1. ## Integral calculus question.

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

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

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!

b). Write down a normal to S at $\displaystyle (x,y,2xy-y)$.
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!

c). At which point is the tangent plane parallel to the x-y plane?
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!

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

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!!

2. Originally Posted by Showcase_22
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!!
If $\displaystyle z = f(x,y)$ then the normal of the tangent plane is given by $\displaystyle {\bf n} = <f_x,f_y,-1>$ where the derivatives are evaluated at some point $\displaystyle P$.

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

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

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

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

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

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

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

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

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

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

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

7. $\displaystyle 2(x-1) + (y-1) - (z+1) = 0$

$\displaystyle 2x+y-z=4$

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

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

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

Is this right?