Calculus 10th Edition

Published by Brooks Cole

Chapter 1 - Limits and Their Properties - 1.1 Exercises: 10

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.

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