Problem 102 - Calculus and its applications

Question

The percentage \(P\) of doctors who prescribe a certain new medicine is $$P(t)=100\left(1-e^{-0.2 t}\right)$$. where \(t\) is the time, in months. a) Find \(P(1)\) and \(P(6)\). b) Find \(P^{\prime}(t)\). c) How many months will it take for \(90 \%\) of doctors to prescribe the new medicine? d) Find \(\lim _{\rightarrow \infty} P(t),\) and discuss its meaning.

Step-by-Step Solution

Answer

a) \(P(1) \approx 18.13\), \(P(6) \approx 70.08\). b) \(P'(t) = 20e^{-0.2t}\). c) \(t \approx 11.51\) months. d) \(\lim_{t \to \infty} P(t) = 100\), meaning all doctors will eventually prescribe the medicine.

Step 1: Calculate P(1)

Substitute 1 for \(t\) in the given formula to find \(P(1)\). \[P(1) = 100 \left( 1 - e^{-0.2 \cdot 1} \right) = 100 \left( 1 - e^{-0.2} \right)\].

Step 2: Calculate P(6)

Substitute 6 for \(t\) in the formula to find \(P(6)\). \[P(6) = 100 \left( 1 - e^{-0.2 \cdot 6} \right) = 100 \left( 1 - e^{-1.2} \right)\].

Step 3: Differentiate P(t)

Differentiate \(P(t)\) with respect to \(t\) to find \(P'(t)\). \[P'(t) = \frac{d}{dt} \left( 100 \left( 1 - e^{-0.2 t} \right) \right) = 100 \cdot 0.2 e^{-0.2 t} = 20 e^{-0.2 t}\].

Step 4: Solve for 90% Administration

To find when \(P(t) = 90\), solve \(90 = 100 \left(1 - e^{-0.2 t}\right)\). Simplifying, you get: \(1 - 0.9 = e^{-0.2t}\), leading to \(0.1 = e^{-0.2t}\). Taking natural logarithms, \(\ln(0.1) = -0.2t\). Thus, \(t = -\frac{\ln(0.1)}{0.2}\).

Step 5: Calculate Limit of P(t)

Evaluate \(\lim_{t \to \infty} P(t)\). Since \(e^{-0.2t}\) approaches 0 as \(t\) goes to infinity, \(P(t) \to 100\left(1 - 0\right) = 100\).

Key Concepts

Differentiation

When we hear about differentiation in calculus, it's all about how we can find the rate at which something changes. Imagine you're on a road trip—differentiation is like your car's speedometer showing how fast you're going at any given moment. Here, we're looking at how the percentage, \( P(t) \), of doctors prescribing a new medicine changes over time.
To find this rate of change, we differentiate our function. In this case, we have a function involving an exponential term. Differentiating \( P(t) = 100 (1 - e^{-0.2t}) \) gives us \( P'(t) = 20 e^{-0.2t} \).
This result tells us how quickly doctors are starting to use this medicine at any point in time. If you look at the equation, since \( e^{-0.2t} \) gets smaller as \( t \) increases, \( P'(t) \) also gets smaller, indicating the rate of new doctors prescribing the medicine slows down over time.

Exponential Functions

Exponential functions are everywhere in the real world—they describe growth or decay processes, like how bacteria multiply or how radioactive materials break down. The function \( P(t) = 100(1 - e^{-0.2t}) \) is an example of an exponential function, which has an \( e \) raised to a power.
Why do we care about exponential functions? They model situations where a quantity grows or decays proportionally to its current size. In our problem, \( e^{-0.2t} \) represents the diminishing group of doctors who haven't yet started prescribing the medicine.
The beauty of exponential functions is their predictability. We can use them to estimate values at different points in time, like finding \( P(1) \) and \( P(6) \), showing us how adoption grows quickly at first and then slows.

Limits

In calculus, limits help us understand what happens to a function as its input gets really large (or small). When solving \( \lim_{t \to \infty} P(t) \), we discover the long-term behavior of our doctor-prescribing function.
Since \( e^{-0.2t} \) approaches zero as \( t \) becomes very large, the limit of \( P(t) \) heads towards \( 100(1 - 0) = 100 \). This means, in the long run, 100% of doctors will eventually prescribe the medicine, assuming the trend continues without external changes.
Understanding limits provides a sneak peek into future outcomes without having to compute each step explicitly. It's a powerful tool to confirm if the assumptions we've made in our model correspond to realistic expectations.

Mathematical Modeling

Mathematical modeling involves creating equations and formulas to represent real-world phenomena. Here, the equation \( P(t) = 100(1 - e^{-0.2t}) \) models how doctors adopt a new medicine over time.
Why is mathematical modeling crucial? It helps us make predictions. By understanding and modeling the factors influencing adoption (like \( t \) being time), we gain insights into the process dynamics. For example, when we set \( P(t) = 90 \), we're solving to find the time it takes for 90% of doctors to prescribe the medicine.
This approach allows healthcare professionals and businesses to anticipate product uptake and plan resources accordingly, often saving time and money by avoiding trial-and-error in the real world. In essence, modeling creates a bridge between abstract math concepts and concrete real-world applications.