/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 60 The body's reaction to medicine ... [FREE SOLUTION] | 91Ó°ÊÓ

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The body's reaction to medicine dose of medicine can sometimes be represented by an equation of the form $$R=M^{2}\left(\frac{C}{2}-\frac{M}{3}\right)$$ where \(C\) is a positive constant and \(M\) is the amount of medicine absorbed in the blood. If the reaction is a change in blood pressure, \(R\) is measured in millimeters of mercury. If the reaction is a change in temperature, \(R\) is measured in degrees, and so on. Find \(d R / d M .\) This derivative, as a function of \(M,\) is called the sensitivity of the body to the medicine. In Section \(4.5,\) we will see how to find the amount of medicine. In Section the body is most sensitive.

Short Answer

Expert verified
The sensitivity as a function of \( M \) is \( MC - M^2 \).

Step by step solution

01

Understand the Problem

We need to find the derivative of the reaction function \( R = M^{2}\left(\frac{C}{2}-\frac{M}{3}\right) \) with respect to the medicine amount \( M \). This derivative \( \frac{dR}{dM} \) represents the sensitivity of the body to the medicine.
02

Apply the Product Rule

The function \( R \) is a product of two functions: \( M^2 \) and \( \left(\frac{C}{2} - \frac{M}{3}\right) \). To find \( \frac{dR}{dM} \), we'll apply the product rule, which states \( \frac{d}{dx}(u\cdot v) = u'v + uv' \). Let \( u = M^2 \) and \( v = \frac{C}{2} - \frac{M}{3} \).
03

Differentiate Each Function

Differentiate \( u = M^2 \) and \( v = \frac{C}{2} - \frac{M}{3} \). For \( u \), the derivative \( u' = 2M \). For \( v \), since \( \frac{C}{2} \) is a constant, its derivative is \( 0 \), and the derivative of \( -\frac{M}{3} \) is \( -\frac{1}{3} \). Thus, \( v' = -\frac{1}{3} \).
04

Substitute into the Product Rule

Using the results from the previous step, substitute into the product rule: \[ \frac{dR}{dM} = 2M \left( \frac{C}{2} - \frac{M}{3} \right) + M^2 \left( -\frac{1}{3} \right) \].
05

Simplify the Derivative

Simplify the expression from the substitution: \[ \frac{dR}{dM} = 2M \left( \frac{C}{2} - \frac{M}{3} \right) - \frac{M^2}{3} \]. Expanding the first term gives \( = M C - \frac{2M^2}{3} \). Combine like terms: \( \frac{dR}{dM} = M C - \frac{3M^2}{3} = M C - M^2 \).
06

Final Expression for Sensitivity

Thus, the derivative \( \frac{dR}{dM} = MC - M^2 \). This represents the sensitivity of the body's reaction to the medicine with respect to the amount \( M \).

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Key Concepts

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

Product Rule
The product rule is an essential tool in calculus for finding derivatives when dealing with functions that are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), the product rule provides a way to differentiate their product. The rule states that the derivative of the product \( u(x) \cdot v(x) \) is given by:
    \( \frac{d}{dx}(u \cdot v) = u'v + uv' \)
Here's how it works:
  • \( u' \) is the derivative of \( u(x) \).
  • \( v' \) is the derivative of \( v(x) \).
  • Multiply \( u' \) by \( v \) and \( u \) by \( v' \), and then add them together.
In the exercise problem, we apply the product rule to the function \( R = M^2 \left( \frac{C}{2} - \frac{M}{3} \right) \). This involves treating \( M^2 \) as \( u(M) \) and \( \frac{C}{2} - \frac{M}{3} \) as \( v(M) \). Differentiating each separately and then applying the product rule gives us the sensitivity of the body's reaction, \( \frac{dR}{dM} \).
Using this approach simplifies complex multiplication problems in calculus by allowing us to focus on smaller, more manageable parts.
Sensitivity Analysis
Sensitivity analysis in this context helps us understand how changes in the amount of medicine absorbed, \( M \), affect the body's reaction, \( R \). Essentially, by calculating \( \frac{dR}{dM} \), we can see how sensitive the body is to changes in \( M \).
For our exercise, the derivative \( \frac{dR}{dM} = MC - M^2 \) tells us:
  • If \( M \) increases by a small amount, \( \Delta M \), the reaction \( R \) changes approximately by \( \Delta R = (MC - M^2) \Delta M \).
  • This expression shows the direct impact of changing medicine amounts on the reaction.
Understanding sensitivity is crucial because it can inform decisions regarding the dosage of medicine. If the sensitivity is high, small changes in \( M \) can cause large reactions, which might be risky in certain medical scenarios.
By conducting sensitivity analysis, healthcare professionals can better tailor treatments to individual patient needs, maximizing effectiveness while minimizing side effects.
Calculus Problem Solving
Calculus problem-solving often involves breaking down equations and applying rules consistently to find derivatives or integrals. For the problem given, it involves straightforward steps that are repeated across many calculus problems:
  • Understand the problem: Identify what you're solving for—in this case, the derivative \( \frac{dR}{dM} \).
  • Simplify and apply rules: Use calculus rules, like the product rule, to tackle complex expressions. Simplifying makes the math more accessible and manageable.
  • Consistency: Ensuring that each step follows logically from the previous ones helps avoid mistakes.
  • Simplify results: Once the calculation is complete, simplify the expression to make it easy to interpret and apply.
  • Check and interpret: Finally, check your work for errors and interpret what the results mean in context.
By practicing these skills and understanding the underlying principles, you can approach calculus problems with confidence. This exercise, focusing on finding a derivative using the product rule, serves as a classic example of how calculus provides tools for analyzing and interpreting changes in real-world contexts, such as medical treatment efficacy.

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