Problem 30 - Applied Calculus

Question

The quantity, \(Q \mathrm{mg}\), of nicotine in the body \(t\) minutes after a cigarette is smoked is given by \(Q=f(t)\). (a) Interpret the statements \(f(20)=0.36\) and \(f^{\prime}(20)=\) \(-0.002\) in terms of nicotine. What are the units of the numbers \(20,0.36\), and \(-0.002 ?\) (b) Use the information given in part (a) to estimate \(f(21)\) and \(f(30)\). Justify your answers.

Step-by-Step Solution

Answer

(a) 0.36 mg at 20 min; decreases by 0.002 mg/min. (b) \(f(21)\approx 0.358\), \(f(30)\approx 0.34\).

Step 1: Understanding the Function at a Point

The given statement \( f(20) = 0.36 \) means that 20 minutes after a cigarette is smoked, the amount of nicotine in the body is 0.36 mg. Here, 20 is measured in minutes and 0.36 is measured in milligrams (mg).

Step 2: Interpreting the Derivative

The derivative \( f'(20) = -0.002 \) represents the rate of change of the nicotine amount at \( t = 20 \) minutes. This means that at 20 minutes, the nicotine level is decreasing at a rate of 0.002 mg per minute. The unit "-0.002" is in mg/min.

Step 3: Estimating f(21)

Using the derivative, we can estimate \( f(21) \) by using linear approximation. Since \( f'(20) = -0.002 \), then \( f(21) \approx f(20) + f'(20) \times (21 - 20) \). This gives \( f(21) = 0.36 + (-0.002) \times 1 = 0.358 \).

Step 4: Estimating f(30)

Using the same method, \( f(30) \) can be estimated. Since the rate of change is \( -0.002 \), then \( f(30) \approx f(20) + f'(20) \times (30 - 20) \). This results in \( f(30) = 0.36 + (-0.002) \times 10 = 0.34 \).

Key Concepts

Derivative

In calculus, a derivative measures how a function changes as its input changes. It's like the function’s rate of change at a specific point. Imagine you're tracking a car's speed: if the car's speed (rate of change) increases, it covers more distance (function value) in the same time period. Similarly, with nicotine levels in the body, the derivative shows how the nicotine amount changes over time. For instance, in our problem, the derivative at 20 minutes is -0.002 mg/min. This means that for every additional minute after 20 minutes, the nicotine level decreases by 0.002 mg.

A few things to remember about derivatives:

• They can be positive, indicating an increase in the function value (imagine a car accelerating).
• They can be negative, indicating a decrease (like our car slowing down).
• The unit of a derivative combines the units of the function's value and its input, shown here as mg/min.

Understanding derivatives helps predict future outcomes by analyzing current trends; a powerful tool for estimating changes.

Linear Approximation

Linear approximation, also known as linearization, provides a method to estimate a function’s value near a known point using its derivative. It’s a technique grounded on the notion that a smooth curve can be approximated as a straight line close to a point you're interested in. Think of how a really small segment of a curve looks nearly straight.

Here’s how it works:

• Start with a known point, say, where you know both the function’s value and its derivative.
• Estimate nearby values using the formula: \( f(a + ext{{step}}) \approx f(a) + f'(a) \cdot ext{{step}} \).

For our nicotine example, to estimate the amount at 21 minutes, starting from 20 minutes:- Use the known value: 20 minutes, 0.36 mg.- Apply the rate of change \( f'(20) = -0.002 \) mg/min.- Calculate for one minute ahead: \( f(21) \approx 0.36 + (-0.002) \times 1 = 0.358 \) mg.
Linear approximation simplifies the process by briefly turning a complex curve into a straightforward line just long enough to get a quick answer.

Rate of Change

The rate of change is a central idea in both calculus and everyday life, revealing how one quantity alters in relation to another. It's everywhere, from speedometers in cars to financial markets indicating growth or loss like a handy compass for direction.

In mathematical functions, the rate of change tells us how a function’s output responds to changes in input. In simple terms, consider how temperature might change through a day, or the way the sale of ice cream might increase on hotter days - both are examples of rates of change.

In the context of our problem, the rate of change \( f'(20) = -0.002 \) mg/min lets us track how swiftly the nicotine level is adjusting at 20 minutes. More precisely:

• A negative value indicates a decreasing function, just like a bank balance going down when spending more than earning.
• A concise measure of how fast the decrease occurs.

This makes the rate of change an invaluable metric for predicting future values based on trends, fueling decisions across various scientific and business strategies.