close to 1 but the approximation usually gets worse as we move further away

# Close to 1 but the approximation usually gets worse

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close to 1, but the approximation usually gets worse as we move further away from 1. 3. (3 pts) The edge of a cube was measured to be 8 cm, with a maximum possible error of 0 . 5 cm. Use a differential to estimate the maximum possible error in computing the volume of the cube. (This next part is ungraded, just for fun. Using a calculator, find the actual error in measuring volume if the radius was really 8 . 5 cm instead of 8 cm, and find the actual error if the radius was actually 7 . 5 cm instead of 8 cm. Compare these errors to the answer you got using differentials). Solution : V = x 3 , so dV = 3 x 2 dx . Here, dx represents the error in the measurement (the difference between the actual edge length and our measured edge length), and dV represents the corresponding approximate error in volume. dV = 3 · 8 2 · 0 . 5 = 96 cm 3 . What was the actual error in volume if we were off by 0 . 5 cm? It depends whether if 8 cm was an overestimate or underestimate. If the actual radius was 7 . 5 cm, then V = 7 . 5 3 = 421 . 88 cm 3 , while our approximation was V = 8 3 = 512 cm 3 . The actual error would be 90 . 12 cm 3 . If the actual radius was 8 . 5 cm, then V = 8 . 5 3 = 614 . 13 cm 3 , while our approximation was V = 512 cm 3 . The actual error would be 102 . 13 cm 3 .
4. (4 points) Alfador produces gear for a racquetball league. It costs him \$0 . 005 per square centimeter to produce racquetballs for the league. In the past, the league used racquetballs with diameter 4 cm, and he produced 30 , 000 racquetballs. However, league regulations now require racquetballs to have diameter 4 . 05 cm.