Consider the surface $S$ in $\mathbb{R}^3$ 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 $(0, 2)$.
Can someone guide me through solving this problem? I am confused on where to start because we didn't cover this in class. Thanks so much
The equation for the tangent plane at the point $(x_0,y_0,z_0)$ is:
$$z-z_0=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$
As an example for point $P_1(0,0)$:
$$f_x = y^2+e^x \implies f_x(0,0)=1$$ $$f_y = 2xy-4y \implies f_y(0,0)=0$$ $$z_0 = f(0,0)=1$$
Thus the equation of the plane through $P_1(0,0)$ is:
$$z-1=1\cdot(x-0)+0 \cdot(y-0) \implies x-z=-1$$
Similary the equation of the plane through $P_2(0,1)$ is:
$$z+1=2\cdot(x-0)-4 \cdot(y-1) \implies 2x-4y-z=-3$$
Finally the equation of the plane through $P_3(0,2)$ is:
$$z+7=5\cdot(x-0)-8 \cdot(y-2) \implies 5x-8y-z=-9$$
Once you have found the equations for the three planes you need to solve the system of linear equations to find the intersection point.
For the solution of the system of equations as an example take a look here: Consistency of a System of linear equations
The final solution should be: P(-2,0,-1).
NOTE check calculation detail by yourself trying to understand each step