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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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