Colby has a square placemat. The area of the

1.) Find the length of the arc of the graph x^4 = y^6 from x = 1 to x = 8.

xxTIMURxx [149]

First, rewrite the equation so that y is a function of x :

$x^4 = y^6 \implies \left(x^4\right)^{1/6} = \left(y^6\right)^{1/6} \implies x^{4/6} = y^{6/6} \implies y = x^{2/3}$

(If you were to plot the actual curve, you would have both $y=x^{2/3}$ and $y=-x^{2/3}$ , but one curve is a reflection of the other, so the arc length for 1 ≤ x ≤ 8 would be the same on both curves. It doesn't matter which "half-curve" you choose to work with.)

The arc length is then given by the definite integral,

$\displaystyle \int_1^8 \sqrt{1 + \left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx$

We have

$y = x^{2/3} \implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac23x^{-1/3} \implies \left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2 = \dfrac49x^{-2/3}$

Then in the integral,

$\displaystyle \int_1^8 \sqrt{1 + \frac49x^{-2/3}}\,\mathrm dx = \int_1^8 \sqrt{\frac49x^{-2/3}}\sqrt{\frac94x^{2/3}+1}\,\mathrm dx = \int_1^8 \frac23x^{-1/3} \sqrt{\frac94x^{2/3}+1}\,\mathrm dx$

Substitute

$u = \dfrac94x^{2/3}+1 \text{ and } \mathrm du = \dfrac{18}{12}x^{-1/3}\,\mathrm dx = \dfrac32x^{-1/3}\,\mathrm dx$

This transforms the integral to

$\displaystyle \frac49 \int_{13/4}^{10} \sqrt{u}\,\mathrm du$

and computing it is trivial:

$\displaystyle \frac49 \int_{13/4}^{10} u^{1/2} \,\mathrm du = \frac49\cdot\frac23 u^{3/2}\bigg|_{13/4}^{10} = \frac8{27} \left(10^{3/2} - \left(\frac{13}4\right)^{3/2}\right)$