## University Calculus: Early Transcendentals (3rd Edition)

$$\dfrac{13}{12}$$
We integrate in order to find the arc length: $$L= \int_{p}^{q} \sqrt {1+(y')^2}$$ $$\\=\int_{1}^{2} \sqrt {1+\dfrac{y^4}{16}-\dfrac{1}{2} +\dfrac {1}{y^4}} \space dy \\=[\dfrac{y^3}{12}-\dfrac{1}{y}]_1^2 \\=[\dfrac{2^3}{12}-\dfrac{1^3}{12})-(\dfrac{1}{2}-\dfrac{1}{1})] \\= \dfrac{13}{12}$$