Problem 31 - Calculus Early Transcendentals

Question

A water tank is shaped like an inverted cone with height \(6 \mathrm{m}\) and base radius \(1.5 \mathrm{m}\) (see figure). a. If the tank is full, how much work is required to pump the water to the level of the top of the tank and out of the tank? b. Is it true that it takes half as much work to pump the water out of the tank when it is filled to half its depth as when it is full? Explain.

Step-by-Step Solution

Answer

Answer: No, it does not take half as much work to pump the water out when the tank is filled to half its depth as when it is full. The ratio of work done in the half-filled tank to a full tank is approximately 0.345, which is not equal to one-half (0.5).

Step 1: Set up the problem

Find the volume of the cone and the height of the water column at any depth. We know that the volume of a cone is given by: \(V = \cfrac{1}{3}\pi r^2 h\). As the radius of the cone decreases linearly with the height, at any height \(x\) of the cone, the radius is related to the height by the equation: \(r = \cfrac{1.5}{6}(6-x)\). Now we can find the volume of the water column of thickness \(\delta x\) at height x: \(\delta V = \pi r^2 \delta x = \pi (\cfrac{3}{4}(6-x))^2 \delta x\).

Step 2: Calculate the work done

The work done to lift this water column is the force times the distance, which equals the mass times gravity times distance: \(W = m*g*(6-x)\). We can relate the mass to volume by multiplying it with the density of water (\(\rho\)): \(m =\rho* \delta V\). So, \(W = \rho * \delta V * g * (6 - x) = \rho * \pi (\cfrac{3}{4}(6-x))^2 \delta x * g * (6 - x)\). To find the total work done to pump the full tank of water, integrate this expression with respect to x from 0 to 6: $$W_{total} = \int_{0}^{6} \rho * \pi (\cfrac{3}{4}(6-x))^2 g (6 - x) dx $$.

Step 3: Evaluate the integral

Evaluating the integral, we get: $$W_{total} = \rho \pi g \int_{0}^{6} (\cfrac{3}{4}(6-x))^2(6-x) dx $$. Let \(u = 6-x\), then \(du = -dx\). The limits of integration will change to 6 for lower and 0 for upper: $$W_{total} = -\rho \pi g \int_{6}^{0} (\cfrac{3}{4}u)^2u du $$. Now solve the integral: $$W_{total} = -\cfrac{27}{64}\rho \pi g \int_{6}^{0} u^3 du $$. $$W_{total} = -\cfrac{27}{64}\rho \pi g [\cfrac{1}{4}u^4]_{6}^{0} $$. $$W_{total} = \cfrac{27}{256}\rho \pi g (6^4)$$. Use \(\rho = 1000 \cfrac{kg}{m^3}\) and \(g = 9.81 \cfrac{m}{s^2}\) to calculate the numerical value: $$W_{full} = \cfrac{27}{256} * 1000 * \pi * 9.81 * 6^4 = 4,054,820\thinspace J.$$. For part (b), we need to find the work done to pump the water out when the tank is filled to half its depth.

Step 4: Calculate work for half-filled tank

When the tank is filled to half its depth, i.e., at x = 3, we need to modify the limits of integration for the work done equation. $$W_{half} = \int_{3}^{6} \rho * \pi (\cfrac{3}{4}(6-x))^2 g * (6 - x) dx$$. Again, substitute \(u = 6-x\) and evaluate the integral. The result is: $$W_{half} = \cfrac{27}{256}\rho \pi g [\cfrac{1}{4}u^4]_{3}^{6} $$. $$W_{half} = \cfrac{27}{256}\rho \pi g (6^4 - 3^4) = 1,398,908 \thinspace J.$$

Step 5: Compare the results

Let's check if the work done to pump water from half-filled tank is half the work for a full tank. Compare the ratio of their values: $$\cfrac{W_{half}}{W_{full}} = \cfrac{1,398,908}{4,054,820} \approx 0.345$$. Since the ratio of work done in the half-filled tank to a full tank is not one-half (0.5), it is not true that it takes half as much work to pump the water when the tank is filled to half its depth as when it is full.

Key Concepts

Integration

Integration is a fundamental concept in calculus, often used to find areas under curves, volumes, and other quantities that accumulate over space or time. In the context of this problem, integration is utilized to calculate the work needed to pump water out of the tank. The total work involves integrating the expression for work over the entire height of the tank. By integrating from zero to the full height of the tank, we consider the cumulative work done in lifting each small volume of water from its initial position to the top of the tank. In this way, integration serves as a powerful tool to sum up infinitesimally small contributions, such as the work needed to move small volumes of water, over a continuous range.

Work and Energy

Work and energy are key concepts in physics, related to force and movement. Work is defined as the force applied to an object times the distance the object moves in the direction of the force. For lifting water out of a tank, the force is the weight of the water, which is the mass times gravity. The work done to pump water is calculated by multiplying this force by the height the water is lifted. Energy, on the other hand, refers to the capacity to do work. In this scenario, the energy needed to pump the water is obtained from the work calculation. Understanding work and energy helps us quantify the effort or power needed to perform tasks like pumping water, emphasizing the practical application of these physics concepts.

Volume of Solids

The volume of solids, such as the inverted conical tank in the exercise, is a vital factor when dealing with physical quantities like water. For a cone, the volume formula is given by \(V = \frac{1}{3} \pi r^2 h\). To manage the volume at varying water levels, it is crucial to understand how the radius of the water surface changes with height, represented by \(r = \frac{1.5}{6}(6-x)\). Knowing how to calculate the volume of the water at different depths allows for precise determination of how much water remains in the tank, essential for accurate work calculations. This understanding aids in applying integration effectively by providing an accurate description of the submerged area's volume.

Physics Applications

Physics applications in this exercise manifest in the practical task of calculating the work required to lift water, demonstrating the use of physics principles in real-life scenarios. This problem combines geometric understanding with physical concepts like gravity and force to solve a tangible situation—removing liquid from a container. By applying formulas related to volume, mass, and work, physics plays an instrumental role in enabling us to design and operate systems efficiently. The accurate computation of work needed to pump water from different levels shows the direct application of theoretical principles in practical engineering tasks, highlighting the relevance of physics in everyday problem-solving.