I am pretty sure the answer to the question in the title—Is solving math problems in LaTeX without paper slower than in paper?—is yes. I work with a computer keyboard in front of me and a pen and pad to my left (yes, I'm left-handed), and when I'm typing up solutions to problems I usually leave the keyboard to scratch on the pad for a while, then go back to the keyboard to LaTeX it up. I can think of some reasons why this might be more true:
- I can think faster than I can type.
- Although in English I type much faster than I write, when writing math I write much faster than I can type LaTeX.
- I can draw pictures by hand much much faster than drawing them in LaTeX.
I admit that this is a totally subjective answer, but I don't know of any colleagues who work first in LaTeX without scribbling on paper or a black/whiteboard first.
As far as tips to be more productive when writing LaTeX, I would suggest to follow good coding practices. AFAIK there are not that many settled conventions on code organization, but some general practices I follow are:
Start each sentence on a new line.
Indent code within environment blocks.
Use braces to enclose groups even if they are only one token. I break this rule in super/subscripts, but adhere to it closely in \frac-tions.
Use the cool package to make many math expression more like macros.
So for instance (not my best example but one that's at hand),
Let $E$ be the solid.
Its volume is $\frac{1}{8} \frac{4}{3}\pi = \frac{\pi}{6}$.
In spherical coordinates it is a wedge $0 \leq \rho\leq 1$, $0 \leq \theta \leq \pi/2$, $0 \leq \phi\leq \pi/2$.
So the moments are
\begin{align*}
M_{yz} = \iiint_E x\,dV &= \int_0^{\pi/2}\int_0^{\pi/2} \int_0^1 (\rho \Sin{\phi}\Cos{\theta})\rho^2\Sin{\phi}\,d\rho\,d\phi\,d\theta \\
&= \int_0^{\pi/2} \Cos{\theta} \,d\theta\cdot\int_0^{\pi/2} \Sin{\phi}^2 \,d\phi\cdot\int_0^{1} \rho^3\,d\rho \\
&= 1 \cdot \frac{\pi}{4} \cdot \frac{1}{4} = \frac{\pi}{16} \\
M_{yz} = \iiint_E y\,dV
&= \int_0^{\pi/2}\int_0^{\pi/2} \int_0^1 (\rho \Sin{\phi}\Sin{\theta})\rho^2\Sin{\phi}\,d\rho\,d\phi\,d\theta \\
&= \int_0^{\pi/2} \Sin{\theta} \,d\theta\cdot\int_0^{\pi/2} \Sin{\phi}^2\,d\phi \cdot \int_0^1 \rho^3 \,d\rho \\
&= 1 \cdot \frac{\pi}{4} \cdot \frac{1}{4} = \frac{\pi}{16} \\
M_{xy} = \iiint_E z\,dV
&= \int_0^{\pi/2}\int_0^{\pi/2} \int_0^1 (\rho \Cos{\phi})\rho^2\Sin{\phi}\,d\rho\,d\phi\,d\theta \\
&= \int_0^{\pi/2} d\theta\cdot \int_0^{\pi/2} \Sin{\phi}\Cos{\phi}\,d\phi \cdot \int_0^1 \rho^3 \,d\rho \\
&= \frac{\pi}{2} \cdot \frac{1}{2} \cdot \frac{1}{4} = \frac{\pi}{16}
\end{align*}
So the coordinates of the centroid are
$\left(\frac{3}{8},\frac{3}{8},\frac{3}{8}\right)$
Even for this one I did work it out on paper first.
Update 2023: Nine years after writing this I got an upvote which gave me the occasion to reread it. My workflow hasn't changed much, except that the "pen and pad" to my left is often replaced with a tablet and stylus. The advantage to this workflow is that diagrams are easier to begin freehand as well.
:D– T. Verron Nov 20 '14 at 16:04