Regarding partial derivatives
Author
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.
I use the symbols Dx, Dy, Dz to represent partial derivatives...
B(x,y) = (1, 0, DxH(x,y,t)) from equation 4b
T(x,y) = (0, 1, DyH(x,y,t)) from equation 5b
So
N(x,y) = B(x,y)xT(x,y) = (1, 0, DxH(x,y,t))x(0, 1, DyH(x,y,t)) =
= (0*DyH(x,y,t) - 1*DxH(x,y,t), 0*DxH(x,y,t) - 1*DyH(x,y,t), 1*1 - 0*0) =
= (-DxH(x,y,t), -DyH(x,y,t), 1) as in equation 6b
B(x,y) = (1, 0, DxH(x,y,t)) from equation 4b
T(x,y) = (0, 1, DyH(x,y,t)) from equation 5b
So
N(x,y) = B(x,y)xT(x,y) = (1, 0, DxH(x,y,t))x(0, 1, DyH(x,y,t)) =
= (0*DyH(x,y,t) - 1*DxH(x,y,t), 0*DxH(x,y,t) - 1*DyH(x,y,t), 1*1 - 0*0) =
= (-DxH(x,y,t), -DyH(x,y,t), 1) as in equation 6b
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.
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.
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.
Author
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?
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?
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.
Author
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)?
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)?
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