I was asked to find the arc length of the curve of the following curve:
$24xy = x^4 + 48$ from $x = 2$ to $x = 4$
This has turned out to be a very difficult problem, I get stuck using the arc length formula with the derivative I have calculated.
$\endgroup$ 12 Answers
$\begingroup$Why is this difficult for you? Follow the right formula and try to find:
$$\sqrt{1+y'^2}=\frac{(x^4+16)}{8x^2},~~(x\neq 0)$$
$\endgroup$ 0 $\begingroup$We can put $$\begin{cases} x&=t \\y&=\frac{1}{24}t^3+2t^{-1}\end{cases}$$ Arc length formula $$s=\int_{2}^{4}{\sqrt{(x')^2+(y')^2}\text{d}t}$$ give us \begin{align*} s&=\int_{2}^{4}{\sqrt{1+\left(\frac{1}{8}t^2-2t^{-2}\right)^2}\text{d}t}\\ &=\int_{2}^{4}{\sqrt{\left(\frac{1}{8}t^2+2t^{-2}\right)^2}\text{d}t}\\ &=\int_{2}^{4}{\left(\frac{1}{8}t^2+2t^{-2}\right)\text{d}t}\\ &=\left[\frac{1}{24}t^3-2t^{-1}\right]_2^4\\ &=\frac{17}{6} \end{align*}
$\endgroup$
