Velocity Field

god_is_atheist

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So after taking the derivative the velocity field with respect to t. I got stuck on partial derivatives.
Im teaching myself.

4txi2t2yi+4xzk4txi-2t^2yi+4xzk[d/dt]4tx+4tx[d/dx](4tx)[d/dt]4tx+4tx[d/dx](4tx)u=4x+4tx(4t)u=4x+4tx(4t)v=2t2y[d/dy]=2t2v=-2t^2y[d/dy]=2t^2w=4xz[d/dx]=4zw=4xz[d/dx]=4zw=4xz[d/dz]=4xw=4xz[d/dz]=4x
I feel this is incorrect and Im working on teaching myself. Any input will be appreciated.[/math]
 
Your notation is a little hard to understand.
Are we to assume that 4txi+4t2yj+4xzk\displaystyle 4txi+4t^2yj+ 4xzk is the velocity field of an object in an xyz-coordinate system? And that x, y, and z are functions of t?
(You have 2 "i"s but I am sure the second was supposed to be "j".)

Then you have to use the "product rule" and "chain rule".
The derivative of 4txi\displaystyle 4txi with respect to t is (4x+4tdxdt)i\displaystyle (4x+ 4t\frac{dx}{dt})i
The derivative of 2t2yj\displaystyle 2t^2yj with respect to t is (4t+2t2dydt)j\displaystyle (4t+ 2t^2\frac{dy}{dt})j
The derivative of 4xzk\displaystyle 4xzk with respect to t is (4zdxdt+4xdzdt)k\displaystyle (4z\frac{dx}{dt}+ 4x\frac{dz}{dt})k.

So the derivative of the vector is
(4x+4tdxdt)i+(4t+2t2dydt)j+(4zdxdt+4xdzdt)k\displaystyle (4x+ 4t\frac{dx}{dt})i+ (4t+ 2t^2\frac{dy}{dt})j+ (4z\frac{dx}{dt}+ 4x\frac{dz}{dt})k.
 
Yes, that is the feild of the xyz coordinate system. I forgot to add that in Latex.
Two "i"s definitely makes a difference. I'm guessing you used \frac to write the problem out.
Thanks for your time.
 
So this DOES have two "i"s? That was not a typo?

Then it is (4xt2t2y)i+4xzk\displaystyle (4xt- 2t^2y)\vec{i}+ 4xz\vec{k} and there is no "j\displaystyle \vec{j}".

The derivative with respect to time, t, is
(4x4ty+4tdxdt2t2dydt)i+(4xdzdt+4zdxdt)k\displaystyle (4x- 4ty+ 4t\frac{dx}{dt}- 2t^2\frac{dy}{dt})\vec{i}+ (4x\frac{dz}{dt}+ 4z\frac{dx}{dt})\vec{k}.
 
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