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One last vector calculus problem

  1. Sep 19, 2009 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Consider a hill that can be modelled by the function

    [tex]z=\frac{14-x^2-y^2}{3}[/tex]

    where the +ve x-axis points south and the +ve y-axis points east. Person A is standing at P=(1,2,3) on the hill.

    i) if A walks due west from P, does he ascend or descend the hill and at what rate?

    ii) at P, what is the direction of the steepest slope? If A climbs a path up the hill in this direction, at what angle (to the horizontal) does he initiallys start to climb at?


    2. Relevant equations

    Not sure what I need

    3. The attempt at a solution

    So since we are travelling east and north (in -ve y and z), I need to keep x fixed.

    So ∂z/∂y= -2y/3

    but since we are moving left, dz/dy = 2y/3 and so at P, the rate is ∂z/∂y= 2/3(2) =4/3 (I don't know what units to put here)

    I am stuck on part ii)
     
  2. jcsd
  3. Sep 20, 2009 #2

    HallsofIvy

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    You should know that the gradient vector, [itex]\nabla z[/itex] points in the direction of fastest increase. What is [itex]\nabla z[/itex]? What is its direction?
     
  4. Sep 20, 2009 #3

    rock.freak667

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    I actually did try that

    f(x,y,z)=14/3-x2/3-y2/3-z

    grad(f)= -2x/3 i - 2y/3 j - k

    at P=(1,2,3)

    grad(f)= -2/3 i - 4/3 j - k
     
  5. Sep 21, 2009 #4

    rock.freak667

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    Wait how come I am to find [itex]\nabla z[/itex] and not [itex]\nabla f [/itex] ?

    EDIT: I am stumped on the angle. I found the unit vector in the direction of the steepest slope to be


    [tex]\frac{-1}{\sqrt{5}} \hat{i} - \frac{2}{\sqrt{5}} \hat{j}[/tex]

    and dot product of this with <0,-1,0> does not give the answer of 56.1
     
    Last edited: Sep 21, 2009
  6. Sep 21, 2009 #5

    Dick

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    You are working with grad(z) because you looking at the rate of change of z as a function of the direction you are taking in the xy plane. You can get 56.1 degrees as the angle of ascent if you take the vector you just got and dot it with grad(z). That gives you the vertical rate of change. The horizontal rate of change is 1, since your direction is a unit vector. Use an arctan to find the angle.
     
  7. Sep 21, 2009 #6

    rock.freak667

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    ok so


    [tex]\left( \frac{-1}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\right) . \left( \frac{-2}{3}, \frac{-4}{3} \right)= \frac{2 \sqrt{5}}{3} cos \theta [/tex]

    which gives θ=90 :confused:
     
  8. Sep 21, 2009 #7

    Dick

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    No, no, no. The dot product of those two vectors is the vertical rate of motion. The horizontal rate of motion is 1, since your xy direction vector is a UNIT vector. That means you have the horizontal leg of a triangle and the vertical leg. I think you want to look at a tan function to find the angle from horizontal, right?
     
  9. Sep 21, 2009 #8

    rock.freak667

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    I understand this now, but one question shouldn't u.grad(f) at P give the directional derivative at P? or is it because I am using grad(z) instead of grad(f) that i get the vertical rate of motion?
     
  10. Sep 21, 2009 #9

    Dick

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    Your f(x,y,z)=0 is the level surface for the hill. Sure, u.grad(f) gives the directional derivative of f. But i) that's 0 in any direction tangent to the surface, which are the only directions you can move without tunneling or flying and ii) that's not the derivative you want you find. The question is asking for rates of change of elevation, which is z. Not f. Are you overtired?
     
  11. Sep 21, 2009 #10

    rock.freak667

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    I swear I forgot to think of this while doing this. Was to preoccupied remembering u.grad(f) instead of what that meant.

    Thanks for your help HallsofIvy and Dick
     
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