# Arc Length Problem

I am currently in the middle of the following problem.

Reparametrize the curve $\vec{\gamma } :\Bbb{R} \to \Bbb{R}^{2}$ defined by $\vec{\gamma}(t)=(t^{3}+1,t^{2}-1)$ with respect to arc length measured from $(1,-1)$ in the direction of increasing $t$.

By reparametrizing the curve, does this mean I should write the equation in cartesian form? If so, I carry on as follows.

$x=t^{3}+1$ and $y=t^{2}-1$

Solving for $t$

$$t=\sqrt[3]{x-1}$$

Thus,

$$y=(x-1)^{2/3}-1$$

Letting $y=f(x)$, the arclength can be found using the formula

$$s=\int_{a}^{b}\sqrt{1+[f'(x)]^{2}}\cdot dx$$

Finding the derivative yields

$$f'(x)=\frac{2}{3\sqrt[3]{x-1}}$$

and

$$[f'(x)]^{2}=\frac{4}{9(x-1)^{2/3}}.$$

Putting this into the arclength formula, and using the proper limits of integration (found by using $t=1,-1$ with $x=t^{3}+1$) yields

$$s=\int_{0}^{2}\sqrt{1+\frac{4}{9(x-1)^{2/3}}}\cdot dx$$

I am now unable to continue with the integration as it has me stumped. I cannot factor anything etc. Is there some general way to approach problems of this kind?

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We have $x=u^3+1$ and $y=u^2-1$. (I changed the names of the parameters because I want to reserve $t$ for the parameter of the endpoint.) Then $\frac{dx}{du}=3u^2$ and $\frac{dy}{du}=2u$. Thus the arclength from $u=0$ to $u=t$ is given by $$\int_0^t \sqrt{\left(\frac{dx}{du}\right)^2 +\left(\frac{dy}{du}\right)^2}\,du.$$ We have used the parametric arclength formula, much easier! The integration starts at $u=0$, since that is the value of the parameter that gives us the point $(1,-1)$.
We end up integrating $\sqrt{9u^4+4u^2}$. Since $u\ge 0$, we can replace this by $u\sqrt{9u^2+4}$. Integrate, making the substitution $w=9u^2+4$. We arrive at $$\frac{1}{27}\left((9t^2+4)^{3/2}-8\right).\qquad (\ast)$$ This is the arclength $s$, expressed as a function of $t$.
We want to parametrize in terms of $s$. So solve for $t$ in terms of $s$, using $(\ast)$. When you solve, there will be two candidate values of $t$. Take the non-negative one, since we started at $t=0$ and were told that t$is increasing. Finally, in the original parametrization, replace$t$by its value in terms of$s$. - Hint: Substitute$(x-1)^{1/3}=t$. Your integral will boil down to $$\int_{-1}^1t\sqrt{4+9t^2}\rm dt$$ Now set$4+9t^2=u$and note that$\rm du=18t~~\rm dt$which will complete the computation. (Note that you need to change the limits of integration while integrationg over$u$.) A Longer way: Now integrate by parts with$u=t$and$\rm d v=\sqrt{4+9t^2}\rm dt$and to get$v$, you'd like to keep$t=\dfrac{2\tan \theta}{3}$- you can substitute$4+9t^2= x$and then proceed.. – zapkm Mar 1 '12 at 3:54 @PradipMishra Thank You. I don't know why I could not think of this! Thank you for the pointer! – user21436 Mar 1 '12 at 4:13 As given curve is not regular when$t=0$and your curve parameter runs from$-1$to$1$, Hence Below is arc lengh parameter of the curve from$0$to$1$. And same will work for$0$to$-1$. What is arc length formula, when curve is given parametric form as in your case $$\gamma (t)= (t^3+1, t^2-1)$$ Arc length formula is $$s(t)= \int_{1}^t\|\gamma'(t)\|dt$$ That is we have $$s(t)=\int_{1}^t\|(3t^2, 2t)\|dt$$ $$s(t)=\int_{1}^t t\sqrt{9t^2+4} dt$$ $$s(t)= \left[\frac{(4+9t^2)^\frac{3}{2}}{27}\right]_{1}^t$$ $$s(t)= \frac{(4+9t^2)^\frac{3}{2}-13^\frac{3}{2}}{27}$$ which gives $$t(s)= \left(\frac{(27s+13^\frac{3}{2})^\frac{2}{3}-4}{9}\right)^\frac{1}{2}$$ Putting the value of$t$in$\gamma(t)$, you will have$\tilde{\gamma}(s)=\gamma(t(s))$arc length parameterization.. - Reparametrizing the curve in terms of arc length from a base point means rewriting the equation of the curve so that it tells you what point is at distance$s$from the base point for any given$s$. It’s easier to find the arc length parametrization directly. Let$s(u)$be the length of the arc from$t=0$(since that’s the value of$t$that yields the point$\langle 1,-1\rangle$) to$t=u; then \begin{align*}s(u)&=\int_0^u\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\ &=\int_0^u\sqrt{(3t^2)^2+(2t)^2}dt\\ &=\int_0^u\sqrt{t^2(9t^2+4)}dt\\ &=\int_0^ut\sqrt{9t^2+4}dt\\ &=\frac1{27}\left[(9t^2+4)^{3/2}\right]_0^u\\ &=\frac1{27}\left((9u^2+4)^{3/2}-8\right)\;. \end{align*} Replaceu$by$t$: the length of the arc from$\langle x(0),y(0)\rangle$to$\langle x(t),y(t)\rangle$is $$s(t)=\frac1{27}\left((9t^2+4)^{3/2}-8\right)\;,$$ so $$t(s)=\left(\frac19(27s+8)^{2/3}-4\right)^{1/2}\;.$$ This gives you the value of$t$that specifies the point on the curve that is$s$units from the initial point$\langle 1,-1\rangle$; to finish the job, you just need to express$x$and$y$as functions of$s$, which is a straightforward substitution into the$x(t)$and$y(t)\$ formulas.