Problem 43 - Calculus for Biology and Medicine

Question

The differential equation $$ \frac{d y}{d x}=k \frac{y}{x} $$ describes allometric growth, where \(k\) is a positive constant. Assume that \(x\) and \(y\) are both positive variables and that \(y=f(x)\) is twice differentiable. Use implicit differentiation to determine for which values of \(k\) the function \(y=f(x)\) is concave up.

Step-by-Step Solution

Answer

The function is concave up for \( k > 1 \).

Step 1: Rewrite the Differential Equation

Rewrite the given differential equation in terms of separable variables: \( \frac{dy}{y} = k \frac{dx}{x} \). This form will help in solving the differential equation later.

Step 2: Solve the Differential Equation

Integrate both sides of the equation obtained in Step 1: \( \int \frac{dy}{y} = \int k \frac{dx}{x} \). This results in the solution: \( \ln|y| = k \ln|x| + C \), where \( C \) is the constant of integration.

Step 3: Express the Solution in Exponential Form

Convert the logarithmic form into an exponential form: \( y = Cx^k \), where \( C = e^{C} \) is a new constant. The function \( y = Cx^k \) represents the general solution to the differential equation.

Step 4: Find the Second Derivative

To determine concavity, find the second derivative of \( y \) with respect to \( x \). Start by finding the first derivative: \( \frac{dy}{dx} = Ck x^{k-1} \), and then differentiate again to find the second derivative: \( \frac{d^2y}{dx^2} = Ck(k-1)x^{k-2} \).

Step 5: Determine Concavity Based on Second Derivative

The function \( y = Cx^k \) is concave up if the second derivative is positive: \( Ck(k-1)x^{k-2} > 0 \). Since \( x > 0 \), \( C eq 0 \), and \( k > 0 \), the condition simplifies to \( k-1 > 0 \) or \( k > 1 \). Therefore, the function is concave up for \( k > 1 \).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves the derivatives of a function. These are crucial in describing how a particular quantity changes over time or space. In our exercise, the given differential equation is \( \frac{d y}{d x}=k \frac{y}{x} \). This means that the rate of change of \( y \) with respect to \( x \) is proportional to the quotient of \( y \) and \( x \). Differential equations are powerful tools in modeling real-world phenomena, such as the way populations grow, chemical reactions unfold, and how different systems evolve over time. They come in various forms, such as ordinary and partial, linear and nonlinear, depending on the number of variables and the highest derivative involved.

Concavity

Concavity refers to the nature of curves on a graph. It tells us how a function bends and is key for understanding the behavior of functions. A function is concave up when the curve is bent upwards, similar to a cup, which means it looks like a "U." This is visually and mathematically represented by the second derivative of the function being positive. For the function \( y=f(x) \) being \( Cx^k \), determining concavity involves evaluating the second derivative, \( \frac{d^2y}{dx^2} = Ck(k-1)x^{k-2} \). The function is concave up if this second derivative is greater than zero. In our specific case, after analysis, we found that for the function to be concave up, \( k \) must be greater than 1.

Implicit Differentiation

Implicit differentiation is a method used to find the derivative of an equation in situations where the equation is not explicitly solved for one variable in terms of the other. This technique is especially useful when dealing with equations that define relationships implicitly rather than explicitly. In our exercise, although the function \( y = Cx^k \) is eventually expressed explicitly, finding derivatives implicitly can offer advantages especially when dealing with multiple variables or when functions are given in more complex forms. It simplifies the process because sometimes explicitly solving for one variable in terms of the other can be complex or impractical.

Separable Variables

The technique of separable variables is a method used to solve differential equations by isolating interactions of different variables on both sides of an equation to make them easier to integrate. For our differential equation, \( \frac{dy}{y} = k \frac{dx}{x} \), it looks rather complex initially. By separating variables, you rewrite it such that all \( y \) terms appear on one side of the equation, and all \( x \) terms are on the other. This makes it possible to integrate both sides independently, allowing the solution to emerge more straightforwardly. Thus, after integration, we find the solution \( y = Cx^k \), demonstrating the power and utility of using separable variables to tackle differential equations.