Problem 83 - Brief Applied Calculus

Question

ATHLETICS: Muscle Contraction The fundamental equation of muscle contraction is of the form \((w+a)(v+b)=c\), where \(w\) is the weight placed on the muscle, \(v\) is the velocity of contraction of the muscle, and \(a, b\), and \(c\) are constants that depend upon the muscle and the units of measurement. Solve this equation for \(v\) as a function of \(w, a, b\), and \(c\).

Step-by-Step Solution

Answer

The function is \(v = \frac{c - (wb + ab)}{w + a}\).

Step 1: Understanding the Equation

Identify the given equation, which is \[(w+a)(v+b)=c\] where the goal is to solve for the variable \(v\).

Step 2: Expand the Equation

Multiply \(w+a\) by \(v+b\). This results in \[wv + wb + av + ab = c.\]

Step 3: Rearrange Terms

Rearrange the equation to isolate terms involving \(v\). Moving terms with \(v\) to one side gives \[wv + av = c - (wb + ab).\]

Step 4: Factor Out v

Factor out \(v\) from the left side: \[v(w + a) = c - (wb + ab).\]

Step 5: Isolate v

Divide both sides of the equation by \(w + a\) to solve for \(v\):\[v = \frac{c - (wb + ab)}{w + a}.\]

Key Concepts

Solving Equations

Solving equations is a fundamental skill in mathematics that involves finding the value of an unknown variable. In the muscle contraction equation provided from the exercise, our task is to solve for the variable \(v\).
This means we need to express \(v\) in terms of the given variables and constants: \(w\), \(a\), \(b\), and \(c\). The initial equation is \((w+a)(v+b)=c\).
To "solve" this equation, we follow a process:

• Identify the goal: Understand which variable you need to isolate, in this case, \(v\).
• Expand the equation: Distribute terms to eliminate parentheses.
• Simplify the equation: Combine like terms and isolate the variable.

Each of these steps is critical to ensure that a clear expression for \(v\) is achieved, providing insights into how it behaves concerning other variables.

Algebraic Manipulation

Algebraic manipulation is an essential part of solving equations. It involves using different operations to rearrange and simplify equations.
In our equation, algebraic manipulation allows us to rearrange terms and isolate \(v\).
Here's how each step plays a role in our process:

• Expansion: Convert \((w+a)(v+b)=c\) into \(wv + wb + av + ab = c\) by distributing the terms.
• Rearrangement: Move all terms involving \(v\) to one side, resulting in \(wv + av = c - (wb + ab)\).
• Factoring: Extract \(v\) from both terms, so \(v(w + a)\) becomes isolated.
• Isolation: Divide by \((w + a)\) to solve for \(v\), yielding \(v = \frac{c - (wb + ab)}{w + a}\).

These steps demonstrate how algebraic manipulation helps transform an equation into a simpler form where the variable of interest is isolated.

Physics of Muscle Contraction

Understanding the physics of muscle contraction through the lens of the given equation helps unravel how muscles function. The equation \((w+a)(v+b)=c\) holds significance because it represents the relationship between different variables that characterize muscle performance.
Muscles contract to generate movement, and this process involves various forces and velocities:

• \(w\) is the external weight or load placed on the muscle.
• \(v\) signifies the velocity at which the muscle contracts.
• \(a\), \(b\), and \(c\) are constants determined by the muscle's properties and the measurement units used.

By isolating \(v\) in terms of \(w\), the equation portrays how changes in weight or load affect the speed at which a muscle can contract.
These relationships are critical in fields such as kinesiology and sports science, where understanding muscle dynamics can inform training techniques and improve athletic performance. Thus, the original mathematical problem is a gateway to deeper comprehension of biological systems.