Past exam of the mathematical methods II course of the University of Oxford / 2019 resit / 2

The printed point is not on the printed surface, so the requested tangent plane does not exist. At $(2,-2,9)$, the left side of the defining equation is $x^{2}y=-8$, whereas the right side is $4z{e}^{x+y}-11=25$. This is a direct counterexample to the premise that the stated point belongs to the surface.

To make the obstruction precise, set $G(x,y,z)=x^{2}y-4z{e}^{x+y}+11$. The specified surface is the level set $G=0$, but $G(2,-2,9)=-33$. Although $\nabla G(2,-2,9)=(-44,-32,-4)$, the resulting plane $11x+8y+z=15$ is tangent to the different level set $G=-33$ through that point. It is not an answer for the printed surface. A tangent plane requires the point to lie on the surface.

If the intended correction preserves $x=2,y=-2$, the surface equation instead forces $z=3/4$. At the corrected point $p_{\ast }=(2,-2,3/4)$,The corresponding conditional tangent plane isThis conditional calculation identifies a possible repair; it does not silently replace the source's coordinate $9$.