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Math Help - Tangent Planes + Linear Approximation

  1. #1
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    Tangent Planes + Linear Approximation

    1. Find the points on the graph of z = x^2e^y at which the tangent plane is parallel to 5x - 2y + \frac{1}2z = 0
    Can someone explain this to me/start this for me?

    2. Suppose that the plane tangent to the surface z = f(x,y) at (-2,3,4) has equations z + 4x + 2y = 2. Estimate f(-2.1, 3.1)
    Same as above; I'm not really sure where I'm supposed to start. Do I integrate -(4x + 2y) + 2 or something?
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  2. #2
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    So here's what I was vaguely told:
    1. Tangent Plane is always z = L( x,y )
    2. Knowing that "think of when they are parallel"

    So if a plane is parallel that just means that it has an extra + c or something too?
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  3. #3
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    1. Use gradient vectors:

    F(x,y,z) = x^2e^y-z = 0
    Let \vec{u_1} be the unit vector of <10,-4,1> (I multiplied all of the components by 2) and let \vec{u_2} be the unit vector of the gradient vector of F(x,y,z)
    Set the components of  \vec{u_2} and \vec{u_1} equal to each other and solve for the values. You get 3 equations and you have 3 variables.
    An alternative method would be to solve both equations (z=x^2e^y and 10x-4y+z=0) for z and set them equal to each other. See where that gets you.

    2. Just plug in -2.1 for x and 3.1 for y in the equation z+4x+2y=2 and solve for z.
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  4. #4
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    Quote Originally Posted by ZeroVector View Post
    So here's what I was vaguely told:
    1. Tangent Plane is always z = L( x,y )
    2. Knowing that "think of when they are parallel"

    So if a plane is parallel that just means that it has an extra + c or something too?
    No, it doesn't.

    Two planes, Ax+ By+ Cz= D (which would be the same as z= (D- Ax- By)/C) and Px+ Qr+ Rz= S (which would be the same as z= (S- Px- Qy)/R) are parallel if and only if their normal vectors, <A, B, C> and <P, Q, R> are parallel- and that means that one is a multiple of the other.

    A plane, Ax+ By+ Cz= D is tangent to a surface f(x, y, z)= constant, if and only if the normal vector to the plane, <A, B, C>, is parallel to the normal vector of the surface which is the same as the gradient of the function that lilaziz1 mentioned: \left<\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y},\frac{\partial f}{\partial z}\right>
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