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Find two angles where the directional derivative is 1 at p0

  1. Feb 18, 2015 #1
    1.

    Given a function f(x,y) at (x0,y0). Find the two angles the directional derivative makes with the x-axis, where the directional derivative is 1. The angles lie in (-pi,pi].

    2.


    f(x,y) = sec(pi/14)*sqrt(x^2 + y^2)
    p0 = (6,6)

    3.

    I use the relation D_u = grad(f) * u, where u is the elementary vector <cos(theta),sin(theta)>.

    grad(f) at (6,6) is <sec(pi/14)/sqrt(2), sec(pi/14)/sqrt(2)>

    Using these we have D_u = 1 = sec(pi/14)/sqrt(2)*(cos(theta) + sin(theta))

    Rearranging:

    cos(theta) + sin(theta) = sqrt(2) * cos(pi/14)

    Isolating theta on LHS by using a relevant angle formula:

    sqrt(2)*cos(theta - gamma) = sqrt(2) * cos(pi/14), where gamma = atan(1). Here, atan(1) can be pi/4 or -3*pi/4.

    using cos^-1 on L- and RHS.

    theta = pi/14 + gamma

    Giving the answers:

    theta1 = pi/14 + pi/4 = 9*pi/28
    theta2 = pi/14 - 3*pi/4 = -19*pi/28

    Somehow these angles are incorrect, but I am unable to locate my error in calculating them. Any help in guidance in the right direction will be greatly appreciated.
     
  2. jcsd
  3. Feb 18, 2015 #2

    Dick

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    Science Advisor
    Homework Helper

    9*pi/28 is corrrect. Things start getting sloppy in the trig part. sqrt(2)(cos(t)+sin(t))=cos(t+pi/4). The other angle for atan(t)=1 doesn't work. It gives negative signs on the sin and cos. And watch out when you use inverse cos. There's more than one angle on (-pi,pi) that has the same cos.
     
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