# Reparametrizing a curve in terms of the arc length

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!

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

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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 $s$.

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Substitution of $u = x^2+1$ should do the job. $du=2xdx$ EDIT: Does the substitution affect the variables of integration? Does $t$ become $t^2+1$? It seems that t is independant of $\tau$ but when substituting we need to change variables of integration? –  fdart17 Feb 14 '11 at 0:22
@Fdart17: That's the one I would use, too. –  Jonas Meyer Feb 14 '11 at 0:23
@Fdart17: Regarding your comment edit, there are 2 ways to do a definite integral by substitution. One would be to find an antiderivative of $3x\sqrt{x^2+1}$ (in terms of $x$), and then apply FTC using the original limits. The other is to change your limits along with the substitution, which means the limits would be $1$ and $t^2+1$ instead of $0$ and $t$; in this case, these limits will be evaluated in an antiderivative of your new integrand, which will be a function of $u$. –  Jonas Meyer Feb 14 '11 at 0:30
@Fdart17: For an example with the same idea, suppose you want to evaluate $\int_1^e\frac{\ln x}{x}dx$. With the substitution $u=\ln x$, you find an antiderivative $\frac{1}{2}(\ln x)^2$ of the integrand, so the integral is $\frac{1}{2}(\ln e)^2-\frac{1}{2}(\ln 0)^2=\frac{1}{2}$. On the other hand, using the substitution to rewrite the integral in terms of $u$ would give $\int_0^1 u du$, which can be computed directly without going back to the original. –  Jonas Meyer Feb 14 '11 at 0:46