Problem 4 - Calculus for Biology and Medicine

Question

4\. 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 of \(w\) with respect to \(t\) when \(t=5\) is \(\frac{17}{20}\).

Step 1: Understand the Problem

We need to find \(\frac{dw}{dt}\), where \(w = f(x,y) = \ln(xy - x^2)\), with \(x(t) = t^2\) and \(y(t) = t\). This involves chain rule and evaluating at \(t=5\).

Step 2: Express w in terms of t

Substitute \(x(t) = t^2\) and \(y(t) = t\) into the function: \[ w = \ln(t^2 \cdot t - (t^2)^2) = \ln(t^3 - t^4) \]

Step 3: Differentiate using the Chain Rule

To find \(\frac{dw}{dt}\), use the chain rule on \(w(t) = \ln(t^3 - t^4)\):\[ \frac{dw}{dt} = \frac{d}{dt}\left[ \ln(u(t)) \right], \text{ where } u(t) = t^3 - t^4 \]differentiate:\[ \frac{dw}{dt} = \frac{1}{u(t)} \cdot \frac{du}{dt} \]

Step 4: Find the derivative of u(t)

Differentiate \(u(t) = t^3 - t^4\):\[ \frac{du}{dt} = 3t^2 - 4t^3 \]

Step 5: Substitute and simplify

Substitute \(u(t)\) and \(\frac{du}{dt}\) back into the equation for \(\frac{dw}{dt}\):\[ \frac{dw}{dt} = \frac{1}{t^3 - t^4} \times (3t^2 - 4t^3) \]

Step 6: Evaluate at \(t = 5\)

Substitute \(t = 5\) into the expression for \(\frac{dw}{dt}\):\[ \frac{dw}{dt}\bigg|_{t=5} = \frac{3 \cdot 25 - 4 \cdot 125}{125 - 625} = \frac{75 - 500}{-500} = \frac{-425}{-500} = \frac{17}{20} \]

Key Concepts

Partial Derivatives

Partial derivatives are derivatives of functions with more than one variable. They help us understand how a function changes as we alter one variable at a time. In calculus, when dealing with a function of two variables like \(f(x, y)\), we compute partial derivatives with respect to each variable, treating the other variable as a constant.

For example, if we have \(f(x, y)=(\ln)(xy - x^2)\), partial derivatives would involve calculations like:

• The partial derivative with respect to \(x\), denoted as \(\frac{\partial f}{\partial x}\), assumes \(y\) is constant.
• Similarly, the partial derivative with respect to \(y\), denoted as \(\frac{\partial f}{\partial y}\), assumes \(x\) is constant.

These derivatives help analyze how the function changes in response to small changes in \(x\) or \(y\). Partial derivatives are foundational in optimization and finding tangent planes to surfaces.

In our exercise, we used the chain rule, which involves these derivatives, to find the rate of change of a function depending on time \(t\), another variable that is not directly in the function \(f\).

Function of Two Variables

A function of two variables, like \(f(x, y)\), assigns a value to every pair \((x, y)\) in its domain. Such functions are vital in multiple areas like physics, engineering, and economics.

The function \(f(x, y) = \ln(xy - x^2)\) has its own domain, determined by the condition that the argument of the natural logarithm must be positive. It maps a pair \((x, y)\) into a real number. This gives us a surface when visualized, with each coordinate point generating a unique height value.

• To explore where such functions increase or decrease, we can analyze their partial derivatives.
• Critical points and the nature of the surface can be assessed using second derivatives, helping us interpret the geometry around a point.

In the exercise, though, we're guided through using the chain rule to see how this surface changes over time based on known functions \(x(t) = t^2\) and \(y(t) = t\).

Derivative of Composite Function

The derivative of a composite function involves using the chain rule in calculus. This rule finds the derivative of a function based on its composition of other functions. It is essential when dealing with derivatives of complex expressions.

• In our initial exercise, \(w = f(x, y)\) can be represented as a composite because \(w\) depends on \(x(t)\) and \(y(t)\), both functions of \(t\).
• When differentiating, we first express \(w\) solely in terms of \(t\), as seen in the transformation to \(w = \ln(t^3 - t^4)\).

From here, we use the chain rule: you first find the derivative of the outer function, followed by multiplying it by the derivative of the inside function.

The solution steps involve substituting \(u(t)\) for \(t^3 - t^4\), and the chain rule guides us to multiply \(\frac{1}{u(t)}\) by \(\frac{du}{dt}\). This simplifies to \(\frac{3t^2-4t^3}{t^3-t^4}\), which we evaluate at \(t=5\), allowing a clear view of how the function could morph through different stages of \(t\). Use this approach anytime you encounter a composite function in calculus.