equation of a tangent plane in two variables

$f(x,y) = x^3y^3+x^2+1$

I'm looking for the equation to the tangent plane to $z=f(x,y)$ at the point $(1,-1)$.

First I find the partial derivatives. $$f_x(x,y) = 3x^2y^3+2x \\ f_x(1,-1) = -1$$

$$f_y(x,y) = 3x^3y^2 \\ f_y(1,-1) = 3$$

Then I get $z = -(x-1)+3(y+1)$. I was informed that the correct answer is $z = 1-(x-1)+3(y+1)$ but I don't know why it's $1-(x-1) ...$ and not $-(x-1)...$

• At the tangent point, the plane and the surface have to touch, so the $z$ value calculated for the plane and the $z$ value calculated for the surface have to be the same. You should be able to figure that out since it seems you've been given $f(x, y)$, although you neglected to share that information with us. May 11 '17 at 3:28
• consider that f(1,-1) = (1)^3(-1)^3 + (1)^2 + 1 = -1 + 1 + 1 = 1. also, for the plane to touch, it must have the same value as the function at that point May 11 '17 at 3:37
• Compare to the similar situation in one variable: the tangent line to $f(x)$ at $x=a$ is $y = f(a) + f'(a)(x-a)$. At $x=a$, we get $y=f(a)$. May 11 '17 at 3:38
• your plane you have defined has the right slope you just haven't gotten the right z value for it yet May 11 '17 at 3:39
• if you move it up in a sense by 1 then it now will touch the function, you have correctly found the slope of hte function but it still needs to touch it May 11 '17 at 3:40

1) Any plane: $ax + by + cz = d$.

We'll determine $a,b,c$ and $d$ make it the tangent plane at $(1,-1,1)$

Consider the surface $F(x,y,z) = f(x,y) -z = 0$ .

$\nabla F(x,y,z)$ is the normal vector to this surface at $(x,y,z)$.

$\nabla F(x,y,z) =$

$=(3x^2y^3 + 2x, 3x^3y^2, -1)$,

at $(1,-1,1)$ the normal vector is:

$n = (-1,3,-1)$.

The normal to the plane $ax + by + cz - d =0$:

$N_p = (a,b,c)$.

Pick : $a = -1; b =3, c = -1$.

For the plane $- x + 3y - z = d$ to pass through $(1,-1,1)$:

$d = -1 -3 -1 = -5$,

Altogether the equation of the tangent plane:

$-x + 3y -z + 5 =0$.