Velocity Field

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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.

[math]4txi-2t^2yi+4xzk[/math][math][d/dt]4tx+4tx[d/dx](4tx)[/math][math]u=4x+4tx(4t)[/math][math]v=-2t^2y[d/dy]=2t^2[/math][math]w=4xz[d/dx]=4z[/math][math]w=4xz[d/dz]=4x [/math]
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 \(\displaystyle 4txi+4t^2yj+ 4xzk\) 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 \(\displaystyle 4txi\) with respect to t is \(\displaystyle (4x+ 4t\frac{dx}{dt})i\)
The derivative of \(\displaystyle 2t^2yj\) with respect to t is \(\displaystyle (4t+ 2t^2\frac{dy}{dt})j\)
The derivative of \(\displaystyle 4xzk\) with respect to t is \(\displaystyle (4z\frac{dx}{dt}+ 4x\frac{dz}{dt})k\).

So the derivative of the vector is
\(\displaystyle (4x+ 4t\frac{dx}{dt})i+ (4t+ 2t^2\frac{dy}{dt})j+ (4z\frac{dx}{dt}+ 4x\frac{dz}{dt})k\).
 
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 \(\displaystyle (4xt- 2t^2y)\vec{i}+ 4xz\vec{k}\) and there is no "\(\displaystyle \vec{j}\)".

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