Problem 20 - Single Variable Calculus

Question

A bucket weighing \(4 \mathrm{lb}\) when empty and attached to a rope of negligible weight is used to draw water from a well that is \(30 \mathrm{ft}\) deep. Initially, the bucket contains \(40 \mathrm{lb}\) of water, but as it is pulled up at a constant rate of \(2 \mathrm{ft} / \mathrm{sec}\), the water leaks out of the bucket at the rate of \(0.2 \mathrm{lb} / \mathrm{sec}\). Find the work done in pulling the bucket to the top of the well.

Step-by-Step Solution

Answer

The work done in pulling the bucket to the top of the well is approximately 1275 ft·lbs.

Step 1: Determine the initial and final weights of the bucket

Initially, the bucket with water weighs = 40 lb (weight of water) + 4 lb (weight of empty bucket) = 44 lb. Since the water leaks at a rate of 0.2 lb/sec, and it takes 30 ft / 2 ft/sec = 15 seconds to reach the top, the leaking water weight at the end is = 0.2 lb/sec × 15 sec = 3 lb. So, the weight of the remaining water in the bucket at the top is 40 lb - 3 lb = 37 lb. Final weight of the bucket at the top of the well = 37 lb (weight of remaining water) + 4 lb (weight of empty bucket) = 41 lb.

Step 2: Calculate the two distances the initial and final weights of the bucket travel up the well

Since the bucket starts from the bottom of the well, the initial weight of 44 lb travels 30 ft up the well. As the bucket reaches the top of the well, the final weight is 41 lb. The final weight also travels 30 ft up the well.

Step 3: Calculate the work done for the two weights

The work done for the initial weight (W1) = initial weight × distance = 44 lb × 30 ft = 1320 ft·lb. The work done for the final weight (W2) = final weight × distance = 41 lb × 30 ft = 1230 ft·lb.

Step 4: Use the Mean Value Theorem to approximate the work done

By the Mean Value Theorem, we can approximate the work done as the average of the initial and final works, since the change in work is constant. Average work done (W) = \(\frac{W_1 + W_2}{2}\) = \(\frac{1320\ \mathrm{ft\cdot lbs} + 1230\ \mathrm{ft\cdot lbs}}{2}\) = \(\frac{2550\ \mathrm{ft\cdot lbs}}{2}\) = 1275 ft·lbs. So, the work done in pulling the bucket to the top of the well is approximately 1275 ft·lbs.

Key Concepts

Mean Value Theorem

The Mean Value Theorem is a widely used concept in calculus that helps approximate the average value of a function within a specified interval. In simple terms, it states that for a continuous and differentiable function on a closed interval, there is at least one point where the instantaneous rate of change (derivative) is equal to the average rate of change over the entire interval. This theorem is very handy in physics and engineering for determining average quantities where precise calculations are challenging due to varying conditions.

When applying the Mean Value Theorem to work calculations, as in the original exercise, it is all about finding the average work done. The inconsistency in the bucket's weight as it is pulled up could normally complicate the calculation. However, by using the theorem, we get a straightforward approach: just average the work calculated for the initial and final weights. In essence, it's a mathematical shortcut that gives a reasonable approximation without needing to calculate every individual change.

Weight Calculation

Understanding how weight varies is crucial when calculating work done. In physics, the weight of an object is the force exerted by gravity on it and is often expressed in pounds (lb) or newtons (N).

In the case of the bucket in the exercise, you start with an initial weight, which includes both the weight of the empty bucket and the water it holds. As the water leaks out, the weight decreases steadily. It's essential first to know the initial weight, which was 44 lb (40 lb of water plus 4 lb of the empty bucket).

• Initial water weight: 40 lb.
• Empty bucket weight: 4 lb.
• Total initial weight: 44 lb.

Then, account for the leak rate of the water. Water leaks at 0.2 lb per second, and since it takes 15 seconds to haul the bucket up, 3 lb of water is lost, leaving 37 lb of water. Hence the final calculated weight, including the bucket, is 41 lb.

Rate of Change

The rate of change is a central idea in calculus that describes how a quantity changes over time. In this context, it refers to how fast the water is leaking from the bucket. The rate of change is steady here: 0.2 lb per second.

Rates of change help to predict how variables evolve, which is especially useful when dealing with dynamic systems where multiple factors vary simultaneously. By determining this rate, you can project how the contents of the bucket progressfully lightens as the bucket is raised.

• Leak rate: 0.2 lb/sec
• Total time to top: 15 seconds
• Water lost: 0.2 lb/sec × 15 sec = 3 lb

Calculating this rate and multiplying it by the time taken to raise the bucket gives you a straightforward way to estimate how the bucket's weight changes as it's hoisted.

Distance Calculation

In physics, calculating distance accurately is crucial in solving work problems. Distance directly influences the amount of work done, as work is defined as the product of force and distance (W = F × d).

In the exercise, the distance is straightforward: it's the height the bucket is lifted, which is 30 feet. Since both the initial and final weights of the bucket travel this entire distance, it plays a pivotal role in determining the work done at each weight stage.

• Initial and final lifting height: 30 ft

By understanding that this distance remains constant during the bucket's ascent, despite the changing weight due to the leaking water, you ensure your calculations for the work done remain accurate and consistent throughout the process.