$r(t) = <2t^2 , 2t^3 >$
Hello everyone, I have been working on this problem for sometime now. I completed the integration (unless I have messed up along the way), but here is what I have now. I am having trouble completing the problem and have been stumped for some time.
$$t = \pm \sqrt(6) * \sqrt{(18s+8)^{2/3}-4}$$
You should show all your work so responders can find your errors. You seem to have made a couple of mistakes. I get $$\vec r=\langle 2t^2,2t^3\rangle$$ $$d\vec r=\langle 4t,6t^2\rangle dt$$ $$ds=\left|\left|d\vec r\right|\right|=\sqrt{16t^2+36t^4}dt=2t\sqrt{4+9t^2}dt$$ $$s=\int ds=\int2t\sqrt{4+9t^2}dt=\frac2{27}\left(4+9t^2\right)^{3/2}+C=\frac2{27}\left(4+9t^2\right)^{3/2}-\frac{16}{27}$$ If we assume the starting point is at $t=0$. Solving for $t$, I get $$t=\frac{\text{sgn}\,s}3\sqrt{\left(\frac{27}2s+8\right)^{2/3}-4}$$ Where $$\text{sgn}\,s=\left.\begin{array}{rl}-1,&s<0\\ 1,&s\ge0\end{array}\right\}$$ I checked by inserting back into the expression for $\vec r$ and differentiating. Where was your mistake?