The height of a surface above ground level is given by $$z(x,y)=e^{6-(x^2+2y^2)}$$ where x and y are the horizontal distances going east and north respectively.
a) A small ball is placed on the surface at the point $x=2$,$y=1$ and is released from rest. in which direction ( in the horizontal plane) will it begin to roll? (Give your answer as a compass bearing)
b) An ant is placed on the surface at $x=2$, $y=1$. give the compass bearing it should follow, initially, in order to follow a level path.
c) Find the equation of the plane tangential to the surface at $x=2$ , $y=1$, in the form, $ax+by+cz=d$
Attempt a) my thought process for this was to apply the $\nabla$ operator on $z(x,y)$ as follows: $$\nabla z=\frac{\partial z}{\partial x}\textbf{i}+\frac{\partial z}{\partial y}\textbf{j}+\frac{\partial z}{\partial z}\textbf{k}$$
$$=-2xe^{6-(x^2+2y^2)}\textbf{i}-4ye^{6-(x^2+2y^2)}\textbf{j}$$
Then at the point $y=1$, $x=2$
$$\nabla z=-2(2)e^{6-((2)^2+2(1)^2)}\textbf{i}-4(1)e^{6-((2)^2+2(1)^2)}\textbf{j}$$
$$\nabla z=-4\textbf{i}-4\textbf{j}$$ Hence the small ball will be moving south-west at compass bearing of $225°$.
I just wanted clarification if my attempt at a) is correct and was confused as to how to approach b) and c), thanks in advance for any help
Thats fine, although in your expression for $\nabla z$ there should be no term $\frac{\partial z}{\partial z}$ (What would that even mean?)
A very useful property of the gradient vector is that if you move in the horizontal plane by small distances $dx,dy$, then the vertical movement is the inner product of the gradient vector with the vector $(dx,dy)$, ie $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial x}{\partial y} dy$$
and your mouse wants to make dz=0. As for the plane, let it be given by $$z=ax+by+c$$ and you want to choose $a,b,c$ so that at the place concerned, this plane has the same gradient and the same elevation as the given surface.