Problem 224 - Calculus 3

Question

For the following exercises, find \(\frac{d f}{d t}\) using the chain rule and direct substitution. $$ f(x, y)=\frac{x}{y}, x=e^{t}, y=2 e^{t} $$

Step-by-Step Solution

Answer

The derivative \( \frac{df}{dt} \) is 0.

Step 1: Identify Function Composition

We have a function \( f(x, y) = \frac{x}{y} \). The variables \(x\) and \(y\) are expressed as functions of \(t\): \(x = e^t\) and \(y = 2e^t\). This means \(f\) can be considered a composition of functions.

Step 2: Apply the Chain Rule

The chain rule for a function \( f(x(t), y(t)) \) states: \( \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \). We need to find \( \frac{\partial f}{\partial x} \), \( \frac{\partial f}{\partial y} \), \( \frac{dx}{dt} \), and \( \frac{dy}{dt} \) to apply this rule.

Step 3: Calculate Partial Derivatives

First, compute \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \):\( \frac{\partial f}{\partial x} = \frac{1}{y} \).\( \frac{\partial f}{\partial y} = -\frac{x}{y^2} \).

Step 4: Compute Derivatives of \(x\) and \(y\)

Since \(x = e^t\), we have \( \frac{dx}{dt} = e^t \).Since \(y = 2e^t\), we have \( \frac{dy}{dt} = 2e^t \).

Step 5: Substitute into Chain Rule Formula

Substitute the derivatives into the chain rule formula:\[ \frac{df}{dt} = \frac{1}{y} \cdot e^t + \left( -\frac{x}{y^2} \right) \cdot 2e^t \].

Step 6: Simplify the Expression

Substitute \(y = 2e^t\) and \(x = e^t\) into the expression and simplify:\[ \frac{df}{dt} = \frac{1}{2e^t} \cdot e^t - \frac{e^t}{(2e^t)^2} \cdot 2e^t \].This simplifies to:\[ \frac{df}{dt} = \frac{1}{2} - \frac{1}{2} = 0 \].

Key Concepts

Partial Derivatives

Let's dive into the world of partial derivatives! Partial derivatives are crucial when dealing with functions of multiple variables. They help us understand how a function changes when we tweak one variable while keeping others constant. In our problem, we have the function \( f(x, y) = \frac{x}{y} \), where both \( x \) and \( y \) are themselves functions of \( t \). To apply the chain rule, we needed partial derivatives of \( f \) with respect to \( x \) and \( y \). First, the partial derivative \( \frac{\partial f}{\partial x} \) treats \( y \) as a constant and results in \( \frac{1}{y} \). That’s because if you increase \( x \) while \( y \) stays the same, \( f \) increases by \( \frac{1}{y} \) times the increase in \( x \).
Next, the partial derivative \( \frac{\partial f}{\partial y} \) is \( -\frac{x}{y^2} \). Here, \( x \) is treated as constant. When you increase \( y \), the value of \( f \) decreases since you are dividing by a larger number, hence the negative sign.
Understanding how these derivatives are calculated helps in grasping how changes in each variable independently affect the function.

Function Composition

Function composition is about building complex functions from simpler ones. In the given exercise, \( f(x, y) = \frac{x}{y} \) is dependent on \( x \) and \( y \) which are each functions of \( t \): \( x = e^t \) and \( y = 2e^t \). This forms a composition of functions since \( f \) depends on \( t \) through its dependencies on \( x \) and \( y \).

• The outer function is \( f(x, y) \).
• Inner functions are \( x(t) \) and \( y(t) \).

Understanding function composition is all about recognizing these layers of dependency. Each layer only knows about the one below it.
In this problem, start by evaluating \( x = e^t \) and \( y = 2e^t \), then use these values to find \( f(x, y) \). This stacked approach is key to unravelling complex problems.

Calculus Problem Solving

Solving calculus problems often involves breaking them down into simpler steps. For instance, solving our problem using the chain rule consists of several clear steps.

• First, identify the function and its dependencies on its variables.
• Next, calculate partial derivatives for each variable.
• Find the derivatives of those variables with respect to \( t \).
• Apply the chain rule formula \( \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} \).
• Finally, substitute and simplify the expression.

In this problem, substituting both \( x = e^t \) and \( y = 2e^t \) into the derivatives and simplifying resulted in \( \frac{df}{dt} = 0 \). Each step builds on the previous, turning a complex problem into a sequence of manageable tasks.
This systematic approach is the essence of calculus problem-solving, enabling us to tackle complex problems with confidence.