Problem 29 - Applied Calculus

Question

For some painkillers, the size of the dose, \(D\), given depends on the weight of the patient, \(W\). Thus, \(D=f(W)\) where \(D\) is in milligrams and \(W\) is in pounds; (a) Interpret the statements \(f(140)=120\) and \(f^{\prime}(140)=3\) in terms of this painkiller. (b) Use the information in the statements in part (a) to estimate \(f(145)\)

Step-by-Step Solution

Answer

For a 140-pound patient, the dose is 120 mg, increasing by 3 mg/pound. Thus, \( f(145) \approx 135 \) mg.

Step 1: Interpret f(140)=120

The expression \( f(140) = 120 \) indicates that for a patient weighing 140 pounds, the required dose of the painkiller is 120 milligrams. This is the function \( f \) evaluated at the weight 140 pounds.

Step 2: Interpret f'(140)=3

The expression \( f'(140) = 3 \) means that at a weight of 140 pounds, the dose's rate of change with respect to weight is 3 milligrams per pound. Thus, if the patient’s weight increases by 1 pound, the dose is expected to increase by 3 milligrams.

Step 3: Estimate f(145) using linear approximation

Using the linear approximation method, \( f(145) \) can be estimated as follows: Since \( f'(140) = 3 \), for every additional pound beyond 140, the dose increases by 3 mg. Therefore, for 5 extra pounds (145 - 140), the dose increase is \( 5 \times 3 = 15 \) mg. Adding this to the dose at 140 pounds: \( f(145) \approx f(140) + 15 = 120 + 15 = 135 \), so \( f(145) \approx 135 \) mg.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that helps us understand how a function changes at any point. It shows the rate at which a function's output changes as its input changes. In the exercise, the differentiation is represented by the derivative, denoted as \( f'(140) \). The derivative at a specific input value provides the slope of the tangent line to the function's curve at that point.
For the painkiller dose scenario, \( f'(140) = 3 \) translates to the dosage increasing by 3 milligrams for every additional pound in the patient's weight. This tells us how sensitive the dose is to weight changes.

• The derivative is a powerful tool because it gives us local behavior insights at any given point.
• It helps in optimizing and adjusting values for various conditions, such as adjusting a medication dose effectively.

By understanding derivatives, you can predict and make informed decisions based on how slight changes in input affect the output.

Function interpretation

Interpreting a function involves understanding what specific values mean in a given context. Functions map inputs to outputs, and comprehension of this relationship is vital for real-world applications.
In our exercise, the statement \( f(140) = 120 \) is interpreted to mean that a patient who weighs 140 pounds should receive 120 milligrams of the painkiller. It's a straightforward application of a function where the input is patient weight, and the output is the medication dosage.

• Knowing how to interpret function values helps translate mathematical notation into meaningful real-life actions.
• Function interpretation allows for effective communication of mathematical results and their implications for practical use.
• It is essential for correctly applying calculated results to scenarios such as medicine dosage determination.

Understanding function values makes it easier to perform tasks that rely on precise and accurate calculations.

Linear approximation

Linear approximation is a useful tool in calculus to estimate the value of a function near a known value. It uses the tangent line at a certain point to approximate the function. This is particularly helpful when we need quick estimates without computing the exact function value.
In the given problem, linear approximation helps estimate \( f(145) \). Since \( f(140) = 120 \) and \( f'(140) = 3 \), we can express this as:\
\[ f(145) \approx f(140) + f'(140) \times (145 - 140) \]
This results in:\
\[ f(145) \approx 120 + 3 \times 5 = 135 \text{ milligrams} \]
Linear approximation makes calculations simpler by allowing:

• Quick assessments based on local behavior around a known point.
• Easy interpretation and communication of approximate values.
• Reduced complexity by avoiding direct computation with potentially complicated functions.

Understanding linear approximation is crucial when precise functions are either unavailable or expensive to compute intricately.