Problem 45 - Calculus and its applications

Question

A monk weighing 170 lb begins a fast to protest a war. His weight after \(t\) days is given by $$W=170 e^{-0.008 t}$$ a) When the war ends 20 days later, how much does the monk weigh? b) At what rate is the monk losing weight after 20 days (before any food is consumed)?

Step-by-Step Solution

Answer

a) Approximately 144.86 lb. b) About -1.16 lb/day.

Step 1: Identify the Given Function

The weight of the monk after \( t \) days is given by the function \( W(t) = 170 e^{-0.008t} \). We are tasked with finding out the monk's weight and the rate of weight change after 20 days.

Step 2: Substitute the Value into the Weight Function

To find the monk's weight after 20 days, substitute \( t = 20 \) into the function: \[ W(20) = 170 e^{-0.008 \times 20} \].

Step 3: Calculate the Exponent Term

First, calculate the product in the exponent: \( -0.008 \times 20 = -0.16 \).

Step 4: Calculate the Exponential Function

Evaluate \( e^{-0.16} \) using a calculator to approximate this value (\( e^{-0.16} \approx 0.8521 \)).

Step 5: Calculate the Monk's Weight

Substitute \( e^{-0.16} \approx 0.8521 \) back into the weight function: \[ W(20) = 170 \times 0.8521 = 144.858 \text{ lb} \]. The monk weighs approximately 144.86 lb after 20 days.

Step 6: Find the Derivative of the Weight Function

The rate of weight loss is given by the derivative of \( W \) with respect to \( t \): \[ \frac{dW}{dt} = \frac{d}{dt}(170 e^{-0.008t}) = 170 \times -0.008 e^{-0.008t} \].

Step 7: Substitute the Value into the Derivative

Now evaluate \( \frac{dW}{dt} \) at \( t = 20 \): \[ \frac{dW}{dt} = 170 \times -0.008 \times e^{-0.16} \approx 170 \times -0.008 \times 0.8521 \].

Step 8: Calculate the Rate of Weight Loss

Compute the product: \( 170 \times -0.008 \times 0.8521 = -1.1577 \). Thus, the rate of weight loss is approximately \(-1.16 \text{ lb/day}\).

Key Concepts

The Derivative and Its Significance

A derivative represents the rate of change of a function. In mathematics, we often symbolize this process as \( \frac{d}{dt} \), where \( t \) is the variable in question. For the given function \( W(t) = 170 e^{-0.008t} \), the derivative \( \frac{dW}{dt} \) will tell us how fast the monk's weight is changing over time. The calculation of the derivative in this scenario can be broken down into simple steps:

• First, identify the derivative of the exponential function, which involves the natural constant \( e \).
• Apply the chain rule, which requires multiplying the derivative of the exponent \(-0.008 \) with the original function itself.

By substituting \( t = 20 \), we found the rate of change at 20 days. It was found to be approximately \(-1.16\) lb per day. This negative value signifies a decrease in weight or weight loss, which is consistent with the expected result of fasting.

Understanding Rate of Change

The rate of change is a fundamental concept in calculus, which helps identify how one quantity varies in relation to another. In the context of the monk's weight function, the rate of change refers to the speed at which his weight decreases over time due to the ongoing fast.

• The derivative \( \frac{dW}{dt} \) calculated earlier gives us the monk's rate of weight loss at any day \( t \).
• On the 20th day, it's helpful to observe the rate as \(-1.16\) pounds per day, which means the monk is losing a little over a pound each day at this point in time.

Breaking it down, the negative sign is crucial as it indicates a loss. We commonly see this interpretation in real-life scenarios across different disciplines, where the rate of change can tell us about growth, decay, or constancy of a particular quantity.

Exponential Functions and Decay

Exponential functions are a special type of mathematical expression where the variable appears in the exponent. In our exercise, the monk’s weight over time is described by the function \( W(t) = 170 e^{-0.008t} \).

• The base of the exponential function here is the number \( e \), approximately 2.718, a constant that appears in many areas of mathematics.
• The negative exponent \(-0.008t\) signifies a decay process, as it causes the function's value to decrease over time.

In practical terms, this type of function is essential for modeling phenomena like radioactive decay, population decline, or, as in our example, weight loss during fasting. It's crucial to not only know how to interpret these functions but also how to adjust them based on different situations, so we correctly model the real-world processes they aim to describe.