## Calculus 10th Edition

(a) $4\sqrt 2$ (b) $\frac{3\sqrt 17+6\sqrt 29+2\sqrt 61+13}{12}$ (c) By creating smaller intervals, the approximation line gets closer to the curve, so the length of the approximation lines better estimate the length of the curve.
(a) Use the distance formula. $$\sqrt {(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$$ $$\sqrt {(5-1)^2+(1-5)^2}$$ $$\sqrt {(4)^2+(-4)^2}$$ $$\sqrt {16+16}$$ $$\sqrt {32}$$ $$4\sqrt {2}$$ (b) Use the distance formula for each interval. $$\sqrt {(2-1)^2+(\frac{5}{2}-5)^2}$$ $$\frac{\sqrt {29}}{2}$$ $$\sqrt {(3-2)^2+(\frac{5}{3}-\frac{5}{2})^2}$$ $$\frac{\sqrt {61}}{6}$$ $$\sqrt {(4-3)^2+(\frac{5}{4}-\frac{5}{3})^2}$$ $$\frac{13}{12}$$ $$\sqrt {(5-4)^2+(1-\frac{5}{4})^2}$$ $$\frac{\sqrt {17}}{4}$$ Add them together to get: $$\frac{3\sqrt 17+6\sqrt 29+2\sqrt 61+13}{12}$$ (c) By creating smaller intervals, the approximation line gets closer to the curve, so the length of the approximation lines better estimate the length of the curve.