Problem 4 - Calculus Early Transcendentals

Question

\(1-6\) Use the Chain Rule to find \(d z / d t\) or \(d w / d t\) $$z=\tan ^{-1}(y / x), \quad x=e^{t}, \quad y=1-e^{-t}$$

Step-by-Step Solution

Answer

\( \frac{dz}{dt} = \frac{-e^{-t} + 2e^{-2t}}{1 + (e^{-t} - e^{-2t})^2} \)

Step 1: Identify the composite function

The function given is \( z = \tan^{-1}(y/x) \). This is a composite function where \( y = 1 - e^{-t} \) and \( x = e^t \). First, express \( z \) in terms of \( t \): \( z = \tan^{-1}\left(\frac{1 - e^{-t}}{e^t}\right) \).

Step 2: Express the argument in terms of a single variable

Simplify the expression \( \frac{y}{x} = \frac{1 - e^{-t}}{e^t} \) to \( e^{-t} - e^{-2t} \) by dividing the numerator and the denominator by \( e^t \).

Step 3: Differentiate with respect to t using the Chain Rule

Use the chain rule: \( \frac{dz}{dt} = \frac{dz}{du} \cdot \frac{du}{dt} \), where \( u = \frac{y}{x} = e^{-t} - e^{-2t} \). First, find \( \frac{dz}{du} = \frac{1}{1+u^2} \) because \( \frac{d}{du}(\tan^{-1}(u)) = \frac{1}{1+u^2} \).

Step 4: Differentiate u with respect to t

Compute \( \frac{du}{dt} \). For \( u = e^{-t} - e^{-2t} \), the derivative is \( \frac{d}{dt}(e^{-t} - e^{-2t}) = -e^{-t} + 2e^{-2t} \).

Step 5: Combine derivatives

Combine the results from Step 3 and Step 4: \( \frac{dz}{dt} = \frac{1}{1+(e^{-t} - e^{-2t})^2} \cdot (-e^{-t} + 2e^{-2t}) \).

Step 6: Simplify the expression

Simplify the expression to find the final derivative: \( \frac{dz}{dt} = \frac{-e^{-t} + 2e^{-2t}}{1 + (e^{-t} - e^{-2t})^2} \).

Key Concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It can be divided into two main branches: differential calculus and integral calculus. In differential calculus, the primary goal is to understand how functions change. We achieve this by finding the derivative, which gives us the rate of change of a function.
Derivatives are found using rules and methods such as the power rule, product rule, quotient rule, and chain rule. These rules help us tackle different types of functions, whether simple or complex, allowing us to analyze their behavior efficiently. Key points in calculus:

• Understanding the concept of a limit, which is foundational in determining derivatives and integrals.
• Applying derivatives to find rates of change, thereby solving real-world problems.
• Using integrals to accumulate quantities, such as finding area under a curve.

Understanding these concepts provides a toolkit for problem-solving in fields ranging from engineering to economics.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with equations not easily solved for one variable in terms of another.
Such equations are termed 'implicit' because they involve a dependent variable and an independent variable intertwined together on one or both sides of the equation. When identifying differentiation problems, if you encounter an equation like a circle, such as \( x^2 + y^2 = r^2 \), it's implicit since \( y \) isn't isolated. Unlike explicit functions, where \( y \) is expressed solely in terms of \( x \), implicit functions require differentiation using both the chain rule and the product rule.
Here's how it works:

• Differentiate both sides with respect to the independent variable. Note the use of the chain rule for \( y \) terms.
• Remember, every time you differentiate \( y \) with respect to \( x \), you add a \( \frac{dy}{dx} \) term, reflecting the derivative of \( y \) in terms of \( x \).
• Solve for \( \frac{dy}{dx} \) by isolating these terms on one side of the equation.

This method enables the differentiation of equations and helps find slopes of tangent lines and other related measures, which might be tricky using regular differentiation.

Composite Function Differentiation

Composite function differentiation is crucial when dealing with nested functions. The chain rule is typically required to differentiate these composite functions.
A composite function is where one function is composed within another, such as \( f(g(x)) \). The core idea is to differentiate the outer function first and then multiply it by the derivative of the inner function. Here's a deeper insight into the process:

• Identify the inner function \( g(x) \) and the outer function \( f(u) \), where \( u = g(x) \).
• Differentiating a composite function means applying the chain rule: \( \frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x) \).
• In practical applications, simplifying expressions before applying the derivative can greatly enhance accuracy and manageability.

By mastering composite function differentiation, students can effectively tackle complex problems involving multiple nested functions. This skill is particularly useful when analyzing real-world processes that occur in layers, such as temperature change over time or growth models of populations.