Problem 6 - Calculus for Biology and Medicine

Question

Let \(f(x, y)=x e^{y}\) with \(x(t)=e^{t}\) and \(y(t)=t^{2} .\) Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=0\).

Step-by-Step Solution

Answer

The derivative \(\frac{dw}{dt}\) at \(t=0\) is 1.

Step 1: Find Partial Derivatives

First, we need to find the partial derivatives of the function \(f(x, y) = x e^y\) with respect to \(x\) and \(y\). These are: \(\frac{\partial f}{\partial x} = e^y\) and \(\frac{\partial f}{\partial y} = x e^y\).

Step 2: Apply the Chain Rule

To find \(\frac{dw}{dt}\), use the chain rule: \(\frac{dw}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\).

Step 3: Derivative of x(t) and y(t)

Compute the derivatives of \(x(t) = e^t\) and \(y(t) = t^2 \) with respect to \(t\). We have \(\frac{dx}{dt} = e^t\) and \(\frac{dy}{dt} = 2t\).

Step 4: Substitute Values into Chain Rule

Substitute the computed derivatives and partial derivatives into the chain rule equation. We get: \(\frac{dw}{dt} = e^y \cdot e^t + x e^y \cdot 2t\).

Step 5: Evaluate at t = 0

Now, substitute \(x(e^t)\) and \(y(t^2)\) back into the equation and evaluate at \(t = 0\): \(x(0) = e^0 = 1\), \(y(0) = 0^2 = 0\), then \(\frac{dw}{dt} = e^0 \cdot e^0 + 1 \cdot e^0 \cdot 2 \cdot 0 = 1 + 0 = 1\).

Key Concepts

Chain Rule

The chain rule is essential when dealing with functions that are compositions of other functions. It allows you to compute the derivative of a composite function by taking derivatives of the inner functions. Imagine a chain where each link depends on the previous one, hence the term 'chain rule.'
For this exercise, you're managing function compositions where one function's output is another's input. We start by differentiating each part of the function in relation to another variable, then multiply to construct the whole. In math terms, if we have a function composed as a chain of functions, say, \(h(t) = g(f(t))\), the chain rule states that:

• \(\frac{dh}{dt} = \frac{dg}{df} \cdot \frac{df}{dt}\)

This rule simplifies the process of finding derivatives for complex compositions by breaking them down into manageable parts. In our exercise, we apply the chain rule by finding the derivatives of the inner functions, \(x(t)\) and \(y(t)\), and combine them with the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\).

Partial Derivatives

When you have a function of several variables, like \(f(x, y) = x e^y\), calculating the derivative involves partial derivatives. Each partial derivative represents the rate of change of the function with respect to one variable while keeping others constant.
In our example, we consider:

• The partial derivative of \(f\) with respect to \(x\), assuming \(y\) remains fixed. This gives us \(\frac{\partial f}{\partial x} = e^y\).
• The partial derivative of \(f\) with respect to \(y\), assuming \(x\) remains unchanged, which results in \(\frac{\partial f}{\partial y} = x e^y\).

Partial derivatives are crucial in multivariable calculus because they help us understand how each variable independently affects the outcome of the function. When used in tandem with the chain rule, they help us find the derivative of a function with respect to time or any other parameter, despite multiple interdependent variables.

Derivatives

The concept of derivatives is fundamental in calculus and describes how a function changes as its input changes. A derivative is the mathematical way of capturing the rate at which something is changing.
For the functions \(x(t) = e^t\) and \(y(t) = t^2\), the derivatives tell us how these functions change with time:

• Derivatives indicate the instantaneous rate of change; for \(x(t)\), \(\frac{dx}{dt} = e^t\) shows how quickly \(x(t)\) rises as \(t\) increases.
• For \(y(t)\), \(\frac{dy}{dt} = 2t\) tells us that \(y(t)\) changes at a rate directly proportional to \(t\).

By understanding these derivatives, we gauge how changes in \(t\) impact variables \(x\) and \(y\). When we combined these with partial derivatives in our solution, we accurately computed the change in \(f(x, y)\) as \(t\) varied. Essentially, derivatives provide a snapshot of how function values evolve, crucial for predicting trends and behaviors in a dynamic system.