Fair enough!

$\displaystyle \bigtriangledown f|_{(1,1,-1)}= (2yi+(2x-1)j) |_{(1,1-1)}=2i+j$

Hence the tangent plane is $\displaystyle 2(x-1)+(y-1)=0 \Rightarrow \ 2x+y=3$

Well that was nice! The rest of the question is tricky though!

This is where I start to have trouble.

The answer I have so far is $\displaystyle x_1=x+2t$, $\displaystyle y_1=y+t$ and $\displaystyle z_1=2xy-y$.

Where $\displaystyle t \in \mathbb{R}$ and $\displaystyle x_1,y_1$ and $\displaystyle z_1$ are the new x,y and z values respectively.

I've been trying to get $\displaystyle z_1$ in terms of $\displaystyle x_1$ and $\displaystyle y_1$, but I can't elminate t!

Call the tangent plane $\displaystyle g(x,y)=2x+y-3$.

Hence $\displaystyle \bigtriangledown g(x,y)=2i+j$.

This is where I start to get a little confused. I was hoping to set $\displaystyle \bigtriangledown g(x,y)$ to 0 since the x-y plane has a gradient of 0. However, $\displaystyle \bigtriangledown g(x,y)$ is a constant function so can never equal 0!

The problem here is that I appear to have three variables: $\displaystyle x,y$ and $\displaystyle z$. This seems to contradict the fact that this is a double integral.

Is z a variable or a constant? If z is a variable then is $\displaystyle z=2xy-y$?

I would appreciate on any of the parts to this question.

If you have any ideas or explanations to any parts, feel free to post!!