#### Answer

(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.

#### Work Step by Step

(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.