math question?

some parts are wrong

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  • Anonymous
    4 months ago

    A and B are OK...

    I’ll use the format <a, b> for ai + bj.

    ____________________

    C.

    The unit vector in the direction of v is u = v/|v| = <-3, 3>/√(-3)² + 3²)

    = 3<-1, 1>/√18 = 3<-1, 1>/(3√2) = (√2/2)<-1, 1>

    The directional derivative in the direction of v is the dot product ∇f•u at P:

    ∇f•u = <-1/2, -1/4>•(√2/2)<-1, 1>

    = (√2/2)((-1/2)(-1) + (-1/4)(1))

    = (√2/2)(1/2 - 1/4)

    = (√2/2)(1/4)

    = √2/8

    ____________________

    D.

    The max. rate of change is simply the magnitude of ∇f

    |∇f | = √((-1/2)² + (-1/4)²) = √(1/4 + 1/16) = √(5/16) = √5/4

    ____________________

    E.

    This is simply the unit vector in the direction of ∇f

    = ∇f/|∇f| = <-1/2, -1/4>/(√5/4)

    = <-√5/8, -√5/16>

  • hfshaw
    Lv 7
    4 months ago

    The directional derivative of a function f(x,y) in the direction of a vector v = a*i + b*j at the point P = (m,n) is given by:

    D_v f(P) = ∇f(m,n) · v,

    i.e., the directional derivative is equal to the dot product of the gradient of f and the directional vector.

    In your case, you have already found that

    ∇f(-1,2) = (-1/2)*i + (-1/4)*j

    Take the dot product of this with the vector v:

    D_v f(P) = (-1/2)*(-3) + (-1/4)*(3) = 3/2 - 3/4 = 3/4. This is the answer for part C.

    The maximum rate of change at a point is simply the magnitude of the gradient at that point, so the answer for part D is:

    |∇f(-1,2)| = sqrt((-1/2)² + (-1/4)²) = sqrt(1/4 + 1/16) = sqrt(5/16) = sqrt(5)/4.

    The direction of the maximum rate of change of a function at a point is given by the gradient vector at that point. To convert this to a unit vector, we normalize by the magnitude of the gradient:

    u = ∇f(-1,2)/|∇f(-1,2)| = [(-1/2)*i + (-1/4)*j]/(sqrt(5)/4) = -2/sqrt(5))*i + (-1/sqrt(5))*j

    This is the answer to part E.

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