Problem 29 - Biocalculus Calculus for the Life Sciences

Question

A manufacturer of lightbulbs wants to produce bulbs that last about 700 hours but, of course, some bulbs burn out faster than others. Let \(F(t)\) be the fraction of the company's bulbs that burn out before \(t\) hours, so \(F(t)\) always lies between 0 and \(1 .\) (a) Make a rough sketch of what you think the graph of \(F\) might look like. (b) What is the meaning of the derivative \(r(t)=F^{\prime}(t) ?\) (c) What is the value of \(\int_{0}^{\infty} r(t) d t ?\) Why?

Step-by-Step Solution

Answer

(a) Graph starts at 0 and approaches 1. (b) r(t) is the burnout rate. (c) Integral equals 1.

Step 1: Understand the Function F(t)

The function \( F(t) \) represents the fraction of lightbulbs that burn out before time \( t \). As \( t \) increases from \( 0 \) hours to infinity, \( F(t) \) increases from \( 0 \) (none burnt out) and approaches \( 1 \) (all have burnt out).

Step 2: Sketch the Graph of F(t)

Since \( F(t) \) starts at 0 when \( t = 0 \) and approaches 1 as \( t \to \infty \), the graph likely begins near the origin and gradually levels off as it approaches horizontal asymptote at \( F(t) = 1 \). This graph might resemble an S-shaped curve (i.e., logistic growth) or an exponential growth curve that plateaus.

Step 3: Interpret the Derivative r(t)

The derivative \( r(t) = F'(t) \) measures the rate at which bulbs burn out at time \( t \). If \( r(t) \) is high at a certain \( t \), many bulbs are burning out at that moment. It gives a sense of how the failure rate changes over time.

Step 4: Evaluate the Integral of r(t)

The integral \( \int_{0}^{\infty} r(t) \, dt \) calculates the total change in \( F(t) \) from \( 0 \) to \( \infty \). This means finding how much \( F(t) \) increases over the entire time span. Since \( F(t) \) starts at 0 and ends at 1, this integral should equal \( 1 \), indicating that eventually all bulbs burn out.

Key Concepts

Lightbulb Lifespan Distribution

The lifespan distribution of lightbulbs from a manufacturer can be represented using the function \( F(t) \), which indicates the fraction of bulbs that burn out before reaching \( t \) hours of usage. This portrayal gives a broad view of the reliability and consistency in the manufacturing process. Beginners might visualize \( F(t) \) starting at zero because no bulbs have burned out at \( t = 0 \), and it progresses to one as time extends towards infinity, illustrating that eventually all the bulbs will have burnt out.

• The shape of \( F(t) \) could demonstrate logistic growth, where there's a slow start, rapid increase, and then a steadying off.
• This graph tells us essential aspects such as the median lifespan and variability among the lightbulbs' longevity.

Understanding this distribution is crucial for manufacturers aiming to improve product quality or targeting specific consumer expectations.

Derivative Interpretation

Interpreting the derivative \( r(t) = F'(t) \) provides insights into the rate at which lightbulbs burn out over time. This derivative isn't just a mathematical operation; it offers a tangible sense of urgency at different phases of the bulb's life.

• When \( r(t) \) is high at any given \( t \), it implies that many bulbs are failing at that moment, indicating a peak burn-out phase.
• A low \( r(t) \) suggests a quiescent period where fewer bulbs burn out, reflecting either durable bulbs or high initial quality.

By analyzing \( r(t) \), we can identify critical periods in the product life cycle, guiding strategic decisions in product development and improvement.

Curve Sketching

Sketching the graph of \( F(t) \) involves visualizing how the fraction of burnt-out bulbs evolves over time. This process can be intuitive and doesn't always require precise computation.

• Begin with the understanding that at \( t = 0 \), no bulbs have burnt out, so \( F(0) = 0 \).
• As \( t \) increases, more bulbs are likely to burn out, making \( F(t) \) rise steadily.
• Eventually, \( F(t) \) approaches 1, suggesting that all bulbs have burnt out as time approaches infinity.

Picture \( F(t) \) forming an S-shaped curve: starting quickly, rising rapidly, then slowly approaching a horizontal line at \( F(t) = 1 \). Curve sketching aids in comprehending complex mathematical relationships through simple graphical interpretations.

Integral Evaluation

Evaluating the integral \( \int_{0}^{\infty} r(t) \, dt \) serves to sum up all changes in \( F(t) \) from 0 to infinity. In practical terms, it measures the cumulative number of burnt-out lightbulbs over their possible lifespan, offering a complete picture of the process.

• This integral equates to the change in \( F(t) \) from its starting value at 0 up to its concluding value at 1.
• Hence, \( \int_{0}^{\infty} r(t) \, dt = 1 \), meaning all bulbs eventually burn out. It's a comprehensive evaluation of the distribution of bulb failures.

This calculation confirms the consistency and reliability of our model, providing confidence in the perceived behavior of the lightbulbs over time.