Problem 17 - Algebra Form and Function

Question

The thrust, \(T,\) in pounds, of a ship's propeller is proportional to the square of the propeller speed, \(R,\) in rotations per minute, times the fourth power of the propeller diameter, \(D,\) in feet. \({ }^{2}\) (a) Write a formula for \(T\) in terms of \(R\) and \(D\). (b) If \(R=300 D\) for a certain propeller, is \(T\) a power function of \(D ?\) (c) If \(D=0.25 \sqrt{R}\) for a different propeller, is \(T\) a power function of \(R\) ?

Step-by-Step Solution

Answer

Answer: Yes, T is a power function of both D and R in both cases (a) and (b).

Step 1: Write a formula for the thrust T

Since the thrust T is proportional to the square of the propeller speed R and the fourth power of the propeller diameter D, we can write this relation as follows: \[T = kR^2D^4\] where k is the proportionality constant.

Step 2: Determine if T is a power function of D (Case a)

Now we are given that \(R = 300D\). We need to determine if this makes T a power function of D by substituting this expression for R in the equation for T. Let's substitute R: \[T = k(300D)^2D^4\] Now simplify the equation: \[T = k90000D^6\] This shows that T is indeed a power function of D since its equation is of the form \(T = aD^n\) where a and n are constants.

Step 3: Determine if T is a power function of R (Case b)

In this case, we are given that \(D = 0.25\sqrt{R}\). Our task is to determine if T is a power function of R. Let's substitute this expression for D in the equation for T: \[T = kR^2(0.25\sqrt{R})^4\] Now simplify the equation: \[T = kR^2(0.0625R^2)^2\] \[T = k0.000244140625R^6\] This shows that T is also a power function of R, since its equation is of the form \(T = bR^m\) where b and m are constants.

Key Concepts

Power Functions

A power function can be understood as a mathematical function of the form \( f(x) = ax^n \), where \( a \) and \( n \) are constants, and \( x \) is a variable. The key characteristic of power functions is the degree of the variable \( x \), which is represented by \( n \).
When analyzing equations like the one given for thrust \( T \), which is expressed in terms of propeller speed \( R \) and diameter \( D \), the formula re-emphasizes the power relations:

• \( T = kR^2D^4 \)
• This showcases a connection between \( T \), \( R \), and \( D \) through power functions.

Additionally, when we consider the substitution \( R = 300D \), we derive \( T = k90000D^6 \), making it evident that thrust \( T \) naturally becomes a power function of \( D \). Similarly, when \( D = 0.25\sqrt{R} \), after simplification, \( T \) becomes \( k0.000244140625R^6 \), illustrating it as a power function of \( R \).
Thus, both cases confirm that the concept of power functions is fundamental to understanding how thrust is correlated with speed and diameter.

Proportionality Concepts

Proportionality is a key concept in mathematics, reflecting how one quantity changes in relation to another. In the context of thrust calculation, the equation \( T = kR^2D^4 \) exemplifies a proportionality where the thrust \( T \) is directly related to the square of the propeller speed \( R \) and the fourth power of the diameter \( D \).
Here, the constant \( k \) acts as the proportionality constant, determining how significantly the change in \( R \) and \( D \) would affect \( T \). Understanding this relationship helps clarify how adjustments in propeller speed and diameter result in variations in thrust:

• If \( R \) doubles, \( R^2 \) increases fourfold, significantly impacting \( T \).
• Similarly, if \( D \) doubles, the effect on \( D^4 \) results in a sixteenfold increase.

Grasping this concept allows one to predict changes in thrust outcomes based on the proportional relationships defined in the formula.

Mathematical Formulas

Mathematical formulas serve as the language of calculation and analysis, providing a structured approach to solve complex problems. The thrust equation \( T = kR^2D^4 \), derived from the given conditions, is a classic example of using a formula to express a relationship between variables in a concise manner.
Formulas encapsulate logical relationships and enable easier computations through substitutions and simplifications. For instance, substituting \( R = 300D \) or \( D = 0.25\sqrt{R} \) into the thrust formula simplifies the relation for specific situations:

• Substituting \( R = 300D \) yields \( T = k90000D^6 \), a formula that directly shows thrust as a power function of \( D \).
• Similarly, substituting \( D = 0.25\sqrt{R} \) results in \( T = k0.000244140625R^6 \), demonstrating a direct power function of \( R \).

Formulas offer precision and predictability, essential for mathematical understanding in engineering and physics where accurate calculation of physical parameters is crucial.