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kwal0203
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Homework Statement
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.
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!
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