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We want to reparametrize the curve
$\displaystyle \vec{r}(t)=$ 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!

2 Answers 2

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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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    @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.2011-02-14