Final answer:
The propagated error in the volume of a set of plastic spheres with a diameter of 11 cm and a manufacturing tolerance of ±3 mm is calculated using the derivative of the volume formula for a sphere. The resulting propagated error is 30.4 π cm³.
Step-by-step explanation:
A set of plastic spheres are to be made with a diameter of 11 cm. The question asks for the propagated error in the volume of these spheres if the manufacturing process is accurate to within ±3 mm. To calculate this, we will use the formula for the volume of a sphere (V = 4/3 πr³) and differentiate it to find the sensitivity of the volume to changes in radius, since error propagation is dependent on this sensitivity. As the diameter is given to be 11 cm, the radius is 5.5 cm. An accuracy of ±3 mm is equivalent to ±0.3 cm in radius.
The differential formula would look like dV = |d(V)/d(r)| × dr, where d(V)/d(r) is the derivative of the volume with respect to the radius. For a sphere, d(V)/d(r) = 4 π r². Substituting the given radius of 5.5 cm, we get d(V)/d(r) = 4 π (5.5 cm)². The propagated error in volume, therefore, is dV = 4 π (5.5²) × 0.3 cm, which simplifies to dV = 30.4 π cm³. The propagated error describes the possible variation in volume due to the manufacturing tolerance.