Problem 23 - Applied Calculus

Question

A tree of height \(y\) meters has, on average, \(B\) branches, where \(B=y-1 .\) Each branch has, on average, \(n\) leaves where \(n=2 B^{2}-B .\) Find the average number of leaves of a tree as a function of height.

Step-by-Step Solution

Answer

The average number of leaves as a function of height is \(2y^3 - 7y^2 + 8y - 3\).

Step 1: Express B in terms of y

According to the problem, the number of branches, \( B \), is given by \( B = y - 1 \). This means, for a tree of height \( y \), the average number of branches can be directly calculated by subtracting 1 from the height.

Step 2: Express n in terms of B

We know that each branch has, on average, \( n = 2B^2 - B \) leaves. Substitute \( B = y - 1 \) from Step 1 into this equation to express \( n \) in terms of \( y \). This gives us \( n = 2(y-1)^2 - (y-1) \).

Step 3: Simplify the expression for n

Expand the expression \( n = 2(y - 1)^2 - (y - 1) \). Calculate \( (y - 1)^2 = y^2 - 2y + 1 \), then substitute it back: \( n = 2(y^2 - 2y + 1) - (y - 1) \). Simplify to get \( n = 2y^2 - 4y + 2 - y + 1 \).

Step 4: Simplify further

Combine like terms to get \( n = 2y^2 - 5y + 3 \). This expression represents the average number of leaves per branch of a tree with height \( y \).

Step 5: Calculate total average leaves

A tree with \( B \) branches and \( n \) leaves per branch has an average total number of leaves given by \( B \times n \). Substitute \( B = y - 1 \) and \( n = 2y^2 - 5y + 3 \) to get \( \, (y-1)(2y^2 - 5y + 3) \).

Step 6: Distribute to find overall function

Expand the expression \( (y-1)(2y^2 - 5y + 3) = y(2y^2 - 5y + 3) - (2y^2 - 5y + 3) \). This results in \( 2y^3 - 5y^2 + 3y - 2y^2 + 5y - 3 \).

Step 7: Simplify final expression

Combine like terms: \( 2y^3 - 7y^2 + 8y - 3 \). This is the average number of leaves as a function of the height of the tree.

Key Concepts

Average Number of Leaves

When studying applied calculus problems, like determining the average number of leaves on a tree, the concept often revolves around finding relationships between different elements. Here, the focus is to make a connection between the tree's height and the number of leaves it bears. To solve the problem, we initially consider how the number of branches and leaves per branch depends on the height.

• Each tree has an average number of branches expressed as \( B = y - 1 \), where \( y \) is the tree's height.
• For each branch, the average number of leaves \( n \) is determined as \( n = 2B^2 - B \).

The goal in these types of problems is to express the total average number of leaves in terms of the tree's height \( y \). By substituting the expressions for \( B \) and \( n \), we can find a comprehensive formula that connects height directly to the average number of leaves.

Function of Height

Expressing quantities as a function of a variable, like height in this case, is a core concept. From the given relations, we begin with determining the function of height. This involves rewriting the equations for branches and leaves entirely in terms of \( y \).
Start by substituting the expression for \( B \), which is \( B = y - 1 \), into the formula for leaves \( n = 2B^2 - B \). This gives us:\[ n = 2(y - 1)^2 - (y - 1) \]The function translates the height directly into the average leaf count per branch.
This algebraic step prepares the way to derive a simplification, which can point to broader relationships, such as polynomial relationships between height and total leaf number.

Polynomial Simplification

Simplifying polynomials is a crucial skill in calculus and algebra. It allows us to transform complex expressions into simpler forms. Here, simplification focuses on calculating the total average number of leaves as a power polynomial.

• Initially, we expanded \( (y-1)(2y^2 - 5y + 3) \) to set up calculation for the total leaves.
• After expansion, the expression was simplified further: \( 2y^3 - 7y^2 + 8y - 3 \).

This final polynomial captures the relationship between tree height \( y \) and total average leaves.
Breaking down the polynomial into simpler terms emphasizes the direct and intuitive link between algebraic processes and real-world biological data. In essence, the ability to simplify such expressions allows one to derive meaningful insights from mathematical models easily.