Problem 15
Question
The average weight, \(W\), in pounds, of an adult is a function, \(W=f(c)\), of the average number of Calories per day, \(c\), consumed. (a) Interpret the statements \(f(1800)=155\) and \(f^{\prime}(2000)=0\) in terms of diet and weight. (b) What are the units of \(f^{\prime}(c)=d W / d c ?\)
Step-by-Step Solution
Verified Answer
(a) 1800 calories corresponds to 155 lbs; at 2000 calories, weight is stable. (b) Units of \(f'(c)\): pounds per calorie.
1Step 1: Interpret f(1800)=155
The function notation \(f(1800)=155\) means that when the average number of calories consumed per day is 1800, the average weight of an adult is 155 pounds. Thus, \(f(1800)\) describes a specific point on the curve where the independent variable (calories) is 1800, and the dependent variable (weight) is 155.
2Step 2: Interpret f'(2000)=0
The notation \(f'(2000) = 0\) indicates that when an adult consumes 2000 calories per day, their weight does not change with a small change in calorie consumption around 2000 calories. In other words, at this point, the average weight is stable, and increasing or decreasing calories from 2000 does not initially affect the weight.
3Step 3: Determine the units of f'(c)=dW/dc
The derivative \(f'(c)=\frac{dW}{dc}\) represents the rate of change of weight with respect to the number of calories consumed. The numerator \(dW\) is measured in pounds and the denominator \(dc\) is measured in calories. Therefore, the units of \(f'(c)\) are pounds per calorie.
Key Concepts
Understanding Derivatives in CalculusExploring Functions and Their Critical RoleRate of Change and Its Applications
Understanding Derivatives in Calculus
Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. They can be thought of as the rate of change of a function at a particular point. To understand derivatives, imagine a curve representing a function. At any point on this curve, a tangent line can be drawn. The slope of this tangent line is the derivative at that point.
For example, consider a situation where the function describes a relationship between two quantities, like in the exercise where weight is a function of calories. Here, the derivative tells us how small changes in the number of calories affect the weight.
For example, consider a situation where the function describes a relationship between two quantities, like in the exercise where weight is a function of calories. Here, the derivative tells us how small changes in the number of calories affect the weight.
- If the derivative is positive, the function is increasing at that point, and thus the weight increases as calories increase.
- If the derivative is negative, the function is decreasing, implying the weight decreases as calories increase.
- A derivative of zero means the function is neither increasing nor decreasing, indicating stability in the weight with regard to calories.
Exploring Functions and Their Critical Role
Functions are mathematical constructs that relate inputs to outputs. They are essential tools for modeling a variety of real-life situations. A function typically involves an independent variable (input) and a dependent variable (output), represented as, for example, \(y = f(x)\).
In the exercise, the function \(W = f(c)\) represents the average weight as a function of the average daily calorie consumption.
In the exercise, the function \(W = f(c)\) represents the average weight as a function of the average daily calorie consumption.
- Each specific calorie intake, \(c\), maps to a specific weight, \(W\).
- Functions provide a clear model for analyzing changes and predicting outcomes by considering the relationship of inputs and outputs.
- They help in explaining complex systems through simple equations.
Rate of Change and Its Applications
The rate of change is a key concept in calculus and essentially describes how one quantity changes in relation to another quantity. It is a crucial concept that connects our understanding of derivatives to real-world problems.
In this exercise, the rate of change is expressed by the derivative \(f'(c) = \frac{dW}{dc}\), which means the rate at which weight changes with respect to calorie consumption. The units of this rate of change are pounds per calorie.
In this exercise, the rate of change is expressed by the derivative \(f'(c) = \frac{dW}{dc}\), which means the rate at which weight changes with respect to calorie consumption. The units of this rate of change are pounds per calorie.
- This rate helps us understand how sensitive weight is to changes in calorie consumption.
- When the rate is zero, weight does not change as calories increase or decrease slightly, indicating a point of equilibrium.
- Understanding the rate of change can help in optimizing dietary plans by indicating how much calorie adjustment is needed for desired weight changes.
Other exercises in this chapter
Problem 14
An industrial production process costs \(C(q)\) million dollars to produce \(q\) million units; these units then sell for \(R(q)\) million dollars. If \(C(2.1)=
View solution Problem 14
Let \(f(x)\) be the elevation in feet of the Mississippi River \(x\) miles from its source. What are the units of \(f^{\prime}(x)\) ? What can you say about the
View solution Problem 15
(a) Graph \(f(x)=x^{2}\) and \(g(x)=x^{2}+3\) on the same axes. What can you say about the slopes of the tangent lines to the two graphs at the point \(x=0 ?\)
View solution Problem 16
The cost, \(C\) (in dollars), to produce \(g\) gallons of a chemical can be expressed as \(C=f(g)\). Using units, explain the meaning of the following statement
View solution