Homework Help: Partial derivative query - guidance needed

1. Nov 11, 2012

petertheta

I have a question but have not seen an example or find anything in my textbooks so would love some advice on how to understand the problem.

Its a theory question on partial derivatives of the second order...

T=T(x,y,z,t) with x=x(t), y=y(t), z=z(t).

Find the second derivative of T wrt t

So, first find the first derivative to give:

$$\frac{dT}{dt} = \frac{\partial T}{\partial x}\frac{dx}{dt} +\frac{\partial T}{\partial y}\frac{dy}{dt}+\frac{\partial T}{\partial z}\frac{dz}{dt}+\frac{\partial T}{\partial t}$$

So I know the second derivative will be:
$$\frac{d}{dt}\left(\frac{dT}{dt}\right)$$

So now my question. How do i take derivatives of products of mixed derivatives and write it in a correct manner, albeit a theoretical one.

Your help means a lot.

PT
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Nov 11, 2012
2. Nov 11, 2012

Ray Vickson

Your dT/dt is a sum of four terms. Do you agree that you can take another d/dt of each term separately, then add the results? So, look at the first term
$$\frac{\partial T}{\partial x} \frac{dx}{dt}.$$ This has the form G(x,y,z)*v(t), where
$$G = \partial T/\partial x, \: v = dx/dt \text{ and } x = x(t), \:y = y(t), \:z = z(t).$$ Just apply your (d/dt) rule to G*v. Similarly for the other terms.

RGV

3. Nov 11, 2012

petertheta

Thanks RGV. So would the first "set" be:

$$\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{dx}{dt}\frac{dx}{dt}$$
or better still...
$$\frac{\partial T}{\partial x}\frac{dx}{dt}+\frac{d^2x}{dt^2}$$

4. Nov 11, 2012

vela

Staff Emeritus
Nope. Try again. I suggest you write the derivative first in terms of G(x,y,z,t) and v(t) and then use the fact that G=dT/dx and v=dx/dt.

5. Nov 11, 2012

petertheta

OK. I'm struggling but I think this might be it...

$$\frac{\partial G}{\partial x}\frac{dx}{dt}+G\frac{dx}{dt}\frac{dx}{dt}$$

6. Nov 11, 2012

vela

Staff Emeritus
Nope. It'll probably help to write out the variable dependencies explicitly. Applying the product rules gives
$$\frac{d}{dt}[G(x,y,z,t)v(t)] = \left(\frac{d}{dt} G(x,y,z,t)\right) v(t) + G(x,y,z,t)\frac{dv}{dt}.$$ Now look at your expression for dT/dt in your original post. You'll get the same sort of thing for dG/dt. I'll let you take it from here.

7. Nov 11, 2012

petertheta

I get a different result everytime i try :(

OK. Third time lucky!! For the first "element"

$$\frac{\partial G}{\partial x}\frac{dx}{dt}v(t)+G\frac{dv}{dt}$$

Then substitute G and v for what is know....

$$\frac{\partial G}{\partial x}\frac{dx}{dt}\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d}{dt}v$$
So
$$\frac{\partial^2 T}{\partial x^2}2\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d^2x}{dt^2}+...$$

8. Nov 11, 2012

vela

Staff Emeritus
You seem to keep missing the point: G is a function of y, z, and t as well as x.

Also, are you really claiming that $\frac{dx}{dt}\times\frac{dx}{dt} = 2\frac{dx}{dt}$?

9. Nov 11, 2012

petertheta

Im only writing for the very first term from the first derivative to see if I'm on the right track how does it look??....

Sorry that was a typo for 2 times dx/dt

10. Nov 11, 2012

petertheta

Ahh, I think I see your point. For the first term in the first derivative, when taking the second derivative I must do it for x,y,z and t. The the same for the second term... etc, etc... Is this correct??

11. Nov 11, 2012

vela

Staff Emeritus
Yup.

12. Nov 11, 2012

petertheta

thats gonna be a big answer! OK. So here we go just for the first term of the first derivative...

$$\frac{\partial^2T}{\partial x\partial x}\frac{dx}{dt}\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d^2x}{dt^2} + \frac{\partial^2T}{\partial x\partial y}\frac{dy}{dt}\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d^2x}{dt^2} + \frac{\partial^2T}{\partial x\partial z}\frac{dz}{dt}\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d^2x}{dt^2} + \frac{\partial^2T}{\partial x\partial t}\frac{dt}{dt}\frac{dx}{dt}+\frac{\partial T}{\partial x}\frac{d^2x}{dt^2}....$$

13. Nov 11, 2012

vela

Staff Emeritus
Almost. It should be
$$\left( \frac{\partial^2T}{\partial x^2}\frac{dx}{dt} + \frac{\partial^2T}{\partial y\partial x}\frac{dy}{dt} + \frac{\partial^2T}{\partial z\partial x}\frac{dz}{dt} + \frac{\partial^2T}{\partial t\partial x}\right)\frac{dx}{dt} + \frac{\partial T}{\partial x}\frac{d^2x}{dt^2}$$ The last term only appears once.