Partial derivatives of f(x,y)

  • #1

Homework Statement


Where [itex]T(x,t)=T_{0}+T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)[/itex]

[itex]\omega = \frac{\Pi}{365}[/itex] and [itex]\lambda[/itex] is a positive constant.

Show that T satisfies [itex]T_{t}=kT_{xx}[/itex] and determine [itex]\lambda[/itex] in terms of [itex]\omega[/itex] and k.

I'm not to sure what is meant by the latter part of "determine [itex]\lambda[/itex] in terms of [itex]\omega[/itex] and k."


Homework Equations





The Attempt at a Solution



So I think I first have to find the partial derivatives of the first order.
[itex]\frac{\partial T}{\partial x}=-\lambda T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+T_{1}e^{-\lambda x}(-\lambda)\cos(\omega t-\lambda x)[/itex]

[itex]\frac{\partial T}{\partial t}=T_{1}e^{-\lambda x}(\omega t)\cos(\omega t-\lambda x)[/itex]


I then work out the second order partial derivative with respect to x and here it gets kind of messy and where I get confused.
[tex]
T_{xx}=-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+T_{1}e^{-\lambda x}(-\lambda)\cos(\omega t-\lambda x)+(-\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)[/tex]

[tex]
T_{xx}=-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-T_{1}e^{-\lambda x}\lambda\cos(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/tex]

[tex]T_{xx}=-2\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)(1-\lambda)[/tex]

[tex]T_{xx}=\lambda T_{1}e^{-\lambda x}(-2\lambda\sin(\omega t-\lambda x)-\cos(\omega t-\lambda x)(1-\lambda)
[/tex]


This looks nothing like the partial derivative of the first order with respect to t...
 

Answers and Replies

  • #2
phyzguy
Science Advisor
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You're just not doing the partial derivatives correctly. For example, when you take the partial derivative with respect to t, the multiplier should be ω, not (ωt). Also, in Txx, each term should be multiplied by λ^2. You missed a factor of λ in the second term. Also, don't forget that (-λ)^2 is equal to λ^2, not -λ^2.
 
  • #3
Ok I've improved it a little but still stuck...

[tex]T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)[/tex]

[tex]T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)-\lambda^{2}T_{1}e^{-\lambda x}\sin(\omega t-\lambda x)+\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/tex]

[tex]T_{xx}=\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)+(\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)=2\lambda^{2}T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/tex]


However, this still doesn't look right to me?

Because if [itex]T_{t}=kT_{xx}[/itex]

then

[itex]\frac{\partial T}{\partial t}=T_{1}e^{-\lambda x}\omega\cos(\omega t-\lambda x)=kT_{xx}=k(2\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/itex] which isn't true
 
  • #4
phyzguy
Science Advisor
4,824
1,778
It is true for certain values of lambda. This basically gives you an equation which defines λ in terms of ω and k, which is what you were asked to provide.
 
  • #5
Ok I see, so it goes like this:

Since[itex]\lambda[/itex]
is a positive constant [itex]=kT_{xx}=k(2\lambda^{2})T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/itex]
and [itex]\frac{\partial T}{\partial t}=\omega T_{1}e^{-\lambda x}\cos(\omega t-\lambda x)[/itex]

For [itex]\frac{\partial T}{\partial t}=kT_{xx}[/itex] to be true [itex] k(2\lambda^{2})=\omega[/itex] therefore [itex]\lambda=\sqrt{\frac{\omega}{2k}}[/itex]


Then [itex]T_{t}=kT_{xx}[/itex]
only when [itex]\lambda=\sqrt{\frac{\omega}{2k}}[/itex]
 
  • #6
phyzguy
Science Advisor
4,824
1,778
You got it!
 

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