B Why is Δv Replaced by du/dx in Series Expansion?

[mech-eng]

Hi, I stamped at a series expansion. It is probably Taylor. Would you explain it? It's in the vid.

https://confluence.cornell.edu/disp...mics+-+Differential+Form+of+Mass+Conservation

I understand equation 1 in the picture but I do not understand 2.
I cannot understand how Δv=du/dx.

Thank you.

 

[jedishrfu]

U is the velocity in the x direction. And they used a Taylor expansion at 2:38 into the video to get the expression you see. They drop the other terms of the series as negligible when $\Delta x$ tends to zero along with the other deltas t and y.

 

[mech-eng]

U is the velocity in the x direction. And they used a Taylor expansion at 2:38 into the video to get the expression you see. They drop the other terms of the series as negligible when $\Delta x$ tends to zero along with the other deltas t and y.

In 2 Δv disappears instead du/dx and Δy appears.

Thank you.

 

[pasmith]

In 2 Δv disappears instead du/dx and Δy appears.

Thank you.

They swapped the order of the first two terms in addition to using \Delta u \approx \frac{\partial u}{\partial x} \Delta x and \Delta v \approx \frac{\partial v}{\partial y} \Delta y