# 11.2 #41

The length of the curve $x=1+3t^2, y=4+2t^3, 0\leq t \leq2$

1. Find dx/dt and dy/dt: dx/dt=6t & dy/dt=6t$^2$
2. Square both: $36t^2$ and $36t^4$, respectively.
3. Length$=\int_{0}^{1} \sqrt{36t^2+36t^4}dt$
4. Factor out 6t: $\int_{0}^{1} \sqrt{t^2+1}dt$
5. u Substitution: $u=t^2+1, du=2tdt, \frac{1}{2}du=tdt$ and integral limits become 1 and 2
6. $L=6\int_{1}^{2}\frac{1}{2}u^\frac{1}{2}du$
7. $L=3\frac{2}{3}u^\frac{3}{2}=2(2\sqrt{2}-1)=4\sqrt{2}-2$

That's not a particularly difficult problem, but it is the first one I've written in LaTeX, which was difficult. I thank the Hamline University Physics Department Latex Equation Editor, although it wouldn't load its renderings properly.