1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial Derivative Problem

  1. Jan 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Find an equation for the path of a particle that starts at P(10,10) and always moves in the direction of maximum temperature increase if the temperature in the plane is T(x,y) = 400-2x^2 -y^2


    2. Relevant equations
    T(x,y) = 400-2x^2 -y^2
    dT/dx = -4x
    dT/dy = -2y
    d^2T/dx^2 = -4
    d^2T/dy^2 = -2
    maximum is at (0,0)=400


    3. The attempt at a solution
    I'm not sure how to solve this because the maximum temperature increase (second derivative) is always -4 in the x and -2 in the y.
     
  2. jcsd
  3. Jan 26, 2014 #2

    CAF123

    User Avatar
    Gold Member

    Yes, the maximum temperature is at the coordinates (0,0), however you are looking for the path that takes you from (10,10) to there via the path of greatest temperature. Have you come across the gradient vector?
     
  4. Jan 26, 2014 #3

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What is the gradient ?
     
  5. Jan 26, 2014 #4
    The gradient plane at (10,10) is -40x-20y+700=z
     
  6. Jan 26, 2014 #5

    CAF123

    User Avatar
    Gold Member

    What is the gradient vector?
     
  7. Jan 26, 2014 #6
    The gradient vector at (10,10):
    dF(x,y)=(-4x,-2x) = (-40, -20)
     
  8. Jan 26, 2014 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Okay, the gradient vector at any point (x, y) is <-4x, -2y> and at (10, 10) the gradient vector is <-40, -20>. But no one asked about the gradient vector at (10, 10)! What is important is that the gradient points in the direction of maximum increase so you want dy/dx to give a curve in the same direction as the gradient. A vector <a, b> has length a in the x-direction and length b in the y direction.

    Of course, the derivative, dy/dx, is the slope of the tangent line so that dy/dx= b/a.
     
  9. Jan 26, 2014 #8


    Oh okay, so once you obtain dy/dx which is 2x/y, how do you make sure it starts at (10,10)?
     
  10. Jan 26, 2014 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    How do you write the equation of a path in general?
     
  11. Jan 27, 2014 #10

    (x,y)=(10,10) + t(2,1)
     
  12. Jan 27, 2014 #11

    pasmith

    User Avatar
    Homework Helper

    That must be wrong: it heads away from the global maximum at the origin as [itex]t[/itex] increases (and doesn't pass through the origin at all).

    You need to solve
    [tex]
    \dot x = \frac{\partial T}{\partial x} = -4x, \\
    \dot y = \frac{\partial T}{\partial y} = -2y
    [/tex]
    subject to [itex]x(0) = y(0) = 10[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Partial Derivative Problem
Loading...