Results 1 to 2 of 2

Math Help - Directional Derivatives/Gradient Vector

  1. #1
    Super Member Aryth's Avatar
    Joined
    Feb 2007
    From
    USA
    Posts
    652
    Thanks
    2
    Awards
    1

    Directional Derivatives/Gradient Vector

    Here goes:

    The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,2,2) is 120^{\circ}.

    (a) Find the rate of change of T at (1,2,2) in the direction toward the point (2,1,3).

    (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by Aryth View Post
    Here goes:

    The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point (1,2,2) is 120^{\circ}
    so T(x,y,z)=\frac{k}{\sqrt{x^2+y^2+z^2}}, for some constant k. since 120=T(1,2,2)=\frac{k}{\sqrt{1+4+4}}=\frac{k}{3}, we get k=360.

    (a) Find the rate of change of T at (1,2,2) in the direction toward the point (2,1,3).
    first you need to find the unit vector in this direction: we have \bold{v}=(2,1,3)-(1,2,2)=(1,-1,1). thus the unit vector is:

    \bold{u}=\frac{\bold{v}}{|\bold{v}|}=\frac{1}{\sqr  t{3}}(1,-1,1). it's also clear that \nabla T(x,y,z)= \frac{360}{(x^2+y^2+z^2)^{\frac{3}{2}}}(-x,-y,-z). now the rate of change of T is:

    D_{\bold{u}}T(1,2,2)=\nabla T (1,2,2) \cdot \bold{u}=\frac{360}{27 \sqrt{3}}(-1,-2,-2) \cdot (1,-1,1)=\frac{-40}{3\sqrt{3}}.

    (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.
    the direction of greatest increase at any point is the direction of \nabla T at that point. but in part (a) we saw that \nabla T(x,y,z)

    and (-x,-y,-z) have the same direction and, clearly, the vector (-x,-y,-z) points toward the origin. Q.E.D.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Directional Derivatives and Gradient Vectors
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 22nd 2011, 06:42 AM
  2. Directional derivatesand the gradient vector
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 14th 2010, 06:12 AM
  3. Replies: 5
    Last Post: October 27th 2009, 04:43 AM
  4. Directional Derivatives and Gradient
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 8th 2009, 01:17 PM
  5. Replies: 1
    Last Post: November 1st 2008, 12:53 PM

Search Tags


/mathhelpforum @mathhelpforum