Reparametrizing a curve in terms of the arc length

Question

We want to reparametrize the curve
$$\displaystyle \vec{r}(t)=<t^3+1, t^2-1, \frac{\sqrt{5}}{2}t^2>$$ in terms of the arc length measured from the point t=0 in the direction of increasing t.

Here is what I tried, but I've hit a snag:
$$\displaystyle \vec{v}(t)=<3t^2, 2t, \sqrt{5}t>$$ $$\displaystyle |\vec{v}(t)|=3t\sqrt{t^2+1}$$

$$\displaystyle s=\int_{0}^{t}3t\sqrt{t^2+1}d\tau=3t^2\sqrt{t^2+1}$$

I think I'm missing something here. but assuming everything is correct, we need to solve:
$$\frac{s^2}{9}=t^4(t^2+1)$$ for $t$, and then we are nearly done. I can't seem to solve for $t$ however, brain fart? Assuming we did, we just plug $t$ in for the expression in terms of $s$ in the original equation and we are done? P.S. this is exam review, not homework!

Thanks for reading!

Answer 1

HINT:

$$\displaystyle s(t)=\int_{0}^{t} \left|\vec{v}(\tau) \right| d\tau=\int_{0}^{t}3\tau \sqrt{\tau^2+1}d\tau$$

Let $\tau^2+1 = x^2$, then $\tau d \tau = x dx$.

Hence, the integral becomes $s(t) = \displaystyle \int_{1}^{\sqrt{1+t^2}}3x^2dx = x^3 \rvert_{1}^{\sqrt{1+t^2}} = (1+t^2)^{3/2} - 1$.

Answer 2

You evaluated the integral incorrectly. You have (as Sivaram also posted before I finished) $$s=\int_0^t 3x\sqrt{x^2+1}dx,$$ which would best be handled by a change of variables. With the correct evaluation, $t$ will only appear once in the equation, making it easier to solve for $t$ in terms of $s$.