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Finding Angles Between Horizontal Planes and Multivariable Functions

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose a mountain is described by the function z = 10x^2 * y − 5x^2 − 4y^2 − x^4 − 2y^4 and that you are standing at the point (1,1,−2). The positive x-axis points east and the positive y- axis points north. If you walk in the northeast direction what angle above the horizontal does your path make?


    2. Relevant equations
    cos(θ) = (v • w)/(|v||w|) (But I'm not sure this is even relevant.)
    z - z0 = fx(x0,y0)(x-x0) + fy(x0,y0)(y-y0)

    3. The attempt at a solution
    I am not convinced that this will work, but I intend to find a tangent plane at (1,1,-2). Then, using the normal vectors I have between the tangent and the xy-plane, I will calculate the angle.

    fx = 20xy - 10x - 4x^3
    fy = 10x^2 - 8y - 8y^3

    Evaluated at (1,1) gives,

    fx = 6
    fy = -6

    so,

    z + 2 = 6(x-1) - 6(y-1), which I rearrange to give,

    0 = 6(x-1) - 6(y-1) - 1(z + 2)

    So, I think I have normal vector <6,-6,-1>, and a normal vector on the xy-plane, <0,0,1>.

    So,

    cos(θ) = (-1)/sqrt(73). So θ = arccos((-1)/sqrt(73)), which is obtuse, so, replacing <0,0,1> with its negative gives θ = arccos((1)/sqrt(73))
     
    Last edited: Oct 15, 2011
  2. jcsd
  3. Oct 16, 2011 #2

    HallsofIvy

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    Science Advisor

    What you need is the directional derivative: if [itex]D_\theta f(x, y)[/itex] is the derivative of f(x,y) in the direction that makes angle [itex]\theta[/itex] with the positive x-axis (so that [itex]\partial f/\partial x= D_0 f[/itex] and [itex]\partial f/\partial y= D_{\theta/2} f[/itex], then
    [tex]D_\theta f= \frac{\partial f}{\partial x} cos(\theta)+ \frac{\partial f}{\partial y} sin(\theta)[/tex]

    Similarly, if [itex]\vec{v}[/itex] is a unit vector the derivative in the direction of that vector is
    [tex]\nabla f \cdot \vec{v}[/tex].
    "Northeast", given that the positive x-axis is east and the positive y-axis north, is at angle [itex]\theta= \pi/4[/itex] which is the same as the direction of unit vector [itex]\vec{v}= (\vec{i}+ \vec{j})/\sqrt{2}[/itex]

    Once you know the "rate of change", "rise over run", in that direction, just take the inverse tangent to get the angle.

    No, [itex]\theta[/itex] is NOT "[itex]arccos(1)/\sqrt{73}[/itex]".

    "fx = 6 fy = -6" should give you the answer immediately.
     
  4. Oct 16, 2011 #3
    So, by the definition of the directional derivative: ∇f ⋅ v⃗. Then my directional derivative should be <6,-6> • <1/sqrt(2),1/sqrt(2)>. This gives a dot product of 0.

    Arctan(0) = 0, which does not seem correct to me.
     
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