🔒 Locked

Regarding partial derivatives

Started by mlt Dec 14, 2008 at 1:34 PM 6 replies 900+ views

Original Post

Dec 14, 2008 at 1:34 PM

I am currently reading about water simulation: http://http.developer.nvidia.com/GPUGems/gpugems_ch01.html When it gets to eq 6b and 7 I am lost. 1) How is it possible to get from 6a to 6b? 2) I assume that the partial derivative with respect to y is missing from eq 7, and its only shown with respect to x since it can be reused with respect to y.

apatriarca

Dec 14, 2008 at 3:12 PM

I use the symbols D_x, D_y, D_z to represent partial derivatives...

B(x,y) = (1, 0, D_xH(x,y,t)) from equation 4b
T(x,y) = (0, 1, D_yH(x,y,t)) from equation 5b
So
N(x,y) = B(x,y)xT(x,y) = (1, 0, D_xH(x,y,t))x(0, 1, D_yH(x,y,t)) =
= (0*D_yH(x,y,t) - 1*D_xH(x,y,t), 0*D_xH(x,y,t) - 1*D_yH(x,y,t), 1*1 - 0*0) =
= (-D_xH(x,y,t), -D_yH(x,y,t), 1) as in equation 6b

smc

Dec 14, 2008 at 3:45 PM

Are the 'X' denoting cross products here?

apatriarca

Dec 14, 2008 at 4:38 PM

The amplitude, the phase, the frequency and the time are usually scalar, so I think in that case the 'x' is the scalar multiplication. But in the equation 6 it's the cross product because B and T are vectors. It's a confusing notation.

smc

Dec 14, 2008 at 4:48 PM

Thanks,

That is what I thought. The notation is a bit confusing. I guess they expect you to see this via the context... which makes sense.

mlt

Dec 15, 2008 at 5:14 PM

Yes the notation is not perfect, but it makes sense when you know whats a vector and whats a scalar.

Another thing. Eq 4.a is the binormalvector. But I don't understand why the partial derivatives in the x-direction gives three coordinates. How are these three coordinates (partial derivatives) derived?

jjd

Dec 16, 2008 at 6:42 AM

Quote:
Original post by mlt
Another thing. Eq 4.a is the binormalvector. But I don't understand why the partial derivatives in the x-direction gives three coordinates. How are these three coordinates (partial derivatives) derived?

The author is just trying to show their working. Eq 4a is the derivative of Eq 3 along the x-axis. So the author is applying the derivative to each of the components, which are then 'evaluated' in 4b.

mlt

Dec 16, 2008 at 8:25 AM

But are the resulting 3 element vector a point or the binormal vector assuming some origo?

It seems that B(x,y) (where x,y indicate a position on the plane) returns a vector of length 3, but is that point or vector from the origo (x,y)?

Topic Locked

This topic has been locked by a moderator. New replies are not allowed.