Problem 5 - Calculus for Biology and Medicine

Question

5\. Let \(f(x, y)=\frac{1}{x}+\frac{1}{y}\) with \(x(t)=t\) and \(y(t)=1-t .\) Find the derivative of \(w=f(x, y)\) with respect to \(t\) when \(t=1 / 2\).

Step-by-Step Solution

Answer

The derivative \( \frac{dw}{dt} \) is 0 when \( t = \frac{1}{2} \).

Step 1: Define the Function

The function given is \( f(x, y) = \frac{1}{x} + \frac{1}{y} \). We have \( x(t) = t \) and \( y(t) = 1-t \), and we need to find the derivative of \( w = f(x, y) \) with respect to \( t \).

Step 2: Apply the Chain Rule

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

Step 3: Partial Derivatives

Calculate the partial derivatives of \( f \):\[ \frac{\partial f}{\partial x} = -\frac{1}{x^2} \]\[ \frac{\partial f}{\partial y} = -\frac{1}{y^2} \]

Step 4: Derivatives of x and y with respect to t

Since \( x(t) = t \) and \( y(t) = 1-t \):\[ \frac{dx}{dt} = 1 \]\[ \frac{dy}{dt} = -1 \]

Step 5: Substitute into Chain Rule Expression

Substitute the results from Steps 3 and 4 into the chain rule expression:\[ \frac{dw}{dt} = \left(-\frac{1}{x^2}\right) \cdot 1 + \left(-\frac{1}{y^2}\right) \cdot (-1) \]

Step 6: Evaluate at t = 1/2

When \( t = \frac{1}{2} \), we have \( x = \frac{1}{2} \) and \( y = \frac{1}{2} \). Substitute these values:\[ \frac{dw}{dt} = \left(-\frac{1}{(\frac{1}{2})^2}\right) \cdot 1 + \left(-\frac{1}{(\frac{1}{2})^2}\right) \cdot (-1) = -4 + 4 = 0 \]

Key Concepts

Understanding the Chain Rule

The chain rule is a powerful technique in calculus, primarily used to differentiate composite functions, meaning functions within functions. When dealing with functions of two variables, like in our exercise, the chain rule helps us compute the derivative of a function that depends indirectly on a parameter, through other functions.

In our scenario, we have a function \( w = f(x, y) \) that depends on \( x(t) \) and \( y(t) \), both functions of \( t \). The chain rule in this context is given by:

• The formula for the chain rule is: \( \frac{dw}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} \)
• This allows us to find how fast \( w \) changes as \( t \) changes, taking into account the rates at which \( x \) and \( y \) change with \( t \).

By using the chain rule, we streamline our process of finding the derivative of \( w \) concerning \( t \).

Derivative with Respect to a Parameter

In this exercise, the goal is to find the derivative of \( w = f(x, y) \) with respect to the parameter \( t \). This involves understanding how our variables \( x \, \text{and} \, y \) change as \( t \) changes. Here, \( t \) is treated as the independent variable.

Calculating the derivative with respect to \( t \), we:

• Differentiate \( x(t) = t \). The derivative \( \frac{dx}{dt} = 1 \) shows that \( x \) increases at a constant rate as \( t \) increases.
• Differentiate \( y(t) = 1-t \). The derivative \( \frac{dy}{dt} = -1 \) indicates that \( y \) decreases at a constant rate with \( t \).

These expressions help us plug into our chain rule, further working with the rates determined by \( dx/dt \) and \( dy/dt \).

Functions of Two Variables

Functions of two variables, such as \( f(x, y) \), are used to describe scenarios where outputs rely on two changing inputs. Here, \( f(x, y) = \frac{1}{x} + \frac{1}{y} \) involves two inputs, \( x \) and \( y \), each depending on the parameter \( t \).

When tackling such functions, partial derivatives become essential tools:

• The partial derivative \( \frac{\partial f}{\partial x} = -\frac{1}{x^2} \) represents how \( f \) changes as \( x \) changes, holding \( y \) constant.
• The partial derivative \( \frac{\partial f}{\partial y} = -\frac{1}{y^2} \) shows how \( f \) changes with adjustments in \( y \), keeping \( x \) fixed.

By understanding these derivatives, we grasp how each variable impacts the function independently, essential for correctly applying the chain rule.