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Math Help - Finding directional derivative, tangent and path of maximal increase

  1. #1
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    Finding directional derivative, tangent and path of maximal increase

    Hi, I have the following question:

    Consider the function f(x,y,z)=x^3+y^4-z^3

    1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

    This is the question I am unsure of.
    However I have:
    grad f=(3x^2)i+(4y^3)j-(3z^2)k



    2) Compute the directional rerivative at point (0,1,1) in the direction of u=i+j+k

    I found this answer to be 1/sqrt(3)

    3) Suppose G(x,y)=f(x,y,0)=x^3+y^4 describes the profit of a company as a measure of two variables. What should be your best strategy to modify y as a function of x in order to maximise the profit if you know that initially x=1, y=1

    I got the path of maximal increase to be y=(y^4)/(3x^2) + (2/3)
    However I am unsure what to say about modifying y.


    Cheers.
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  2. #2
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    Quote Originally Posted by BlueEagle View Post
    Hi, I have the following question:

    Consider the function f(x,y,z)=x^3+y^4-z^3

    1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

    This is the question I am unsure of.
    However I have:
    grad f=(3x^2)i+(4y^3)j-(3z^2)k



    2) Compute the directional rerivative at point (0,1,1) in the direction of u=i+j+k

    I found this answer to be 1/sqrt(3)
    The directional derivative at a point p = (0,1,1) in the direction u = (1,1,1) is just \nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.
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  3. #3
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    Quote Originally Posted by JakeD View Post
    The directional derivative at a point p = (0,1,1) in the direction u = (1,1,1) is just \nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.
    Aren't you supposed to use the unit vector of u?
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  4. #4
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    Quote Originally Posted by JakeD View Post
    The directional derivative at a point p = (0,1,1) in the direction u = (1,1,1) is just \nabla f(p) \cdot u = (0,4,-3)^T(1,1,1) = 1.
    Quote Originally Posted by BlueEagle View Post
    Aren't you supposed to use the unit vector of u?
    OK, I agree. I gave the definition of the directional derivative along u instead of the directional derivative in the direction of u.
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  5. #5
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    Quote Originally Posted by BlueEagle View Post
    Hi, I have the following question:

    Consider the function f(x,y,z)=x^3+y^4-z^3

    1) Find the points on the surface f(x,y,z)=1 that have horizontal tangent plane (tangent plane parallel to the xy plane).

    This is the question I am unsure of.
    However I have:
    grad f=(3x^2)i+(4y^3)j-(3z^2)k
    A tangent plane parallel to the xy plane has unit normal (0,0,1). The gradient is normal to the tangent plane. So grad f must be a multiple of (0,0,1), implying x = y = 0, f(0,0,z) = 1, -z^3 = 1 and thus z = -1. Hence (0,0,-1) is the desired point.
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