1. ## Integral calculus question.

let S be the graph of $f(x,y)=2xy-y$ over the square $[0,2] \times [0,2]$, ie. $S:=\{(x,y,2xy-y)| 0 \leq x \leq 2, 0 \leq y \leq 2 \}$.
Fair enough!

a). Find an equation of the tangent plane to S at $(1,1,-1).$
$\bigtriangledown f|_{(1,1,-1)}= (2yi+(2x-1)j) |_{(1,1-1)}=2i+j$

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

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

b). Write down a normal to S at $(x,y,2xy-y)$.
This is where I start to have trouble.

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

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

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

c). At which point is the tangent plane parallel to the x-y plane?
Call the tangent plane $g(x,y)=2x+y-3$.

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

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

d). Evaluate $\int \int_S \frac{z}{\sqrt{1+2x^2+2y^2-2x}}~ dA$

The problem here is that I appear to have three variables: $x,y$ and $z$. 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 $z=2xy-y$?

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!!

2. Originally Posted by Showcase_22
Fair enough!

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

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

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 $x_1=x+2t$, $y_1=y+t$ and $z_1=2xy-y$.

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

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

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

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

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

The problem here is that I appear to have three variables: $x,y$ and $z$. 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 $z=2xy-y$?

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!!
If $z = f(x,y)$ then the normal of the tangent plane is given by ${\bf n} = $ where the derivatives are evaluated at some point $P$.

3. $

{\bf n} = <(f_x,f_y,-1>
$
What does this mean?

4. Originally Posted by Showcase_22
What does this mean?
You mean $< \;\cdot \; , \;\cdot \;, \;\cdot \;>$ - It's a vector.

5. Ah, I thought it spanned something (that's my excuse anyway.......).

Here $P=(x,y,2xy-y)$.

So ${\bf n}=< f_x,f_y,-1>=$? (obtained by making $x=t$ and rearranging $2x+y=3$).

What are the precise meanings of $f_x$ and $f_y$?

6. Originally Posted by Showcase_22
Ah, I thought it spanned something (that's my excuse anyway.......).

Here $P=(x,y,2xy-y)$.

So ${\bf n}=< f_x,f_y,-1>=$? (obtained by making $x=t$ and rearranging $2x+y=3$).

What are the precise meanings of $f_x$ and $f_y$?
They are partial derivatives. So ${\bf n}=< f_x,f_y,-1>=<2y,2x-1,-1>$ and at $P(1,1,-1)$ ${\bf n}=<2,1,-1>$ so the tangent plane is

$
2(x-1) + (y-1) - (z+1) = 0
$

7. $

2(x-1) + (y-1) - (z+1) = 0
$

$2x+y-z=4$

For c), I need to find when this is parallel to the x-y plane.

The x-y plane has an equation $z=0$.

$z=2x+y-4 \Rightarrow \ x=2, y=0$ so the point where the tangent plane is parallel to the x-y plane is $(2,0,0)$.

Is this right?