I'm really at the end of my wits on this problem. Basically I'm trying to find arc length. The vector-valued function is: $R=\langle t,\sqrt{t}\rangle$ and $t\ge0$.
We're looking for the length of the arc between $0\leq t\leq3$. I know that you basically have to integrate the magnitude of the tangent vector over the interval specified, and to this end
I got as far as the $\int_0^3 \sqrt(1+1/4t)dt.$ On that note, I cannot for the life of me figure out how to integrate this. I've tried factoring, simplifying, substitution (which seemed to make it worse), and anything else I could think of. Am I going about this the wrong way? Is there an easier method? Any help would be greatly appreciated.
Hint
Start with a change of variable such that $$\sqrt{\frac{1}{4 t}+1}=u$$ which means $$t=\frac{1}{4 \left(u^2-1\right)}$$ $$dt=-\frac{u}{2 \left(u^2-1\right)^2}du$$ Then $$\int \sqrt{\frac{1}{4 t}+1}~dt=-\int \frac{u^2}{2 \left(u^2-1\right)^2}~du$$ Now use partial fraction decomposition. The result is quite simple.
I am sure that you can take from here.