Problem 85
Question
The circumference \(C,\) in centimeters, of a healing wound is approximated by $$ C(r)=6.28 r $$ where \(r\) is the wound's radius, in centimeters. a) Find the rate of change of the circumference with respect to the radius. b) Find \(C^{\prime}(4)\) c) Explain the meaning of your answer to part (b).
Step-by-Step Solution
Verified Answer
a) The rate of change is 6.28.
b) \( C'(4) = 6.28 \).
c) The circumference increases by 6.28 cm per cm of radius.
1Step 1: Understanding the Formula
The given formula for circumference is \( C(r) = 6.28 r \). This represents a linear function where the circumference \( C \) is directly proportional to the radius \( r \). The coefficient 6.28 is the proportional constant, representing the rate at which the circumference changes with respect to the radius.
2Step 2: Finding the Rate of Change
To find the rate of change of the circumference with respect to the radius, we differentiate the function \( C(r) = 6.28r \) with respect to \( r \). The derivative \( C'(r) \) is calculated as follows:\[ C'(r) = \frac{d}{dr}[6.28r] = 6.28 \]This indicates that with each one-unit increase in the radius, the circumference increases by 6.28 centimeters.
3Step 3: Calculate \(C'(4)\)
To find \( C'(4) \), use the derivative we found in the previous step. Since \( C'(r) = 6.28 \) is constant, \( C'(4) \) is simply:\[ C'(4) = 6.28 \]
4Step 4: Interpretation of \(C'(4)\)
The value \( C'(4) = 6.28 \) means that when the radius of the wound is 4 centimeters, the circumference of the wound changes at a rate of 6.28 centimeters for each additional centimeter increase in the radius.
Key Concepts
DifferentiationLinear FunctionDerivative
Differentiation
Differentiation is a fundamental concept in calculus that is used to find the rate at which one quantity changes with respect to another. In simple terms, it helps us calculate the slope or steepness of a curve at any given point. In the context of our exercise, differentiation is used to determine how the circumference of the wound changes as the radius changes.
When we differentiate a function, we apply specific rules to find this rate of change. The function given in the exercise is linear, meaning it is of the form \( C(r) = 6.28r \), where \( C(r) \) represents the circumference and \( r \) is the radius. Differentiating this function with respect to \( r \) gives \( C'(r) = 6.28 \).
This derivative tells us that for every one-unit increase in the radius, the circumference increases by 6.28 centimeters. Differentiation is, therefore, valuable for understanding relationships and predicting behavior of quantities that change relative to each other.
When we differentiate a function, we apply specific rules to find this rate of change. The function given in the exercise is linear, meaning it is of the form \( C(r) = 6.28r \), where \( C(r) \) represents the circumference and \( r \) is the radius. Differentiating this function with respect to \( r \) gives \( C'(r) = 6.28 \).
This derivative tells us that for every one-unit increase in the radius, the circumference increases by 6.28 centimeters. Differentiation is, therefore, valuable for understanding relationships and predicting behavior of quantities that change relative to each other.
Linear Function
A linear function is an algebraic equation in which the quantities involved display a linear or constant rate of change. It can be expressed in the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept. However, our current focus is the rate of change.
In the exercise, the formula \( C(r) = 6.28r \) represents a linear function because the circumference \( C \) changes at a constant rate with respect to the radius \( r \). The proportionality constant, 6.28, is equivalent to the slope \( m \) in the general form of a linear equation. This slope indicates how quickly \( C \) increases as \( r \) increases.
Linear functions are simple yet powerful tools in mathematics because they model situations with a constant rate of change. Understanding them can provide insight into real-world problems where variables change linearly, such as distance-speed-time calculations or the growth of savings over time.
In the exercise, the formula \( C(r) = 6.28r \) represents a linear function because the circumference \( C \) changes at a constant rate with respect to the radius \( r \). The proportionality constant, 6.28, is equivalent to the slope \( m \) in the general form of a linear equation. This slope indicates how quickly \( C \) increases as \( r \) increases.
Linear functions are simple yet powerful tools in mathematics because they model situations with a constant rate of change. Understanding them can provide insight into real-world problems where variables change linearly, such as distance-speed-time calculations or the growth of savings over time.
Derivative
The derivative is a key tool in calculus that measures how a function changes as its input changes. In our exercise, the derivative \( C'(r) \) is used to find the rate at which the circumference of a wound changes as its radius changes.
Mathematically, the derivative is often represented as \( f'(x) \) and is computed using differentiation rules. For linear functions, the derivative is straightforward to calculate. The derivative of \( C(r) = 6.28r \) results in the constant \( 6.28 \), reflecting the constant rate of change in the relationship between circumference and radius.
This constant derivative is significant: it implies that no matter the current size of the wound, each additional centimeter in radius will consistently add 6.28 centimeters to the circumference. Understanding the derivative helps us make predictions and infer relationships between the variables involved.
Mathematically, the derivative is often represented as \( f'(x) \) and is computed using differentiation rules. For linear functions, the derivative is straightforward to calculate. The derivative of \( C(r) = 6.28r \) results in the constant \( 6.28 \), reflecting the constant rate of change in the relationship between circumference and radius.
This constant derivative is significant: it implies that no matter the current size of the wound, each additional centimeter in radius will consistently add 6.28 centimeters to the circumference. Understanding the derivative helps us make predictions and infer relationships between the variables involved.
Other exercises in this chapter
Problem 84
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