Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, find dw/dt by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=\ln (x / y) ; x=\tan t, y=\sec ^{2} t $$

Step-by-Step Solution

Answer

\( \frac{dw}{dt} = \csc t - 2\tan t \) in terms of \( t \).

Step 1: Differentiate w with respect to x and y

The function is given as \[ w = \ln \left( \frac{x}{y} \right) \]. Let's use the quotient rule for logarithms to rewrite this as \[ w = \ln(x) - \ln(y) \]. Differentiate both terms using the chain rule: \[ \frac{dw}{dx} = \frac{1}{x} \] and \[ \frac{dw}{dy} = -\frac{1}{y} \].

Step 2: Differentiate x and y with respect to t

Given \( x = \tan t \) and \( y = \sec^2 t \), differentiate each with respect to \( t \):- \( \frac{dx}{dt} = \sec^2 t \) (since the derivative of \( \tan t \) is \( \sec^2 t \))- \( \frac{dy}{dt} = 2\sec t \cdot \sec t \tan t = 2\sec^2 t \tan t \) (since the derivative of \( \sec^2 t \) is \( 2\sec t \cdot \sec t \tan t \))

Step 3: Apply the Chain Rule

Use the chain rule to find \( \frac{dw}{dt} \) combining the partials and the derivatives we found:\[ \frac{dw}{dt} = \frac{dw}{dx} \cdot \frac{dx}{dt} + \frac{dw}{dy} \cdot \frac{dy}{dt} \]Substitute \( \frac{dw}{dx} = \frac{1}{x} \), \( \frac{dx}{dt} = \sec^2 t \), \( \frac{dw}{dy} = -\frac{1}{y} \), and \( \frac{dy}{dt} = 2\sec^2 t \tan t \) into this equation.

Step 4: Substitute x and y in terms of t

Since \( \frac{dw}{dt} = \frac{1}{x} \cdot \sec^2 t - \frac{1}{y} \cdot 2\sec^2 t \tan t \), substitute \( x = \tan t \) and \( y = \sec^2 t \) into the expression:\[ \frac{dw}{dt} = \frac{1}{\tan t} \cdot \sec^2 t - \frac{2\sec^2 t \tan t}{\sec^2 t} \]Simplify the terms.

Step 5: Simplify the Expression

Now simplify \[ \frac{1}{\tan t} \cdot \sec^2 t - 2\tan t \]. The term \( \frac{1}{\tan t} \cdot \sec^2 t \) simplifies to \( \frac{\sec^2 t}{\tan t} = \cot t \cdot \sec t = \csc t \). Hence, the expression becomes \[ \frac{dw}{dt} = \csc t - 2 \tan t \].

Key Concepts

Logarithmic Differentiation

Logarithmic differentiation is a powerful technique useful for differentiating complex functions, especially when the function comprises products, quotients, or powers that are difficult to manipulate directly. In the context of this exercise, we have the function \( w = \ln \left( \frac{x}{y} \right) \). The use of natural logarithms simplifies the differentiation by transforming such expressions using logarithmic rules:

• The quotient rule states that \( \ln(a/b) = \ln(a) - \ln(b) \).
• This allows us to express the function as \( w = \ln(x) - \ln(y) \).

By differentiating each term separately, we obtain partial derivatives:

• \( \frac{dw}{dx} = \frac{1}{x} \)
• \( \frac{dw}{dy} = -\frac{1}{y} \)

This method assists in simplifying the differentiation process, making logarithmic differentiation crucial for tackling such problems easily.

Trigonometric Functions

Trigonometric functions often appear in calculus problems, providing a unique challenge due to their periodic nature and the requirement to leverage specific differentiation rules. In this exercise, both \( x \) and \( y \) are expressed in terms of trigonometric functions:

• \( x = \tan t \)
• \( y = \sec^2 t \)

To differentiate these functions with respect to \( t \), we apply the following rules:

• The derivative of \( \tan t \) is \( \sec^2 t \).
• The derivative of \( \sec^2 t \) utilizes the chain rule and is calculated as \( 2\sec^2 t \tan t \).

Understanding the properties and derivatives of these trigonometric functions is essential for identifying changes in variables effectively.

Partial Derivatives

Partial derivatives are used when functions have multiple variables, allowing us to focus on how a function changes with respect to one variable at a time while keeping others constant. In the exercise, the function \( w \) depends on both \( x \) and \( y \), leading us to compute the partial derivatives:

• \( \frac{dw}{dx} = \frac{1}{x} \) indicates how \( w \) changes with \( x \).
• \( \frac{dw}{dy} = -\frac{1}{y} \) indicates how \( w \) changes with \( y \).

Once these partial derivatives are known, the chain rule is applied to combine them with the derivatives of \( x \) and \( y \) with respect to \( t \). Ultimately, these operations culminate in the expression for \( \frac{dw}{dt} \), offering a comprehensive view of how the overall function \( w \) changes with \( t \). This highlights the importance of partial derivatives in handling complex multivariable functions.