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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.7.42b
Capitolo 6, Problema 6.7.42b

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).
b. If the length is doubled, is the required work doubled? Explain.

Guida verificata passo dopo passo
1
Step 1: Understand the problem context. The trough has a semicircular cross section with radius 0.25 m and length 3 m. The work required to empty the trough depends on the volume of water and the distance it must be lifted.
Step 2: Recall that work is calculated as the integral of force times distance. Here, force is related to the weight of the water, which depends on the volume of water and the density of water.
Step 3: The volume of water in the trough is the area of the semicircular cross section multiplied by the length. The area of a semicircle is given by \(\frac{1}{2} \pi r^2\), so the volume is \(V = \frac{1}{2} \pi (0.25)^2 \times 3\).
Step 4: If the length is doubled, the volume of water doubles because the cross-sectional area remains the same but the length doubles. Therefore, the total weight of the water doubles.
Step 5: Since the work is the integral of weight times the distance the water is lifted, and the lifting distance does not change with length, doubling the length doubles the volume and weight, thus doubling the required work.

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Work Done by a Variable Force

Work is the integral of force over distance. When emptying a trough, the force varies with the depth of water, so the work is calculated by integrating the weight of water lifted at each depth multiplied by the distance it is lifted.
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Work Done On A Spring (Hooke's Law)

Volume and Cross-Sectional Area

The volume of water in the trough depends on the cross-sectional area and the length. Doubling the length doubles the volume, which directly affects the total weight and thus the work required to empty the trough.
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Introduction to Cross Sections

Linear Scaling of Work with Length

Since the cross-sectional shape and height remain constant, the work required to empty the trough scales linearly with its length. Doubling the length doubles the volume and weight of water, so the total work required is also doubled.
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06:22
Introduction To Work
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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.

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