Emptying a water trough A water trough has a semicircular cross section with a radius of 0.25 m and a length of 3 m (see figure).
c. If the radius is doubled, is the required work doubled? Explain.

Verified step by step guidance

1. Step 1: Understand the problem context. The water trough has a semicircular cross section with radius \(r = 0.25\) m and length \(L = 3\) m. The work required to empty the trough depends on the volume of water and the height it must be lifted.
2. Step 2: Recall the formula for work done in lifting water. Work is the integral of force times distance. The force is the weight of the water, which depends on the volume and density, and the distance is the height the water is lifted.
3. Step 3: Express the volume of water in terms of the radius. The cross-sectional area of the semicircle is \(A = \frac{1}{2} \pi r^2\), so the volume is \(V = A \times L = \frac{1}{2} \pi r^2 L\).
4. Step 4: Analyze how doubling the radius affects the volume and the height the water must be lifted. Doubling the radius changes the cross-sectional area by a factor of \$4\( (since area depends on \)r^2$), and the height the water must be lifted also changes proportionally to the radius.
5. Step 5: Conclude that the work required is not simply doubled when the radius is doubled because work depends on both the volume (which scales with \(r^2\)) and the lifting height (which scales with \(r\)), so the total work scales with \(r^3\).

Key Concepts

Work Done by a Variable Force

Work in calculus is often calculated as the integral of a force over a distance. When emptying a trough, the force varies with the depth of the water, so the work is found by integrating the weight of water lifted at each layer times the distance it is lifted.

Volume and Cross-Sectional Area of a Semicircle

The volume of water in the trough depends on the semicircular cross-sectional area multiplied by the length. The area of a semicircle is (1/2)πr², so doubling the radius increases the area, and thus the volume, by a factor of four.

Relationship Between Radius and Work Required

Doubling the radius affects both the volume of water and the height it must be lifted. Since volume scales with the square of the radius and the lifting height scales linearly, the total work does not simply double but increases by a larger factor.