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Math Help - Equation of the plane tangent to the surface

  1. #1
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    Equation of the plane tangent to the surface

    Hi,

    I tried to this way:

    <br />
5u^2 = 5
    <br />
u = 1 <br />

    <br />
5v^2 = 45
    <br />
v = sqrt(9) <br />

    when u = 1 and v = -3
    <br />
2u - v^2 = -7
    <br />
2(1) - (-3)^2 = -7 <br />

    when u = 1 and v = 3
    <br />
2u - v^2 = -7
    <br />
2(1) - (3)^2 = -7 <br />

    Therefore, P lies on the surface.. now how do you find the equation? please help...

    Problem:



    Find the equation of the plane tangent to the surface





    at the point P that is (approximately) P(5,-7,45).





    many thanks
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  2. #2
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    Quote Originally Posted by althaemenes View Post
    Hi,

    I tried to this way:

    <br />
5u^2 = 5
    <br />
u = 1 <br />

    <br />
5v^2 = 45
    <br />
v = sqrt(9) <br />

    when u = 1 and v = -3
    <br />
2u - v^2 = -7
    <br />
2(1) - (-3)^2 = -7 <br />

    when u = 1 and v = 3
    <br />
2u - v^2 = -7
    <br />
2(1) - (3)^2 = -7 <br />

    Therefore, P lies on the surface.. now how do you find the equation? please help...

    Problem:









    Find the equation of the plane tangent to the surface











    at the point P that is (approximately) P(5,-7,45).





    many thanks
    First we take the partial to find to vectors tangent to the surface.

    \vec r(u,v) = 5u^2 \vec i +(2u-v^2)\vec j+5v^2 \vec k

    r_u(u,v)=10u \vec i + 2 \vec j

    r_v(u,v)=-2v \vec j +10v \vec k

    r_u(1,3)=10(1) \vec i + 2 \vec j =10 \vec i +2 \vec j

    r_v(1,3)=-2(3) \vec j +10(3) \vec k=-6 \vec j + 30 \vec k

    Now we take the cross product of these two vectors to get a Normal vector to the surface.

    Can you finish from here?

    Good luck.
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  3. #3
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    reply

    Quote Originally Posted by TheEmptySet View Post
    First we take the partial to find to vectors tangent to the surface.

    \vec r(u,v) = 5u^2 \vec i +(2u-v^2)\vec j+5v^2 \vec k

    r_u(u,v)=10u \vec i + 2 \vec j

    r_v(u,v)=-2v \vec j +10v \vec k

    r_u(1,3)=10(1) \vec i + 2 \vec j =10 \vec i +2 \vec j

    r_v(1,3)=-2(3) \vec j +10(3) \vec k=-6 \vec j + 30 \vec k

    Now we take the cross product of these two vectors to get a Normal vector to the surface.

    Can you finish from here?

    Good luck.
    The cross product is = 60i - 300j + 12k

    Therefore, equation at (5,-7,45) =

     60(x-5)-14( y-(-7))+12(z-45) = 0
     60x-300-14y+98+12z-540 = 0
     60x+14y+12z-742 = 0

    Now, how do you format it in z(x,y) form? should I take z to the other side of the equation?

    Please help
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by althaemenes View Post
    The cross product is = 60i - 300j + 12k

    Therefore, equation at (5,-7,45) =

     60(x-5)-14( y-(-7))+12(z-45) = 0
     60x-300-14y+98+12z-540 = 0
     60x+14y+12z-742 = 0

    Now, how do you format it in z(x,y) form? should I take z to the other side of the equation?

    Please help
    Yes that is how you would isolate z to get a function of x and y.
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  5. #5
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    Thanks!!

    Thanks...
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