Problem 6 - Calculus for Biology and Medicine

Question

6\. 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 of \( w = f(x, y) \) with respect to \( t \) at \( t = 0 \) is 1.

Step 1: Understanding the problem

We need to find the derivative of the function \( w = f(x, y) = xe^y \) with respect to \( t \), at \( t = 0 \), given \( x(t) = e^t \) and \( y(t) = t^2 \). This involves using the chain rule since \( w \) is expressed in terms of \( x \) and \( y \), both of which are functions of \( t \).

Step 2: Applying the chain rule

To find \( \frac{dw}{dt} \), we use the chain rule: \( \frac{dw}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \). Calculate the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), then multiply them by their respective derivatives of \( x(t) \) and \( y(t) \) with respect to \( t \).

Step 3: Calculating partial derivatives

The partial derivative of \( f \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = e^y \), and with respect to \( y \) is \( \frac{\partial f}{\partial y} = xe^y \).

Step 4: Finding derivatives of x(t) and y(t)

Differentiate \( x(t) = e^t \) with respect to \( t \) to get \( \frac{dx}{dt} = e^t \). Differentiate \( y(t) = t^2 \) to get \( \frac{dy}{dt} = 2t \).

Step 5: Substitute and simplify

Substitute the partial derivatives and \( \frac{dx}{dt} \), \( \frac{dy}{dt} \) into the chain rule expression:\[ \frac{dw}{dt} = e^y \cdot e^t + xe^y \cdot 2t \].

Step 6: Evaluating at t = 0

At \( t = 0 \), \( x(0) = e^0 = 1 \) and \( y(0) = 0^2 = 0 \). Substitute \( x(0) = 1 \), \( y(0) = 0 \), and their derivatives into the expression:\[ \frac{dw}{dt} = e^0 \cdot e^0 + 1 \cdot e^0 \cdot 0 = 1 + 0 = 1 \].

Key Concepts

The Chain Rule in Calculus

The chain rule is an essential concept in calculus, especially when dealing with composite functions. In essence, it provides a method to find the derivative of a function based on its dependency on other functions or variables. If you have a function that depends on intermediate variables, and these variables themselves depend on another variable, the chain rule helps us determine the overall rate of change.
In our exercise, the function \( w = f(x, y) = xe^y \) depends on two variables \( x \) and \( y \), which in turn relate to \( t \). This is where the chain rule comes in handy. By breaking the problem down into parts, we can find the derivative of \( w \) with respect to \( t \) efficiently. Here’s how:
- First, calculate the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). This examines how \( f \) changes as each variable, \( x \) or \( y \), changes independently.
- Next, differentiate \( x(t) \) and \( y(t) \) with respect to \( t \). This shows how each of these variables changes over time with \( t \).
Finally, the chain rule formula \( \frac{dw}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \) brings these pieces together to give the full derivative of \( w \) with respect to \( t \).
This powerful rule reveals how interconnected variables impact the rate of change in multivariable functions.

Understanding Partial Derivatives

Partial derivatives are derivatives of functions with multiple variables where we only allow one variable to change at a time, keeping the others constant. This allows us to focus on the effect of that one variable's change on the function. Consider a function like \( f(x, y) = xe^y \). Even though \( f \) has two variables, we first consider only one changing, treating the other as constant.

• For the partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), treat \( y \) as constant. This yields \( e^y \), because \( f = xe^y \) changes linearly with \( x \).
• For \( \frac{\partial f}{\partial y} \), we treat \( x \) as a constant. So, the derivative is \( xe^y\), since differentiation of \( e^y \) with respect to \( y \) keeps an accompanying constant \( x \).

By calculating these, we precisely measure how a function's value shifts with every small change in each variable, offering insights into the function's behavior when multiple variables are involved.

Derivatives with Respect to a Parameter in Composite Functions

When working with composite functions, especially in fields like biology, it’s common to have functions that entail several layers of dependency, as seen in our example. Here, \( x(t) = e^t \) and \( y(t) = t^2 \) define how the variables \( x \) and \( y \) depend on another parameter, \( t \). This makes them composite functions with \( t \) as the parameter.
To understand this better:

• Calculate \( \frac{dx}{dt} \) from \( x(t) = e^t \). You'll find that the derivative is \( e^t \), illustrating how \( x \) changes over \( t \).
• Similarly, finding \( \frac{dy}{dt} \) from \( y(t) = t^2 \) provides \( 2t \), showing the variability of \( y \) as \( t \) alters.

This approach sheds light on the interconnected equation and reveals how each layer—obscured by multiple functional relationships—contributes to the total rate of change. By tracing each variable back to \( t \), a deeper understanding of the system's dynamics emerges. This is crucial in fields like biology, where many phenomena rely on such interconnected relationships.