Help With Partial Differentiation & Integration

[tshafer]

I know I should know this... it looks so ridiculously easy. In the course of getting d'Alembert's wave equation solution, we get the following equation:

[tex]2cp'\left(x\right)=cf'\left(x\right)+g\left(x\right)[/tex]

The primes are derivatives wrt t. Then we re-order the equation and "integrate the relation" to get an expression for p:

[tex]p\left(\xi\right)=\frac{1}{2}f\left(\xi\right)+\frac{1}{2c}\int^{\xi}_{0}g\left(s\right)ds[/tex]

I have to be missing something very, very simple. How can I differentiate [tex]p\left(x\right)[/tex] wrt [tex]t[/tex] then integrate wrt something (x, I suppose, in this case) and recover the original function p? Or am I NOT recovering it... just something new named [tex]p\left(\xi\right)[/tex]? Thanks for the help!

Tom

 

[tiny-tim]

Hi Tom!

[tex]\int^{\xi}_{0}f'(x)dx\ =\ \left[f(x)\right]^{\xi}_{0}\ =\ f(\xi)\ -\ f(0)[/tex]

 

[tshafer]

Alright, cool. Why is that true, though? If I have f(x) = 3x^2, differentiate wrt t and integrate wrt x from 0 to xi I don't think I will get 3(xi)^2?

Tom

 

[tiny-tim]

f'(x) = 6x.

[tex]\int^{\xi}_{0} 6x dx\ =\ [3x^2]^{\xi}_{0}\ =\ 3\xi^2[/tex]

Integration is the opposite of differentiation …

that's how it works!​

 

[yenchin]

The fancy name of what tiny-tim show you is "Fundamental Theorem of Calculus". You may wish to read up more about it :)

 

[tshafer]

Yes... when it was covered waaaay back in Calc I, it was something more like:
[tex]\frac{d}{dt}\int^{x}_{a}f\left(t\right)dt=f\left(x\right)-f\left(a\right)[/tex].

I was just concerned with differentiating wrt t AND integrating wrt x:
[tex]\int^{\xi}_{0}\frac{df\left(x\right)}{dt}dx[/tex]

I felt like I had variables flying everywhere, hehe. Thanks!

Tom

 

[NoMoreExams]

Yes... when it was covered waaaay back in Calc I, it was something more like:
[tex]\frac{d}{dt}\int^{x}_{a}f\left(t\right)dt=f\left(x\right)-f\left(a\right)[/tex].

I was just concerned with differentiating wrt t AND integrating wrt x:
[tex]\int^{\xi}_{0}\frac{df\left(x\right)}{dt}dx[/tex]

I felt like I had variables flying everywhere, hehe. Thanks!

Tom

Your first formula is incorrect I believe. You should only be left with f(x) and I believe you want [tex] \frac{d}{dx} [/tex] instead of [tex] \frac{d}{dt} [/tex]. The "general" version of this

Let

[tex] F(x) = \int_{g(x)}^{h(x)} f(t) dt [/tex]

Then

[tex] F'(x) = f(h(x)) h'(x) - f(g(x)) g'(x) [/tex]

 

[tshafer]

alright... thanks.

now, here's the thing... this all makes sense, except in my problem.

I have [tex]f(x) = 3x^{2}[/tex] and [tex]\frac{df}{dt}=f'\left(x\right) = 6x\frac{dx}{dt}[/tex]

[tex]\int^{\xi}_{0}6x\frac{dx}{dt}dx[/tex] = ??

or am I waaay confused?

 

[NoMoreExams]

If you are doing implicit differentiation you should get [tex] 6x \frac{dx}{dt}[/tex] (minor error).

 

[tshafer]

My bad, just in a hurry. Still... [tex]\frac{dx}{dt}dx[/tex]??

 

[tshafer]

Ok, I am an idiot... these aren't t-derivatives, I guess.

[tex]u\left(x,t\right)=p\left(x+ct\right)+q\left(x-ct\right)=p\left(\xi\right)+q\left(\eta\right)[/tex]

[tex]\frac{\partial u}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial \xi}\right|_{t=0}\frac{\partial \xi}{\partial t}\right|_{t=0} + \frac{\partial u}{\partial \eta}\right|_{t=0}\frac{\partial \eta}{\partial t}\right|_{t=0}[/tex]

but [tex]\xi=\left(x+ct\right)\right|_{t=0}=x[/tex], same for [tex]\eta[/tex]. So...

[tex]\frac{\partial u}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial x}\frac{\partial \xi}{\partial t}\right|_{t=0} + \frac{\partial u}{\partial x}\frac{\partial \eta}{\partial t}\right|_{t=0}=\frac{\partial u}{\partial x}\cdot c + \frac{\partial u}{\partial x}\cdot \left(-c\right)[/tex]

So [tex]p'\left(x\right)=\frac{dp}{dx}[/tex], not [tex]\frac{dp\left(x\right)}{dt}[/tex].

Does that seem correct?
Tom

 

[HallsofIvy]

Alright, cool. Why is that true, though? If I have f(x) = 3x^2, differentiate wrt t and integrate wrt x from 0 to xi I don't think I will get 3(xi)^2?

Tom

Surely that's not what you meant! If you differentiate f(x) wrt t, you get 0. Integrating that with respect to x still gives 0.

 

[tshafer]

Exactly... yet my professor was adamant. I'm sure now that he didn't understand my question.

 

[HallsofIvy]

alright... thanks.

now, here's the thing... this all makes sense, except in my problem.

I have [tex]f(x) = 3x^{2}[/tex] and [tex]\frac{df}{dt}=f'\left(x\right) = 6x\frac{dx}{dt}[/tex]

[tex]\int^{\xi}_{0}6x\frac{dx}{dt}dx[/tex] = ??

or am I waaay confused?

Your second statement is incorrect. It should be
[tex]\int^{t_i}_0 6x \frac{dx}{dt}dt= \int^{x_i}_0 6x dx= 3x_i^2[/itex]
Where [itex]x_i= x(t_i)[/itex]

 

[tshafer]

right... this is why i was so confused. i thought my prof was saying to differentiate wrt t, then integrate wrt x, not t.