Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(d w / d t\) 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: Identify Dependencies

First, we need to determine what variables depend on others. The function to differentiate is \( w = e^x \sin y + e^y \sin x \), and both \( x \) and \( y \) depend on \( t \) as given by \( x = 3t \) and \( y = 2t \).

Step 2: Apply the Chain Rule

We use the Chain Rule to express \( \frac{dw}{dt} \). The Chain Rule tells us: \[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \]. We need to find each of these partial derivatives.

Step 3: Derivative of w with respect to x

Calculate \( \frac{\partial w}{\partial x} \): For \( w = e^x \sin y + e^y \sin x \), the derivative with respect to \( x \) is: \[ \frac{\partial w}{\partial x} = e^x \sin y + e^y \cos x \].

Step 4: Derivative of w with respect to y

Calculate \( \frac{\partial w}{\partial y} \): For \( w = e^x \sin y + e^y \sin x \), the derivative with respect to \( y \) is: \[ \frac{\partial w}{\partial y} = e^x \cos y + e^y \sin x \].

Step 5: Derivative of x and y with respect to t

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): Since \( x = 3t \), we have \( \frac{dx}{dt} = 3 \). Since \( y = 2t \), we have \( \frac{dy}{dt} = 2 \).

Step 6: Substitute and Simplify

Substitute the partial derivatives and derivatives of \( x \) and \( y \) into the Chain Rule formula to find \( \frac{dw}{dt} \):\[\frac{dw}{dt} = \left(e^x \sin y + e^y \cos x\right) \cdot 3 + \left(e^x \cos y + e^y \sin x\right) \cdot 2\]Substituting \( x = 3t \) and \( y = 2t \) into the equation:\[\frac{dw}{dt} = 3\left(e^{3t} \sin (2t) + e^{2t} \cos (3t)\right) + 2\left(e^{3t} \cos (2t) + e^{2t} \sin (3t)\right)\]

Step 7: Simplify Further

Let's expand each term to make the equation simpler:\[ \frac{dw}{dt} = 3e^{3t} \sin (2t) + 3e^{2t} \cos (3t) + 2e^{3t} \cos (2t) + 2e^{2t} \sin (3t)\]This expression is our final answer for \( \frac{dw}{dt} \) in terms of \( t \).

Key Concepts

Partial Derivatives

A partial derivative represents the rate at which a function changes as one of its variables changes, while keeping the other variables constant. When working with functions of multiple variables, like our given function \[ w = e^x \sin y + e^y \sin x \]we need to individually explore the effect of changing just one variable at a time.

• For the function turning concerning variable \( x \), we calculate \( \frac{\partial w}{\partial x} \) with all other variables held fixed. This gives an insight into how \( w \) changes just by modifying \( x \).
• Similarly, for changes in \( y \), \( \frac{\partial w}{\partial y} \) is calculated.
  • \( \frac{\partial w}{\partial x} = e^x \sin y + e^y \cos x \)
  • \( \frac{\partial w}{\partial y} = e^x \cos y + e^y \sin x \)

Recognizing the contribution of each variable independently through its partial derivative is crucial for thorough function analysis, allowing the precise application of further calculus techniques like the Chain Rule.

Calculus Problem Solving

Calculus problem solving often involves breaking down a problem into smaller, more manageable pieces. By strategically approaching each part of the problem, like derivatives and substitutions, we can solve complex calculations step-by-step.

• Identify all functions and their dependencies. In our problem, the function \( w \) is directly dependent on two variables, \( x \) and \( y \), each of which further depends on \( t \).
• Apply suitable calculus principles when approaching each segment of the problem. For instance, the Chain Rule is selected due to the indirect dependency of \( w \) on \( t \) through \( x \) and \( y \).
• Simplify each computation after applying derivatives or substituting variables to prevent errors and miscalculations. Each derivative or computation leads to the final solution after simplification.

Effective problem-solving in calculus requires a methodological strategy, use of appropriate rules, and simplification for a clear and accurate solution.

Differentiation Techniques

Differentiation techniques in calculus are tools we use to find the derivative of functions. These include using rules such as the Chain Rule and knowing when to deploy each to solve problems efficiently.The Chain Rule is particularly important when dealing with functions where a variable depends on other variables. It allows us to take derivatives where a function has nested dependencies like in our problem:\[ \frac{dw}{dt} = \frac{\partial w}{\partial x} \frac{dx}{dt} + \frac{\partial w}{\partial y} \frac{dy}{dt} \]Applying the Chain Rule involves:

• Calculating partial derivatives \( \frac{\partial w}{\partial x} \) and \( \frac{\partial w}{\partial y} \).
• Determining the derivatives of \( x \) and \( y \) with respect to \( t \): \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = 2 \).
• Combining these findings using the Chain Rule formula to get \( \frac{dw}{dt} \).

Learning different techniques for differentiation ensures one is prepared to tackle a range of potential calculus problems effectively.