Problem 16
Question
Tree Height The height, \(h,\) in feet of a certain tree increases at a rate that is inversely proportional to time, \(t\) in years. The height of the tree is 4 feet at the end of 2 years and reaches 30 feet at the end of 7 years. a. Write a differential equation describing the rate of change of the height of the tree. b. Give a particular solution for this differential equation. c. How tall will the tree be in 15 years? What will happen to the height of this tree over time?
Step-by-Step Solution
Verified Answer
The differential equation is \( \frac{dh}{dt} = \frac{k}{t} \). The height in 15 years is approx. 50 feet. The height increases indefinitely.
1Step 1: Formulate the Differential Equation
Since the rate of increase of height is inversely proportional to time, the differential equation can be written as: \( \frac{dh}{dt} = \frac{k}{t} \), where \( k \) is a constant of proportionality.
2Step 2: Solve the Differential Equation to Find General Solution
Integrate the differential equation \( \frac{dh}{dt} = \frac{k}{t} \) to find \( h \). \[ h = \int \frac{k}{t} dt = k \ln|t| + C \]Here, \( C \) is the constant of integration.
3Step 3: Find Constants Using Initial Conditions
Use the given initial conditions to find \( k \) and \( C \). - At \( t = 2 \), \( h = 4 \): \( 4 = k \ln(2) + C \)- At \( t = 7 \), \( h = 30 \): \( 30 = k \ln(7) + C \)Solve these equations simultaneously to find \( k \) and \( C \).
4Step 4: Determine Constant Values
Subtract the first equation from the second:\[ 30 - 4 = k(\ln(7) - \ln(2)) \]\[ 26 = k \ln\left(\frac{7}{2}\right) \]Solve for \( k \):\[ k = \frac{26}{\ln\left(\frac{7}{2}\right)} \]Substitute back to find \( C \) using \( 4 = k\ln(2) + C \).
5Step 5: Particular Solution
Substitute \( k \) and \( C \) back into the equation:\[ h = \frac{26}{\ln\left(\frac{7}{2}\right)} \ln|t| + C \]
6Step 6: Calculate Height After 15 Years
Using the particular solution, calculate \( h \) for \( t = 15 \):\[ h = \frac{26}{\ln\left(\frac{7}{2}\right)} \ln(15) + C \]
7Step 7: Behavior of the Tree Height Over Time
Observe that as \( t \to \infty \), the \( \ln(t) \) term grows without bound, so the tree height will continue to increase indefinitely, but at a slower rate due to the nature of the logarithmic function.
Key Concepts
Inverse ProportionalityIntegrationInitial ConditionsTree Growth Modeling
Inverse Proportionality
When something is said to be inversely proportional, it means that one quantity increases while the other decreases at a rate such that their product is constant. In the context of the given exercise, this means the rate of growth of the tree's height decreases as time goes on. Mathematically, if we say the rate of change of height, \( \frac{dh}{dt} \), is inversely proportional to the time \( t \), we express this relationship with the equation:
- \( \frac{dh}{dt} = \frac{k}{t} \)
Integration
Integration is a mathematical process used to find functions given their derivatives. In this exercise, integrating the differential equation helps us find a function for the tree's height over time. Starting with the equation:
- \( \frac{dh}{dt} = \frac{k}{t} \)
- \( h = \int \frac{k}{t} dt = k \ln|t| + C \)
Initial Conditions
Initial conditions are the known starting values of a system being modeled. They are crucial for solving differential equations uniquely. In our case, we're given two points in time where the height of the tree was measured:
- At \( t = 2 \) years, \( h = 4 \) feet
- At \( t = 7 \) years, \( h = 30 \) feet
- \( 4 = k \ln(2) + C \)
- \( 30 = k \ln(7) + C \)
Tree Growth Modeling
Tree growth modeling involves creating mathematical models to predict how a tree's height changes over time. In this exercise, the tree's growth is expressed through a logarithmic function derived from the differential equation. This model implies that the tree grows fast initially but slows down over time, reflecting the natural resource constraints as the tree becomes larger.
- The derived equation: \( h = \frac{26}{\ln\left(\frac{7}{2}\right)} \ln|t| + C \) captures the tree's growth pattern.
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