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Multivariable Calculus 1 Problem

  1. Jun 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Let P be the tangent to the graph of g(x,y) = 8-2x^2-3y^2 at the point (1, 2, -6). Let f(x,y) = 4-x^2-y^2. Find the point on the graph of f which has tangent plane parallel to P.


    2. Relevant equations
    g(x,y) = 8-2x^2-3y^2 at (1, 2, -6)
    f(x,y) = 4-x^2-y^2

    3. The attempt at a solution
    I found P, which is P = 4x+12y-34
     
    Last edited by a moderator: Jun 29, 2015
  2. jcsd
  3. Jun 29, 2015 #2

    Mark44

    Staff: Mentor

    How did you find the plane P? Note that P is just a name for the plane - P should not be in the equation of the plane.
     
  4. Jun 29, 2015 #3
    Took df/dx, df/dy, and I know that g(x,y) = 8-2x^2-3y^2. I also followed the general formula, which is: f(x,y)+[df/dx(x,y)](x-x0)+[df/dy(x,y)](y-y0). In the derivatives, I plugged in the point.
     
  5. Jun 29, 2015 #4

    HallsofIvy

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    You mean, I presume, z= 4x+ 12y- 34. But that is incorrect. The first, g, graph is z= 8- 2x^2- 3y^2 or 2x^2- 3y^2+ z= 8.

    Mod note: some text removed as too much help.
     
    Last edited by a moderator: Jun 29, 2015
  6. Jun 29, 2015 #5
    Hallsoflvy, no that is the equation for P, not z. z is g(x,y).
     
  7. Jun 29, 2015 #6

    Mark44

    Staff: Mentor

    What HallsOfIvy was saying is that your equation of the plane is incorrect.

    This is really confusing.

    To find the equation of the plane that is tangent to the graph of g, you should be working with the partial derivatives ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}##, not "df/dx" and "df/dy" as you wrote. And in general formula should involve g and its two partial derivatives, not f.

    So try again for the equation of the tangent plane. The equation should involve terms with x, y, and z, and should NOT have P in it.
     
  8. Jun 29, 2015 #7
    Mark, I mean the partial derivative, not the actual full one. I just don't know how to type it in here. P is just a generic variable for the equation.
     
  9. Jun 29, 2015 #8
    Regardless of the equation of P, the point at which the plane tangent to f is parallel to P will be where the x and y partial derivatives of f are equal to those of g at the point (1,2,-6), right?
     
  10. Jun 29, 2015 #9
    Albuser, yes.
     
  11. Jun 29, 2015 #10
    So why not simply take the partial derivatives of g at that point, find the general partial derivatives of f for any x and y and set the two equal to each other?
     
  12. Jun 29, 2015 #11

    Mark44

    Staff: Mentor

    And those are what I wrote, partial derivatives of g, ##\frac{\partial g}{\partial x}## and ##\frac{\partial g}{\partial y}##. If you're working with g, you shouldn't write the partials of the general formula in terms of f. That was my point.
    P is NOT a variable. It is only a symbolic name for the plane. You could call the plane Fred instead of P, but the equation of the plane will not involve Fred or P.
     
  13. Jun 29, 2015 #12

    Mark44

    Staff: Mentor

    The OP has already found the partial derivatives, and has attempted to find the equation of the tangent plane, but has an error in his equation of the plane.
     
  14. Jun 29, 2015 #13
    Albuser, I have already found the equation tangent to g. However, I'm not sure how I can make the partial derivatives equal to g and solve from there. It would be really difficult and I would not get a concrete answer.
     
  15. Jun 29, 2015 #14

    Mark44

    Staff: Mentor

    You found an equation, but as has already been pointed out, you have an error in it.
     
  16. Jun 29, 2015 #15
    Okay so can you guys point out the error?
     
  17. Jun 29, 2015 #16

    Mark44

    Staff: Mentor

    Actually, your equation looks OK. I'm not sure what HallsOfIvy was talking about.

    The equation for P is z = 4x + 12y - 34

    A normal for this plane is <4, 12, -1>
    What do you get for the gradient of your other function, z = f(x, y) = 4 - x2 - y2?

    The goal here is to find the point on the surface z = f(x, y) so that the gradient is equal to <4, 12, -1>.
     
  18. Jun 29, 2015 #17
    Mark, this is exactly where I got stuck: on where to go after I found the gradient of g(x,y); and how I can connect that to f(x,y) in order to find what the point f is. The gradient of f(x,y) is:

    ∂f/∂x = -2x
    ∂f/∂y = -2y
    f(x,y) = z = 4 - 2x2 - 2y2

    That is as far as I got and then I didn't know how to proceed.

    All I know is that the general form is

    f(x0, y0) + [∂f/∂x(x0, y0)] (x - x0) + [∂f/∂y(x0, y0)] (y - y0)
     
    Last edited: Jun 29, 2015
  19. Jun 29, 2015 #18

    HallsofIvy

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    No, you are the one who is wrong. You wrote P= 4x+ 12y- 34. That is impossible. P, on the left, is a plane while 4x+ 23y- 34, on the right, is a number. The cannot be equal, they are different kinds of things.

    What is true is that the plane, P, is given by the equation z= 4x+ 12y- 34.
     
  20. Jun 29, 2015 #19
    Halls, that is what he said. I guess I just didn't make myself clear. Sorry about that.
     
  21. Jun 29, 2015 #20

    Mark44

    Staff: Mentor

    What you wrote is incomplete. It is supposed to be an equation, but since there is no =, it's not an equation.

    For a function z = f(x, y), the tangent plane at the point (x0, y0) is
    z = f(x0, y0) + [∂f/∂x(x0, y0)] (x - x0) + [∂f/∂y(x0, y0)] (y - y0)

    A normal to your tangent plane at (1, 2, -6) is <-4, -12, 1>, and these numbers represent the partials of g with respect to x, and y, and the partial of z with respect to z, respectively. For the tangent plane to the graph of f, you want the partials to be equal to these numbers.
     
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