Problem 218 - Calculus Volume 3

Question

Use the information provided to solve the problem. If \(w=x y^{2}, x=5 \cos (2 t), \quad\) and \(\quad y=5 \sin (2 t)\), find \(\frac{\partial w}{\partial t}\).

Step-by-Step Solution

Answer

\( \frac{\partial w}{\partial t} = 250 \sin(2t) \cos(4t) \)

Step 1: Write the expression for w

Given that \( w = x y^2 \), substitute \( x = 5 \cos(2t) \) and \( y = 5 \sin(2t) \) into the equation:\[ w = (5 \cos(2t))(5 \sin(2t))^2 \]

Step 2: Simplify the expression for w

Simplify the expression from Step 1:\[ w = (5 \cos(2t))(25 \sin^2(2t)) = 125 \cos(2t) \sin^2(2t) \]

Step 3: Differentiate w with respect to t

To find \(\frac{\partial w}{\partial t}\), apply the product rule of differentiation to \( w = 125 \cos(2t) \sin^2(2t) \): - Differentiate \( \cos(2t) \) with respect to \( t \):\[ \frac{d}{dt} \cos(2t) = -2 \sin(2t) \]- Differentiate \( \sin^2(2t) \) with respect to \( t \):\[ \frac{d}{dt} \sin^2(2t) = 2\sin(2t)\cos(2t)\]Apply the product rule:\[ \frac{\partial w}{\partial t} = 125 \left( \cos(2t) \cdot 2\sin(2t)\cos(2t) + \sin^2(2t) \cdot (-2 \sin(2t)) \right) \] Simplify the expression.

Step 4: Simplify \\ \( \frac{\partial w}{\partial t} \\) expression

Simplify further:\[ \frac{\partial w}{\partial t} = 125 \left( 2 \cos(2t) \sin(2t) \cdot \cos(2t) - 2 \sin^3(2t) \right) \]\[ = 250 \cos^2(2t) \sin(2t) - 250 \sin^3(2t) \]Factor out \( 250\sin(2t) \):\[ \frac{\partial w}{\partial t} = 250 \sin(2t)(\cos^2(2t) - \sin^2(2t)) \]

Step 5: Use a trigonometric identity

Recognize \( \cos^2(2t) - \sin^2(2t) \) is equivalent to \( \cos(4t) \) using the double angle identity:\[ \frac{\partial w}{\partial t} = 250 \sin(2t) \cos(4t) \]

Key Concepts

Product Rule in Differentiation

The product rule is a fundamental concept in calculus, specifically when dealing with the differentiation of functions that are products of two or more differentiable functions. In essence, if you have a function that is a product, say \( u(t) \) and \( v(t) \), then its derivative is not just the derivative of each function taken separately. Instead, there’s a specific way to combine them:

• For a product \( uv \), the derivative \((uv)'\) is \( u'v + uv' \).

When solving problems that require finding the derivative of a product, such as the one in the original exercise, the product rule is applied to each part appropriately. The crucial part is recognizing when you need to employ this rule—it activates any time you see the multiplication of two variable-dependent functions.
In our given problem, we have \( w = 125 \cos(2t) \sin^2(2t) \), a product involving \( \cos(2t) \) and \( \sin^2(2t) \). Applying the product rule allowed us to efficiently differentiate each component and then sum their derivatives, showcasing the practicality of this powerful tool.

Trigonometric Identities Simplified

Trigonometric identities are vital tools in calculus, especially when dealing with the differentiation of trigonometric functions. They are equations involving trigonometric functions that are true for every value of the variable within the domain. These identities simplify expressions and make it easier to perform calculations, such as differentiation.

• One important identity used is the double angle identity. The double angle formula for cosine, \( \cos(2x) = \cos^2(x) - \sin^2(x) \), reappears in various problems, as seen in the original exercise.

These identities allow us to transform expressions into newer forms that might be more manageable. In Step 5 of the original solution, the identity \( \cos^2(2t) - \sin^2(2t) = \cos(4t) \) was critical in expressing \( \frac{\partial w}{\partial t} \) more compactly. This simplification is not only useful for simplification but can also reveal more about the behavior of the function that might not be evident in unsimplified forms. Understanding and applying these identities can be a key to solving complex trigonometric calculus problems.

Differentiation: Unlocking Change

Differentiation is one of the foundational operations in calculus. At its core, it is the process of finding the derivative of a function, which represents the rate of change of the function's output value with respect to changes in its input value. This is crucial in many fields, from physics to economics, where understanding how one quantity changes in relation to another is invaluable.

• The derivative tells us how the function dynamically behaves; for example, whether it is increasing or decreasing and how steeply.
• In partial differentiation, we focus on how a multi-variable function changes with respect to one variable, keeping others constant.

In the exercise tackled, we found \( \frac{\partial w}{\partial t} \), the rate at which \( w \) changes with respect to time \( t \), while considering variables \( x \) and \( y \) as functions of \( t \). By differentiating each part according to its dependence on \( t \), we capture the nuanced temporal changes in \( w \), illustrating the power of derivatives as tools for exploring and modeling dynamic change. This foundational calculus concept is indispensable for tackling more complex mathematics and real-world problem-solving.