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Math Help - Multivariable Calculus - Minimum value along a curve

  1. #1
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    Multivariable Calculus - Minimum value along a curve

    3. Let g(x,y)=x2+y3.
    (a) Find the directional derivative of
    g at the point (−3, 2) in the direction (3, 5).

    (b) Determine the tangent line to the curve g(x, y) = g(4, −2) at the point (4, −2).
    (c) Find the tangent plane to the surface
    z = g(x, y) at the point (−1, 1, 2).
    (d) What is the minimum of g along the curve x2 + y2 = 4/9?






    Hey there! Looking for help on part (d) but don't know if the other information is relevant so I just posted the whole question.
    Any help is appreciated!

    - Maedbh
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  2. #2
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    Re: Multivariable Calculus - Minimum value along a curve

    You could eliminate the x^2 in g(x,y) and differentiate with respect to y but you are probably supposed to be using Lagrange multipliers. It's a very easy example. Check your notes. You can use the other method as a check.
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  3. #3
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    Re: Multivariable Calculus - Minimum value along a curve

    You are asked to find the max or min of g(x,y)= x^2+ y^3 on the curve f(x,y)= x^2+ y^2= 4/9. Look at the two gradients: \nabla g= 2x\vec{i}+ 3y^2\vec{j} and \nabla f= 2x\vec{i}+ 2y\vec{j}. \nabla g points in the direction of greatest increase of g so you can find a maximum value by moving in the direction of \nabla g until it is 0. But, here, you can't move in that direction because you must stay on the curve f(x,y)= 4/9. What you can do is move along the curve in the direction of the projection of \nabla g on the curve- again until that projection is 0. That happens when \nabla g is perpendicular to the curve. And since \nabla f is perpendicular to the curve f(x,y)= constant, you want those two gradients parallel- that is, one is a multiple of the other. You want \nabla g= \lambda \nabla f for some constant \lambda. That is the "Lagrange multiplier" a tutor refers to. Surely, you seen that in class or your textbook.
    Last edited by HallsofIvy; December 13th 2012 at 07:37 AM.
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