Consider the surface $S$ in $\Bbb R^3$ given by the graph of $f(x,y)$. Find the intersection point of the tangent planes.

Question

Consider the surface $S$ in $\Bbb R^3$ given by the graph of $f(x,y) = xy^2 − 2y^2 + e^x$.

Calculate the intersection point of the tangent planes to $S$ above $(x, y) = (0, 0)$, $(0, 1)$, and $(0, 2)$.

I am confused about what this question is asking. Would I have to choose a tangent plane above that point and see where it intersects $S$? Thanks

Answer 1

• The normal vector for $\{(x,y,xy^2-2y^2+e^x) \in \mathbb{R}^3\}$ is given by $N(x,y)= (f_x(x,y),f_y(x,y),-1)$.

• If you want to find the tangent plane above $(0,0)$ then that's the tangent plane at $(0,0,1)$. The plane is given by $N(0,0) \cdot (x-0,y-0,z-1) = 0$.

• Similarly for a plane above $(x_0,y_0)$ you have,

$$\textbf{Plane}: N(x_0,y_0) \cdot (x-x_0,y-y_0, z-f(x_0,y_0))$$

• Once you get the equations of these three planes, you'll have a system of three equations with three unknowns say,

$$ a_{11} x + a_{12} y + a_{13} z = b_1$$

$$ a_{21} x + a_{22} y + a_{23} z = b_2$$

$$a_{31} x + a_{32} y + a_{33} z = b_3$$