# Thread: Equation of the tangent plane to the given surface

1. ## Equation of the tangent plane to the given surface

Find an equation of the tangent plane to the given surface at the specified point.
z = ex2 - y2, (1, -1, 1)

2. Originally Posted by acg716
Find an equation of the tangent plane to the given surface at the specified point.
z = ex2 - y2, (1, -1, 1)
What does ex2 mean? You have to be a whole lot clearer in stating your equations if you expect to get effective help.

3. i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2

4. Originally Posted by acg716
i'm really sorry! when i posted it it looked correct, but e is raised to x^2-y^2
First get the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.

Now construct the vector $\frac{\partial z}{\partial x} \, \vec{i} + \frac{\partial z}{\partial y} \, \vec{j} - \vec{k}$.

Substitute $x = 1$ and $y = -1$ into this vector. This gives a normal of the form $a \vec{i} + b \vec{j} - \vec{k}$ to the surface at $(1, -1, 1)$.

The equation of the tangent plane is therefore $ax + by - z = d$.

To get the value of $d$, substitute $x = 1$, $y = -1$ and $z = 1$.