Problem 47
Question
Average leaf width, \(w\) (in \(\mathrm{mm}\) ), in tropical Australia is a function of the average annual rainfall, \(r\) (in \(\mathrm{mm}\) ), so \(w=f(r) .\) We have \(f^{\prime}(1500)=0.0218.\) (a) What are the units of the \(1500 ?\) (b) What are the units of the \(0.0218 ?\) (c) About how much difference in average leaf width would you find in two forests whose average annual rainfalls are near 1500 mm but differ by 200 mm?
Step-by-Step Solution
Verified Answer
1500 mm, 0.0218 has no units, and the difference is about 4.36 mm.
1Step 1: Determine Units for 1500
The given value 1500 is used in the function input as \( r \), representing the average annual rainfall in tropical Australia. Therefore, the units for 1500 are in \( \mathrm{mm} \) of rainfall.
2Step 2: Determine Units for 0.0218
The given value \( f'(1500) = 0.0218 \) represents the derivative of the leaf width function with respect to rainfall. Since \( f(r) \) has units of \( \mathrm{mm} \) (leaf width) and \( r \) has units of \( \mathrm{mm} \) (rainfall), the derivative \( f'(r) \) represents \( \frac{\mathrm{mm}}{\mathrm{mm}} \). Therefore, the units for 0.0218 are dimensionless or \( \mathrm{mm/mm} \), indicating the change in leaf width per change in rainfall.
3Step 3: Calculate Change in Leaf Width
To estimate the change in average leaf width for a change in rainfall \( \Delta r = 200 \ \mathrm{mm} \): Use the linear approximation formula: \[ \Delta w = f'(1500) \times \Delta r \]Plug in the values:\[ \Delta w = 0.0218 \times 200 \]\[ \Delta w = 4.36 \ \mathrm{mm} \]Therefore, the difference in average leaf width for the two forests is approximately 4.36 mm.
Key Concepts
Average Annual RainfallLeaf Width FunctionLinear ApproximationUnits of Derivatives
Average Annual Rainfall
Average annual rainfall is an important metric, especially in areas like tropical Australia, where it influences many aspects of the environment. It represents the total rain that falls in a particular region over the course of a year. In this exercise, 1500 mm is a specific value of average annual rainfall.
Understanding this concept is essential because it directly impacts other variables in environmental studies. Here, we focus on how it determines the variations in leaf width, a crucial factor in ecosystem studies.
The units for the 1500 in the exercise are millimeters (mm), which signify the depth of water that would accumulate in the environment. This is crucial for calculating the derivative or changes in other parameters, like leaf width, which helps in predicting and understanding ecological responses.
Understanding this concept is essential because it directly impacts other variables in environmental studies. Here, we focus on how it determines the variations in leaf width, a crucial factor in ecosystem studies.
The units for the 1500 in the exercise are millimeters (mm), which signify the depth of water that would accumulate in the environment. This is crucial for calculating the derivative or changes in other parameters, like leaf width, which helps in predicting and understanding ecological responses.
Leaf Width Function
The Leaf Width Function, denoted here as \(w=f(r)\), describes the relationship between leaf width and average annual rainfall. The function demonstrates how leaf size adapts to different rainfall amounts.
It is important to understand that this function is an abstraction of real-world behavior, making it more manageable and allowing predictions about changes in leaf width based on changes in rainfall.
- Input: Average annual rainfall \(r\)
- Output: Average leaf width \(w\)
It is important to understand that this function is an abstraction of real-world behavior, making it more manageable and allowing predictions about changes in leaf width based on changes in rainfall.
Linear Approximation
Linear approximation is a method used in calculus to estimate the value of a function near a given point using its derivative. In this problem, the derivative \(f'(1500)\) is used to estimate changes in leaf width for variations in rainfall.
The formula for linear approximation is \( \Delta w = f'(r) \times \Delta r \) and is used to determine the estimated change in the function output for a small change in input.
Here, \(f'(1500) = 0.0218\) tells us that for each 1 mm increase in rainfall, there's an approximate increase of 0.0218 mm in leaf width. Using the linear approximation:
The formula for linear approximation is \( \Delta w = f'(r) \times \Delta r \) and is used to determine the estimated change in the function output for a small change in input.
Here, \(f'(1500) = 0.0218\) tells us that for each 1 mm increase in rainfall, there's an approximate increase of 0.0218 mm in leaf width. Using the linear approximation:
- \( \Delta r = 200 \, \mathrm{mm} \)
- \( \Delta w = 0.0218 \times 200 = 4.36\, \mathrm{mm} \)
Units of Derivatives
The units of a derivative are a crucial aspect of calculus, serving as a measure of how one quantity changes concerning another. In this problem, it is about understanding how the leaf width changes per unit change in rainfall.
The derivative \(f'(1500) = 0.0218\) represents the rate of change of leaf width with respect to rainfall. Since both leaf width and rainfall are measured in millimeters, the units of \(f'(r)\) simplify to being dimensionless or, more specifically, mm/mm.
This means for every millimeter change in rainfall, the leaf width changes by a factor of 0.0218. Understanding units helps ensure the interpretability and correctness of the calculated values, making it easier to convey scientific findings. It's noteworthy that dimensionless results often signify a normalized rate of change, vital for comparing different datasets.
The derivative \(f'(1500) = 0.0218\) represents the rate of change of leaf width with respect to rainfall. Since both leaf width and rainfall are measured in millimeters, the units of \(f'(r)\) simplify to being dimensionless or, more specifically, mm/mm.
This means for every millimeter change in rainfall, the leaf width changes by a factor of 0.0218. Understanding units helps ensure the interpretability and correctness of the calculated values, making it easier to convey scientific findings. It's noteworthy that dimensionless results often signify a normalized rate of change, vital for comparing different datasets.
Other exercises in this chapter
Problem 47
Use algebra to evaluate the limits. $$\lim _{h \rightarrow 0} \frac{1 /(1+h)-1}{h}$$
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Explain what is wrong with the statement. A function for which \(f^{\prime}(0)=0\) but \(f^{\prime \prime}(0) \neq 0\)
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True or false? Give an explanation for your answer. If \(f^{\prime \prime}(x)>0\) then \(f^{\prime}(x)\) is increasing.
View solution Problem 48
Use algebra to evaluate the limits. $$\lim _{h \rightarrow 0} \frac{1 /(1+h)^{2}-1}{h}$$
View solution