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 $\displaystyle \frac{\partial z}{\partial x}$ and $\displaystyle \frac{\partial z}{\partial y}$.

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

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

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

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