Find the directional derivative of the function f(x,y,z)=𝑧^2 -x𝑦^2 at the point P(2,1,3)  in the direction of the vector v = < -1, 2, 2 >.?

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  • 4 weeks ago
    Favorite Answer

    f(x, y, z) = z² – xy²

    grad(f(x, y, z)) = <∂f/∂x, ∂f/∂y, ∂f/∂z>

    = <-y², -2xy, 2z>

    At the point P(2 ,1, 3) the gradient vector, g, is:

    g = grad(f(2, 1, 3)) = <-1², -2*2*1, 2*3> = <-1, -4, 6>

    The unit vector in the direction v=<-1, 2, 2> is:

    u = v/||v|| = <-1, 2, 2>/√((-1)² + 2² + 2²) = <1, 2, 2>/√9

    = ±(1/3)<-1, 2, 2>

    I will use the positive (+) solution.

    Now find the dot product of g and u:

    g•u = <-1, -4, 6>•(1/3)<-1, 2, 2>

    = [(-1)*(-1) + (-4)2 + 6*2]/3

    = (1 - 8 + 12)/3 = 5/3

  • Ian H
    Lv 7
    4 weeks ago

    Explanation of general concept for a hill drawn on a map in contours.

    If we imagine standing at a particular point on the side of a hill,

    facing along a level curve, (a contour), the slope would be zero.

    The maximum gradient, grad f(x, y) at that point would be vector perpendicular to the level curve and this is like the steepest slope downhill.

    At the same point, but facing in any other direction, (specified by a UNIT vector), the slope will be some component of grad f(x, y), calculated as the dot product of the two vectors. That is known as the directional derivative D.

    f(x, y ,z) = z^2 – xy^2

    We can apply the same method with these 3 variables

    grad[f(x, y, z)] = <∂f/∂x, ∂f/∂y, ∂f/∂z>

    grad f = <-y^2, -2xy, 2z> ....... and this depends on position

    At the particular point P(2 ,1, 3) the substituted values give

    grad[f(2, 1, 3)] = <-1², -2*2*1, 2*3>

    grad f at P is <-1, -4, 6>

    We now need to calculate the UNIT vector u associated with v <-1, 2, 2>

    √(1^2 + 2^2 + 2^2) = √(9) = 3, so,

    u = (1/3)<-1, 2, 2>

    Next recall that the result of calculating the dot product <a, b, c>•<p, q, r>

    is the scalar ap + bq + cr

    D = grad[f(2, 1, 3)] • u = (1/3)<-1, -4, 6><-1, 2, 2>

    D = (1/3)(1 - 8 + 12) = 5/3

    Steve4Physics found this first.

  • rotchm
    Lv 7
    4 weeks ago

    Just apply the formula you saw for this.

    ∇f • u, where u is the unit vector along your given direction; u = v/|v|.

    Or, did you see a different formula for this?

    Show your steps here and we will help you through.

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. 

  • Vaman
    Lv 7
    4 weeks ago

    Gradients are directional derivatives.

    Along x axis, df/dx =-y^2, y axis , df/dy=-2y x, along z axis df/dz=2z.

    P(2,1,3) will be (-1, -4, 6). So you get p dot v= (1-8+12)=5. This must be the answer.

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