalcIII/GradientVectorTangentPlane.aspx

Answer 10

so how do you find what Fz is? and x,y,z

Answer 11

xyz are just fillins for the plane equation; kinda like y = mx + b, asking what y and x are is futile.

Answer 12

since z = f(x,y)
0 = f(x,y) - z

we can say that g(x,y,z) = f(x,y) - z to rename things to find Gx, Gy, Gz with

Gx = Fx
Gy = Fy
Gz = -1, which we can use to fill in the z spot on the normal vector

Answer 13

is the answer z-9=-10(x+1)+2(y-2) ??

Answer 14

maybe ...

given a point: (-1,2,9)

the plane is Fx(x+1) + Fy(y-2) + Fz(z-9) = 0

Fx = 2(-1)(2) -6(-1)^2 = -10
Fy = (-1)^2 +1 = 2
Fz = -1, just becuase

-10(x+1) + 2(y-2) - (z-9) = 0

and since leading terms are somehow bad if they are negative ....

10(x+1) - 2(y-2) + (z-9) = 0

they also dont like it all bottled up like that and expect you to expand it all out.

also, doble check my work to make sure i didnt do any simple mistakes :/