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Chapter 1: Problem 88

McNeil and associates \(^{19}\) showed that for small loblolly pine trees \(V(H)=0.0000837 H^{3.191}\), where \(V\) is the volume in cubic meters and \(H\) is the tree height in meters. Find \(V(10)\). Explain what this means. Find \(V(1)\), and explain what this means. What happens to \(V\) when \(H\) doubles? Graph \(V=V(H)\).

Short Answer

Expert verified
\( V(10) \approx 0.1296 \) cubic meters, \( V(1) = 0.0000837 \) cubic meters; volume increases rapidly as height increases.

Step by step solution

01

Understand the Function

The function given is \( V(H) = 0.0000837 H^{3.191} \). This function represents the volume of a loblolly pine tree in cubic meters, where \( H \) is the height of the tree in meters. Our task is to find \( V(10) \) and \( V(1) \) and understand the behavior of \( V \) when \( H \) changes.
02

Calculate \( V(10) \)

To find \( V(10) \), substitute \( H = 10 \) into the formula:\[V(10) = 0.0000837 \times 10^{3.191}\]Calculate the exponent:\[10^{3.191} \approx 1548.816\]Now, multiply by 0.0000837:\[V(10) \approx 0.0000837 \times 1548.816 \approx 0.1296\]Thus, the volume of the tree when \( H = 10 \) meters is approximately 0.1296 cubic meters.
03

Interpret \( V(10) \)

A tree with a height of 10 meters has a volume of approximately 0.1296 cubic meters. This gives us a reference for understanding the size of a small loblolly pine tree at this height.
04

Calculate \( V(1) \)

To find \( V(1) \), substitute \( H = 1 \) into the formula:\[V(1) = 0.0000837 \times 1^{3.191} = 0.0000837\]Since \( 1^{3.191} = 1 \), the volume \( V(1) \) is exactly 0.0000837 cubic meters.
05

Interpret \( V(1) \)

A tree with a height of 1 meter has a volume of 0.0000837 cubic meters. This indicates that even at a very small size, the volume is quite small, reinforcing that the formula models small trees.
06

Analyze Behavior as \( H \) Doubles

Let's consider how the volume changes when \( H \) doubles. If \( H = 1 \) becomes \( H = 2 \), then:\[V(2) = 0.0000837 \times 2^{3.191}\]Calculate the exponent:\[2^{3.191} \approx 9.053\]Thus,\[V(2) \approx 0.0000837 \times 9.053 \approx 0.000757\]The volume increases significantly more than twice the original when the height doubles, indicating that volume increases more rapidly than linearly with height.
07

Graph the Function \( V(H) \)

To graph \( V(H) \), plot the function \( V(H) = 0.0000837H^{3.191} \) over a range of \( H \) values. Observe that the curve rises sharply, displaying the power nature of the function where a slight increase in height results in a much larger increase in volume.

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Key Concepts

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

Volume of a Tree
Understanding the volume of a tree involves practical applications of calculus and mathematical functions such as the one given for loblolly pine trees. In the equation \(V(H) = 0.0000837 H^{3.191}\), each component has a purpose:
  • \(V(H)\) represents the volume measured in cubic meters.
  • \(H\) is the tree's height in meters.
  • The constant \(0.0000837\) is a scaling factor that adjusts the outcome based on empirical data.
  • The exponent \(3.191\) is derived from observational data and reflects how volume increases with height.
The relationship here is a power function, meaning volume increases rapidly as height increases. For example, at \(H = 10\), the volume is around \(0.1296\) cubic meters, providing insight into the physical space a tree occupies. If our height is reduced to \(H = 1\), the volume is \(0.0000837\) cubic meters, showcasing how small trees have significantly less volume but still occupy measurable space.
Exponential Functions
In the context of calculating the volume of a tree, exponential functions play a critical role due to their non-linear nature. While the function \(V(H) = 0.0000837 H^{3.191}\) is not purely exponential, it shares characteristics due to the exponent applied to \(H\):
  • The volume \(V\) does not increase linearly with height \(H\); instead, it does so at an accelerated rate.
  • As shown, doubling \(H\) does not simply double \(V\); for example, when \(H\) increases from \(1\) to \(2\), \(V(2)\approx 0.000757\), a larger increase than doubling \(0.0000837\).
  • This behavior illustrates why understanding the concept of power (or exponential-like) functions is vital in real-world applications, as they often describe phenomena that grow rapidly over time or size.
Grasping this helps in predicting and understanding growth patterns in natural scenarios, such as plant or animal growth.
Graphing Functions
Graphing the function \(V(H) = 0.0000837H^{3.191}\) offers a visual representation of how tree volume varies with height. When graphing:
  • Start by choosing a range for \(H\), perhaps from \(0\) to \(20\), to see significant changes.
  • Plot points for various \(H\) values, such as \(1, 2, 5, 10, 15,\) and \(20\).
  • Evaluate \(V(H)\) at these points and note how quickly the volume increases, especially for taller trees.
The graph will show a curve that rises sharply, a hallmark of power functions. This is crucial for interpreting data visually and confirming that the theoretical function behaves as expected when applied to real heights. Whether predicting how trees grow or just understanding mathematical relationships, graphs are indispensable tools for analysis.

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