Problem 3 - 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 \(t\). $$ w=e^{x} \sin y+e^{y} \sin x ; x=3 t, y=2 t $$

Step-by-Step Solution

Answer

\( \frac{dw}{dt} = 3e^{3t} \sin 2t + 3e^{2t} \cos 3t + 2e^{3t} \cos 2t + 2e^{2t} \sin 3t \).

Step 1: Understand the Problem

We need to find \( \frac{dw}{dt} \), where \( w = e^x \sin y + e^y \sin x \), and the variables \( x \) and \( y \) are functions of \( t \), specifically \( x = 3t \) and \( y = 2t \). We will use the chain rule to differentiate \( w \) with respect to \( t \).

Step 2: Apply the Chain Rule

The chain rule states that \( \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} \). Thus, we must find the partial derivatives \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \), as well as the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Step 3: Find Partial Derivatives

For \( \frac{\partial w}{\partial x} \), treat \( y \) as a constant:\[\frac{\partial w}{\partial x} = \frac{d}{dx}(e^x \sin y) + \frac{d}{dx}(e^y \sin x) = e^x \sin y + e^y \cos x\]For \( \frac{\partial w}{\partial y} \), treat \( x \) as a constant:\[\frac{\partial w}{\partial y} = \frac{d}{dy}(e^x \sin y) + \frac{d}{dy}(e^y \sin x) = e^x \cos y + e^y \sin x\]

Step 4: Differentiate x and y with respect to t

Since \( x = 3t \) and \( y = 2t \), we find:\[\frac{dx}{dt} = 3\]\[\frac{dy}{dt} = 2\]

Step 5: Substitute the Derivatives into the Chain Rule Formula

Substitute into the chain rule formula:\[\frac{dw}{dt} = \left(e^x \sin y + e^y \cos x\right) \cdot 3 + \left(e^x \cos y + e^y \sin x\right) \cdot 2\]

Step 6: Substitute x and y in Terms of t

Substitute \( x = 3t \) and \( y = 2t \) into the formula:\[\frac{dw}{dt} = (e^{3t} \sin 2t + e^{2t} \cos 3t) \cdot 3 + (e^{3t} \cos 2t + e^{2t} \sin 3t) \cdot 2\]

Step 7: Simplify the Expression

Combining and simplifying terms, the final expression becomes:\[\frac{dw}{dt} = 3e^{3t} \sin 2t + 3e^{2t} \cos 3t + 2e^{3t} \cos 2t + 2e^{2t} \sin 3t\]

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in calculus, especially when dealing with functions of multiple variables. In this exercise, we are working with the function \(w = e^x \sin y + e^y \sin x \). The task is to find \(\frac{\partial w}{\partial x}\) and \(\frac{\partial w}{\partial y}\).

• Keeping Variables Constant: Treat one variable as constant while differentiating with respect to the other. For \(\frac{\partial w}{\partial x}\), treat \(y\) as constant. For \(\frac{\partial w}{\partial y}\), treat \(x\) as constant.
• Compute \(\frac{\partial w}{\partial x}\): \[ \frac{\partial w}{\partial x} = \frac{d}{dx}(e^x \sin y) + \frac{d}{dx}(e^y \sin x) = e^x \sin y + e^y \cos x \]
• Compute \(\frac{\partial w}{\partial y}\): \[ \frac{\partial w}{\partial y} = \frac{d}{dy}(e^x \sin y) + \frac{d}{dy}(e^y \sin x) = e^x \cos y + e^y \sin x \]

Understanding these derivatives allows you to apply the chain rule effectively later in solving the problem.

Differentiation

Differentiation, in the context of this problem, is about finding how a function changes as its inputs change. Here, we are interested in how the function \(w\) changes concerning time \(t\). This involves using the derivatives of \(x(t)\) and \(y(t)\), which are both functions of \(t\).

• Differentiate \(x\) and \(y\): Since \(x = 3t\) and \(y = 2t\), differentiate each with respect to \(t\): \[ \frac{dx}{dt} = 3 \] \[ \frac{dy}{dt} = 2 \]

These straightforward derivatives go hand-in-hand with the partial derivatives found earlier, assisting in applying the chain rule for finding \(\frac{dw}{dt}\). Differentiation, especially taking derivatives concerning one variable while considering other variables, is a key step in calculus problem solving.

Calculus Problem Solving

Calculus problem solving often involves tying together several concepts and techniques. In this exercise, you apply the chain rule, which is crucial when dealing with composite functions. The chain rule facilitates the differentiation of \(w\) with respect to \(t\). It's essentially a formula to connect the partial derivatives with the change of \(x\) and \(y\) with respect to \(t\):

• Apply the Chain Rule: \[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} \]
• Substitute and Solve: Using the values: \[ \frac{dw}{dt} = (e^{3t} \sin 2t + e^{2t} \cos 3t) \cdot 3 + (e^{3t} \cos 2t + e^{2t} \sin 3t) \cdot 2 \]
• Simplify: Finally, the simplified expression is: \[ \frac{dw}{dt} = 3e^{3t} \sin 2t + 3e^{2t} \cos 3t + 2e^{3t} \cos 2t + 2e^{2t} \sin 3t \]

By carefully applying the chain rule and practicing calculus problem solving, you can handle complex problems involving multiple variables and their derivatives effectively.