Problem 4 - Calculus for Biology and Medicine

Question

Let \(f(x, y)=\ln \left(x y-x^{2}\right)\) with \(x(t)=t^{2}\) and \(y(t)=t\). Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=5\).

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} \) at \( t=5 \) is \( \frac{17}{20} \).

Step 1: Define the Problem

We need to find \( \frac{dw}{dt} \), where \( w = f(x(t), y(t)) = \ln(xy-x^2) \). Given that \( x(t) = t^2 \) and \( y(t) = t \), we must use the chain rule for derivatives because \( w \) is a function of \( x \) and \( y \), which themselves are functions of \( t \).

Step 2: Apply the Chain Rule

According to the chain rule, \( \frac{dw}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \). We need to find each component: the partial derivatives of \( f \) with respect to \( x \) and \( y \), and the derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \).

Step 3: Find Partial Derivatives

Let's calculate the partial derivatives. First, \( \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \ln(xy - x^2) = \frac{y - 2x}{xy - x^2} \). Next, \( \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \ln(xy - x^2) = \frac{x}{xy - x^2} \).

Step 4: Find Derivatives of x(t) and y(t)

Now we need \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). Using the definitions given, \( x(t) = t^2 \), so \( \frac{dx}{dt} = 2t \). Similarly, \( y(t) = t \), so \( \frac{dy}{dt} = 1 \).

Step 5: Substitute and Simplify

Substitute the derivative expressions into the chain rule formula: \[ \frac{dw}{dt} = \left(\frac{y - 2x}{xy - x^2}\right) \cdot 2t + \left(\frac{x}{xy - x^2}\right) \cdot 1. \] For simplicity, substitute \( x = t^2 \) and \( y = t \) at this stage.

Step 6: Evaluate Derivative at t=5

Substituting \(x=25\) and \(y=5\) (plugging in \(t=5\)), we have:- \(xy - x^2 = 25 \cdot 5 - 25^2 = 125 - 625 = -500\),- \(\frac{y - 2x}{xy - x^2} = \frac{5 - 50}{-500} = \frac{-45}{-500} = \frac{9}{100}\),- \(\frac{x}{xy - x^2} = \frac{25}{-500} = -\frac{1}{20}\).Substitute back: \[ \frac{dw}{dt} = \frac{9}{100} \times 10 + \left(-\frac{1}{20}\right) = \frac{90}{100} - \frac{5}{100} = \frac{85}{100} = \frac{17}{20}. \]

Step 7: Conclusion

The derivative \( \frac{dw}{dt} \) at \( t=5 \) is \( \frac{17}{20} \).

Key Concepts

Understanding the Chain Rule

The Chain Rule is a fundamental tool in calculus used for differentiating composite functions. Imagine a scenario where one quantity depends on another, which in turn depends on a third quantity. This is where the Chain Rule becomes essential, allowing us to differentiate these linked functions effectively.

In the context of our problem, the function \( w = \ln(xy - x^2) \) depends on both \( x(t) \) and \( y(t) \). Since \( x \) and \( y \) are themselves functions of \( t \), the Chain Rule offers a strategy for calculating \( \frac{dw}{dt} \) by breaking it down into simpler parts:

• \( \frac{\partial f}{\partial x} \frac{dx}{dt} \)
• \( \frac{\partial f}{\partial y} \frac{dy}{dt} \)

This breakdown allows us to differentiate each piece directly, then sum the results. This rule is particularly powerful in handling complex scenarios, ensuring precision and simplifying the differentiation of multi-variable functions.

Exploring Partial Derivatives

Partial derivatives involve differentiating a multi-variable function with respect to one variable at a time, treating other variables as constants. This concept is crucial when dealing with functions of multiple variables, like \( f(x, y) = \ln(xy - x^2) \), where the behavior of one variable influences another.

In our exercise, we find the partial derivatives of \( f \) with respect to \( x \) and \( y \):

• \( \frac{\partial f}{\partial x} = \frac{y - 2x}{xy - x^2} \)
• \( \frac{\partial f}{\partial y} = \frac{x}{xy - x^2} \)

These expressions help in understanding how changes in \( x \) or \( y \) alone affect the function \( f \). Such calculations are invaluable in multi-dimensional analysis, providing insight into the sensitivities of each independent variable.

Grasping Differentiation

Differentiation is a core concept in calculus used to determine the rate at which a quantity changes. It lies at the heart of understanding motion, growth, and other dynamic systems. In essence, differentiation gives us the derivative, representing how a function changes as its input changes.

In the example provided, we differentiate \( x(t) = t^2 \) and \( y(t) = t \) with respect to \( t \), yielding:

• \( \frac{dx}{dt} = 2t \)
• \( \frac{dy}{dt} = 1 \)

These derivatives demonstrate how \( x \) and \( y \) change as \( t \) varies. Thus, differentiation not only helps us find rates of change but also enables us to predict future behavior in responsive systems based on their current state.