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Math Help - Integral calculus question.

  1. #1
    Super Member Showcase_22's Avatar
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    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!!
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    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} = <f_x,f_y,-1> where the derivatives are evaluated at some point P.
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    Super Member Showcase_22's Avatar
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    <br /> <br />
{\bf n} = <(f_x,f_y,-1><br />
    What does this mean?
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  4. #4
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    Quote Originally Posted by Showcase_22 View Post
    What does this mean?
    You mean < \;\cdot \; , \;\cdot \;, \;\cdot \;> - It's a vector.
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  5. #5
    Super Member Showcase_22's Avatar
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    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>=<t, 3-2t,-1>? (obtained by making x=t and rearranging 2x+y=3).

    What are the precise meanings of f_x and f_y?
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  6. #6
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    Quote Originally Posted by Showcase_22 View Post
    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>=<t, 3-2t,-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

     <br />
2(x-1) + (y-1) - (z+1) = 0<br />
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  7. #7
    Super Member Showcase_22's Avatar
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    <br /> <br />
2(x-1) + (y-1) - (z+1) = 0<br />

    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?
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