MHB When Should Parametric Equations Be Used to Calculate Curve Length?

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Parametric equations can be useful for calculating curve length when the relationship between variables is complex or when dealing with curves that are not easily expressed in Cartesian form. In the discussion, the arc length of the function f(x) = (1/3)(x^2 + 2)^(3/2) over the interval [0, a] was explored. The derivative f'(x) was found to be x√(x^2 + 2), leading to the integral for arc length. The integral simplifies to s = (a^3/3) + a, demonstrating that parametric forms may complicate the process unnecessarily. Ultimately, using Cartesian equations is often more straightforward for this type of calculation.
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Find the arc length of
$$f (x)=(1/3)(x^2 +2)^{3/2}$$
On the interval [0, a]

The parametric I got

$$y=t$$
$$x=\sqrt{(3t)^{2/3}-2}$$

I proceeded but didnt get the answer of

$$a+\frac{a^3}{3}$$
 
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Re: Lenght of a curve

I wouldn't bother with parametrization...:)

We have:

$$f(x)=\frac{1}{3}\left(x^2+2\right)^{\frac{3}{2}}$$

Hence:

$$f'(x)=x\sqrt{x^2+2}$$

And so the arc-length $s$ will be given by:

$$s=\int_0^a\sqrt{1+\left(x\sqrt{x^2+2}\right)^2}\,dx$$

Can you proceed?
 
$$\displaystyle
s=\int_0^a\sqrt{1+\left(x\sqrt{x^2+2}\right)^2}\,dx
=\int_{0}^{a} \left({x}^{2}+1\right)\,dx
=\frac{a^3}{3}+a
$$

When do we use parametrics for length of curve
 
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