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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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4 Answers

Your integration could be done. However, there is a much easier way. Calculate the arc-length using the parametric version of the curve.

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 $s$.

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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 $t=\dfrac{2\tan \theta}{3}$

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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
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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..

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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

$$\begin{align*}s(u)&=\int_0^u\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\\ &=\int_0^u\sqrt{(3t^2)^2+(2t)^2}dt\\ &=\int_0^u\sqrt{t^2(9t^2+4)}dt\\ &=\int_0^ut\sqrt{9t^2+4}dt\\ &=\frac1{27}\left[(9t^2+4)^{3/2}\right]_0^u\\ &=\frac1{27}\left((9u^2+4)^{3/2}-8\right)\;. \end{align*}$$

Replace $u$ 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.

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