Problem 122 The reaction \(R\) of the body t... [FREE SOLUTION] (2024)

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Chapter 1: Problem 122

The reaction \(R\) of the body to a dose \(Q\) of medication is often representedby the general function $$R(Q)=Q^{2}\left(\frac{k}{2}-\frac{Q}{3}\right)$$ where \(k\) is a constant and \(R\) is in millimeters of mercury \((\mathrm{mmHg})\)if the reaction is a change in blood pressure or in degrees Fahrenheit\(\left(^{\circ} \mathrm{F}\right)\) if the reaction is a change in temperature.The rate of change \(d R / d Q\) is defined to be the body's sensitivity to themedication. a) Find a formula for the sensitivity. b) Explain the meaning of your answer to part (a).

Short Answer

Expert verified

Sensitivity formula: \( \frac{dR}{dQ} = kQ - Q^2 \). It shows sensitivity initially increases with dose but decreases at higher doses.

Step by step solution

01

Understand the Given Function

The given function is \[ R(Q) = Q^2 \left( \frac{k}{2} - \frac{Q}{3} \right) \]This function represents the body's reaction to a dose \(Q\).

02

Differentiate the Function

To find the body's sensitivity, we need to calculate the rate of change of \(R\) with respect to \(Q\) which involves differentiation.\[ \frac{dR}{dQ} = \frac{d}{dQ} \left( Q^2 \left( \frac{k}{2} - \frac{Q}{3} \right) \right) \]

03

Apply the Product Rule

Use the product rule for differentiation, which states \[ \frac{d}{dx} [f(x) \, g(x)] = f(x) \, \frac{dg(x)}{dx} + g(x) \, \frac{df(x)}{dx} \]Let \( f(Q) = Q^2 \) and \( g(Q) = \frac{k}{2} - \frac{Q}{3} \).

04

Differentiate Each Part

First, differentiate \(f(Q) = Q^2\):\[ \frac{df(Q)}{dQ} = 2Q \]Next, differentiate \(g(Q) = \frac{k}{2} - \frac{Q}{3}\):\[ \frac{dg(Q)}{dQ} = -\frac{1}{3} \]

05

Combine Derivatives Using Product Rule

Plug the derivatives back into the product rule formula:\[ \frac{dR}{dQ} = Q^2 \left( -\frac{1}{3} \right) + \left( \frac{k}{2} - \frac{Q}{3} \right) 2Q \]

06

Simplify the Expression

Simplify the expression from Step 5:\[ \frac{dR}{dQ} = -\frac{Q^2}{3} + 2Q \left( \frac{k}{2} \right) - 2Q \left( \frac{Q}{3} \right) \]\[ \frac{dR}{dQ} = -\frac{Q^2}{3} + kQ - \frac{2Q^2}{3} \]Combine like terms:\[ \frac{dR}{dQ} = kQ - Q^2 \left( \frac{1}{3} + \frac{2}{3} \right) \]\[ \frac{dR}{dQ} = kQ - Q^2 \frac{3}{3} \]\[ \frac{dR}{dQ} = kQ - Q^2 \]

07

Provide Final Answer

The formula for the sensitivity \( \frac{dR}{dQ} \) is:\[ \frac{dR}{dQ} = kQ - Q^2 \]

08

Interpret the Result

The result \( \frac{dR}{dQ} = kQ - Q^2 \) shows the sensitivity of the body to the medication dose. Sensitivity depends on the dose \(Q\) and a constant \(k\). When the dose increases, the sensitivity increases initially and then decreases when the quadratic term \( -Q^2 \) starts to dominate.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

differentiation

Differentiation is the mathematical process of finding the rate at which a function changes at any point. In this exercise, we needed to differentiate the reaction function \(R(Q)\) with respect to the dose \(Q\). By doing this, we can determine how the body’s reaction changes as the medication dose changes. Differentiation gives us the sensitivity formula, which describes the body's response to different doses of medication in terms of blood pressure or temperature variation.

rate of change

The rate of change tells us how one quantity changes in relation to another quantity. For example, in our exercise, the rate of change is \(dR/dQ\), representing how the body's reaction \(R\) changes with respect to the dose \(Q\) of the medication. It essentially shows how sensitive the body is to small changes in the dose. This concept is crucial because understanding the rate of change helps in determining the optimal dose of medication for maximum effectiveness without adverse effects.

product rule

When differentiating functions that are the product of two other functions, we use the product rule. The product rule states that if you have a function \(R(Q) = f(Q) \times g(Q)\), then its derivative \(R'(Q)\) is given by \(\frac{d}{dQ} [f(Q) \times g(Q)] = f(Q) \times \frac{dg(Q)}{dQ} + g(Q) \times \frac{df(Q)}{dQ}\). In our exercise, we set \(f(Q) = Q^2\) and \(g(Q) = \frac{k}{2} - \frac{Q}{3}\). Applying the product rule allowed us to find the derivative and thus understand how the body’s sensitivity changes with dose.

medication response

The medication response describes how the body reacts to a given dose of medicine. In our case, it’s represented by the function \(R(Q) = Q^2(\frac{k}{2} - \frac{Q}{3})\). This function essentially captures the physiological changes (such as changes in blood pressure or temperature) in response to varying doses. By analyzing the derived sensitivity formula \(dR/dQ = kQ - Q^2\), we can see that the body’s response is not linear. Initially, the response increases with increasing dosage up to a certain point, after which the response starts to decline as the dose continues to increase. This non-linear behavior is vital for medical practice, ensuring that patients receive safe and effective doses.

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Problem 122 The reaction \(R\) of the body t... [FREE SOLUTION] (3)

Most popular questions from this chapter

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Problem 122 The reaction \(R\) of the body t... [FREE SOLUTION] (2024)

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