UV partial derivatives of a cylinder shape primitive

Question

I am new to rendering and I would like to calcualte UV partial derivatives of a cylinder shape, which is parameterized by radius $R$ and length $L$.
Using cylindrical coordinate mapping, for a surface point $mathbf{p}=(x,y,z)$, the UV map (normalized to $[0,1]$) is:
$$
(u,v) = Big(frac{phi}{2pi}, frac{z}{L}Big)qquad text{where} quad phi = arctan frac{y}{x}
$$
Then I compute the derivatives with respect to $mathbf{p}$, that is:
$$
begin{aligned}
frac{partial u}{partial mathbf{p}} & = frac{1}{2pi R^2} Big( -y, x, 0Big) = frac{1}{2pi R} Big( -sinphi, cosphi, 0Big)\
frac{partial v}{partial mathbf{p}} &= Big( 0, 0, frac{1}{L}Big)
end{aligned}
$$
Or equivalently:
$$
begin{aligned}
frac{partial mathbf{p}}{partial u} & = 2pi R Big( -frac{1}{sinphi}, frac{1}{cosphi}, 0Big)\
frac{partial mathbf{p}}{partial v} &= Big( 0, 0, LBig)
end{aligned}
$$
Everything looks good until now.
Question
In pbrt-3, the relevant part of cylinder.cpp is: (i) UV same as above; (ii) $partial mathbf{p}/partial u$ however is different:
Float u = phi / phiMax;
Float v = (pHit.z - zMin) / (zMax - zMin);

// Compute cylinder $dpdu$ and $dpdv$
Vector3f dpdu(-phiMax * pHit.y, phiMax * pHit.x, 0);
Vector3f dpdv(0, 0, zMax - zMin);

What am I missing here?

Nathan Reed · Accepted Answer

I think it's easiest to get the tangent frame by writing the forward mapping from $(u,v)$ to $mathbf{p}$:
$$
    mathbf{p}(u,v) = begin{bmatrix} R , cos (2pi u) \ R , sin (2pi u) \ vL end{bmatrix}
$$
Then you can see that the derivatives are:
$$
    frac{partialmathbf{p}}{partial u} = begin{bmatrix} -2pi R , sin(2pi u) \ 2pi R , cos(2pi u) \ 0end{bmatrix} = begin{bmatrix} -2pi y \ 2pi x \ 0end{bmatrix}, qquad
    frac{partialmathbf{p}}{partial v} = begin{bmatrix} 0 \ 0 \ L end{bmatrix}
$$
which matches the pbrt code.
I think your derivation went wrong by assuming that $partialmathbf{p}/partial u$ was the component-wise reciprocal of $partial u / partial{mathbf{p}}$. It's not so simple, because the latter object is really the gradient of $u$, where $u$ is now considered as a scalar field defined over all space (not just on the cylinder). To invert this properly we really need another coordinate running perpendicular to $u$ in the $xy$ plane, so that we have a locally invertible mapping; then we can take its Jacobian matrix and invert that. We can use the radius $R$ as such a coordinate. Then we have the inverse mapping,
$$
begin{aligned}
    u(x,y) &= frac{1}{2pi} arctan(y,x) \
    R(x,y) &= sqrt{x^2 + y^2}
end{aligned}
$$
whose Jacobian is:
$$
    frac{partial(u,R)}{partial(x,y)} =
    begin{bmatrix}
        tfrac{partial u}{partial x} & tfrac{partial u}{partial y} \
        tfrac{partial R}{partial x} & tfrac{partial R}{partial y}
    end{bmatrix} =
    begin{bmatrix}
        -y/(2pi R^2) & x/(2pi R^2) \
        x/R & y/R
    end{bmatrix}
$$
and the inverse of that is:
$$
    frac{partial(x,y)}{partial(u,R)} =
    -2pi R begin{bmatrix}
        y/R & -x/(2pi R^2) \
        -x/R & -y/(2pi R^2)
    end{bmatrix} =
    begin{bmatrix}
        -2pi y & x/R \
        2pi x & y/R
    end{bmatrix}
$$
from which you can then read off $partial{(x,y)}/partial u$ as the first column, which matches the result obtained for $partialmathbf{p}/partial u$ above.

UV partial derivatives of a cylinder shape primitive

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