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Multivariable calculus question

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    A function [itex]f(x,y)=\mathbb{R}^{2}\rightarrow \mathbb{R}[/itex] is defined by:

    [itex]f(x,y)=5y^{2}-x^{2}[/itex]

    (i) Find its gradient vector and evaluate it at the point [itex](x,y)^{T}=(1,1)^{T}[/itex]. Find the rate of change of the function in the direction [itex](2,1)^{T}[/itex] at the point [itex](1,1)^{T}[/itex]

    (ii) In what direction is the rate of change of the function at the point [itex](x,y)^{T}=(2,0)^{T}[/itex] equal to [itex]-2[/itex]?

    Is there a direction for which the rate of change at the point [itex](x,y)^{T}=(2,0)^{T}[/itex] is equal to [itex]-5[/itex]? Find the greatest rate of decrease of the function at this point.

    3. The attempt at a solution

    (i)

    [itex]f(x,y)=5y^{2}-x^{2}[/itex]

    [itex]f_{x}(x,y)=-2x=-2[/itex] at (1,1)

    [itex]f_{y}(x,y)=10y=10[/itex] at (1,1)

    The unit vector in the direction of [itex](2,1)^{T}[/itex] is [itex](2,1)^{T}=\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}[/itex]

    To find the rate of change in the direction of the unit vector above I:

    [itex]<\bigtriangledown f,u>=f_{x}u_{1}+f_{y}u_{2}=-2\cdot \frac{2}{\sqrt{5}}+10\cdot \frac{1}{\sqrt{5}}=\frac{6}{\sqrt{5}}[/itex]

    (ii)

    at [itex](2,0)^{T}[/itex] the gradient vector is:

    [itex]\bigtriangledown f=(-4,0)[/itex]

    so knowing [itex](\bigtriangledown f,u)=-2[/itex] gives

    [itex]-2=-4\cdot u_{1}+0\cdot u_{2}[/itex]

    [itex]-2=-4\cdot u_{1}[/itex]

    [itex]u_{1}=\frac{1}{2}[/itex]

    so the direction in which the rate of change of the function is [itex]u=(\frac{1}{2},0)[/itex]

    (iia) for the part where it asks about the rate of change being [itex]-5[/itex] I used the same method and got [itex]u=(\frac{5}{4},0)[/itex].

    Is this all correct?

    but now I don't know how to 'Find the greatest rate of decrease of the function at this point.'

    Any help appreciated thanks!
     
    Last edited: Mar 2, 2013
  2. jcsd
  3. Mar 2, 2013 #2

    vela

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    Actually, you don't know that ##u_2=0##. You want to use the fact that ##\vec{u}## is a unit vector to determine ##u_2##.

     
  4. Mar 2, 2013 #3
    Oh I see, so:

    [itex]\parallel u \parallel=1=\sqrt{\frac{1}{2}\cdot \frac{1}{2}+u_{2}^{2}}[/itex]

    [itex]\frac{1}{4}+u_{2}^{2}=1^{2}[/itex]

    [itex]u_{2}^{2}=1-\frac{1}{4}[/itex]

    [itex]u_{2}=\sqrt{\frac{3}{4}}[/itex]

    [itex]u_{2}=\pm \frac{1}{2}\sqrt{3}[/itex]

    [itex]u=(\frac{1}{2},\pm \frac{1}{2}\sqrt{3})[/itex]
     
  5. Mar 2, 2013 #4

    vela

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    Yup, so can you see now that the first answer to iia is wrong?
     
  6. Mar 2, 2013 #5
    I think so:

    [itex]u=(\frac{1}{2},\pm \frac{1}{2}\sqrt{3})[/itex]

    [itex]\bigtriangledown f=(-4,0)[/itex]

    so,

    [itex](\bigtriangledown f,u)=-4\cdot \frac{1}{2}+0[/itex]

    [itex]u=(-2,0)[/itex]???
     
  7. Mar 3, 2013 #6

    vela

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    For iia?

    The gradient evaluated at the point is (-4,0), and now you're dotting a unit vector into it. When is the dot product the greatest? What is the biggest value ##(-4,0)\cdot\vec{u}## can take on when ##\vec{u}## is a unit vector?
     
  8. Mar 3, 2013 #7
    I'm not sure. When is the dot product the greatest?

    You mean when [itex]\left \| \bigtriangledown f \right \|=\sqrt{-4^{2}+0}[/itex]

    [itex]\left \| \bigtriangledown f \right \|=4[/itex]?

    then,

    [itex]-\left \| \bigtriangledown f \right \|=-4[/itex]?

    so,

    [itex]-4[/itex] is the minimum and there is no direction that can make the rate of change [itex]-5[/itex]
     
    Last edited: Mar 3, 2013
  9. Mar 4, 2013 #8

    vela

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    Yeah, pretty much.

    The dot product of two vectors ##\vec{A}## and ##\vec{B}## is given by ##\vec{A}\cdot\vec{B} = \|\vec{A}\| \|\vec{B}\| \cos \theta##, where ##\theta## is the angle between them. Clearly, the product must be between ##-\|\vec{A}\| \|\vec{B}\|## and ##\|\vec{A}\| \|\vec{B}\|##. In this problem, this means that the directional derivative must be between -4 and +4.
     
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