Find an equation for the tangent plane to the surface $2xz^2+3xy-4x=7$ at the point $(1,-1,2)$, by considering a gradient of a suitable scalar field.

Question

This is what I've done, followed by a few questions I don't undestand;

$f(x, y, z) = 2xz^2-3xy-4x=7$

Surface is at 0, So therefore we set our equation to zero $2xz^2-3xy-4x-7=0$

We then take partials and evaluate them at the point.

$F_x = x-2, F_x(1,-1,2) = 1-2=3$

$F_y = 1, F_y(1,-1,2) = 1$

$F_z = 2z, F_z(1,-1,2) = 2(2) = 4$

Now the tangent plane is $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

=$-3(x-1)+1(y+1)+4(z-1) = 0$

=$-3x+y+4z-6=0$

My questions;

1. I found this method of computing online and it was done for a vector. Now given this is a scalar field would it still work?

2. In the end we have $-3x+y+4z-6=0$, what exactly do I need to solve for? Should I bring the equation back in the original form where it was equal to a number? Or do I solve for z, x or y?

3. Why is surface at zero and why do we take partials?

Answer 1

I think you made a mistake in finding the gradient vector. From there on, given you have to find the scalar equation of the plane, you can do this way -

$f(x, y, z) = 2xz^2-3xy-4x-7 = 0$ (as called out in comments, the function in the title seems wrong.

$f_x = 2z^2 - 3y - 4$

$f_y = -3x$

$f_z = 4xz$

As the gradient vector is orthogonal to the surface at a given point, it is also orthogonal (normal) to the tangent plane at that point.

Then at point $(1, -1, 2)$, normal vector to the plane will be $(7, -3, 8)$

The general equation of the plane or the scalar equation is given by,

$a(x-x_0) + b(y-y_0) + c(z-z_0) = 0$

where $(a, b, c)$ is the normal vector to the plane and $(x_0, y_0, z_0)$ is a point on the plane.

So the equation of the plane is -

$7(x-1) - 3(y+1) + 8(z-2) = 0$

i.e. $7x - 3y + 8z = 26$

Answer 2

Consider the function $F(x, y, z) = 2xz^2-3xy-4x-7.$ The level surfaces of $F$ are perpendicular to the vector field $\nabla F=(F_x,F_y,F_z).$ Thus, the equation of the tangent plane of a level surface at $(x_0,y_0,z_0)$ is given by $$F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0.$$ In your case, it is $$7(x-1)-3(y+1)+8(z-2)=0.$$

In the particular case where we have $z=f(x,y)$ then the tangent plane at $(x_0,y_0)$ is given by $z=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0).$ Just consider $F(x,y,z)=z-f(x,y)$ and apply what was said in the paragraph above.