Problem 30
Question
Allometric Equations Suppose that two quantities, \(y\) and \(x\) are related by a power law: $$ y=k x^{a} $$ where \(k\) and \(a\) are both constants. \(x\) grows with time at a rate \(d x / d t\) (a) Explain why \(\frac{1}{x} \frac{d x}{d t}\) can be thought of as the relative rate of growth of \(x\). (b) Show that the relative rates of growth of \(y\) and \(x\) are related by an equation: $$ \frac{1}{y} \frac{d y}{d t}=\frac{a}{x} \frac{d x}{d t} $$
Step-by-Step Solution
Verified Answer
(a) It's the normalized rate of change of \(x\).
(b) The relative growth rate of \(y\) is \(a\) times that of \(x\): \(\frac{1}{y} \frac{d y}{d t} = \frac{a}{x} \frac{d x}{d t}\).
1Step 1: Interpret the Relative Rate of Growth
The expression \( \frac{1}{x} \frac{d x}{d t} \) represents the relative rate of growth of \( x \) because it measures how \( x \) changes over time relative to its current size. It's the rate of change of \( x \) (\( \frac{d x}{d t} \)) normalized by \( x \) itself, illustrating the proportion by which \( x \) grows over time.
2Step 2: Differentiate the Allometric Equation
Start with the given power law \( y = kx^a \). To find the growth rate of \( y \), differentiate \( y \) with respect to time \( t \). Using the chain rule, we have:\[ \frac{d y}{d t} = \frac{d}{d t} (kx^a) = k a x^{a-1} \frac{d x}{d t} \]
3Step 3: Calculate Relative Growth Rate of y
We express the relative growth rate of \( y \) in a similar form to that of \( x \). This is accomplished by dividing \( \frac{d y}{d t} \) by \( y \):\[ \frac{1}{y} \frac{d y}{d t} = \frac{1}{kx^a} (k a x^{a-1} \frac{d x}{d t}) \]Cancelling terms:\[ = \frac{a}{x} \frac{d x}{d t} \]
4Step 4: Conclude the Relationship
Thus, the expression \( \frac{1}{y} \frac{d y}{d t} = \frac{a}{x} \frac{d x}{d t} \) shows that the relative rate of growth of \( y \) is \( a \) times the relative rate of growth of \( x \). Therefore, as \( x \) grows, \( y \) grows at a rate proportional to \( x \)'s growth multiplied by the constant \( a \).
Key Concepts
Power LawRelative Rate of GrowthChain Rule DifferentiationProportional Growth
Power Law
Power laws are mathematical relationships where one quantity varies as a power of another. In the context of allometric equations, a power law describes the relationship between two quantities, like how "y" is related to "x" using the formula \(y = kx^a\). Both "k" and "a" are constants here. Generally, power laws reveal how changes in one quantity affect another.
- "k" is a constant multiplier that scales the relationship.
- "a" determines how the dependent variable "y" scales with changes in "x".
Relative Rate of Growth
Relative rate of growth measures how a quantity changes concerning its size. The formula \(\frac{1}{x} \frac{dx}{dt}\) simplifies the rate at which "x" changes over time. This rate is divided by "x" to show the proportion of change relative to its current value.
- It gives an insightful look into the dynamics of growth, avoiding absolute changes.
- This concept is crucial in judging efficiency, growth dynamics, and other performance metrics of systems.
Chain Rule Differentiation
The chain rule is a core principle in calculus, used to find the derivative of compositions of functions. In the context of the power law equation, to differentiate \(y = kx^a\) with respect to time "t", the chain rule is utilised. This method provides a way to find how the rate of change of "y" depends on "x".
- First, take the derivative of the outer function \(kx^a\) with respect to "x".
- Multiply this by the derivative of "x" with respect to time \(\frac{dx}{dt}\).
Proportional Growth
Proportional growth describes scenarios where changes in one quantity cause corresponding changes in another. Within the equation \(\frac{1}{y} \frac{dy}{dt} = \frac{a}{x} \frac{dx}{dt}\) from the original exercise, "y" grows at a rate that is scaled by the constant "a" multiplied by the rate of "x".
- The constant "a" influences how strongly "y" responds to shifts in "x".
- It indicates direct proportionality but with a specific multiplier which is critical in growth systems.
Other exercises in this chapter
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