Problem 22 - Calculus Early Transcendentals

Question

Find the following derivatives. \(z_{s}\) and \(z_{t},\) where \(z=\sin x \cos 2 y, x=s+t,\) and \(y=s-t\)

Step-by-Step Solution

Answer

Question: Given the function \(z = \sin x \cos 2y\) with \(x = s + t\) and \(y = s - t\), find the partial derivatives of \(z\) with respect to \(s\) and \(t\). Answer: The partial derivatives of \(z\) with respect to \(s\) and \(t\) are given by: \[ z_s = (\cos x)(\cos 2y) - 2(\sin x)(\sin 2y) \] and \[ z_t = (\cos x)(\cos 2y) + 2(\sin x)(\sin 2y) \].

Step 1: Differentiate z with respect to x and y.

Using the product rule, find the partial derivatives of \(z = \sin x \cos 2y\) with respect to \(x\) and \(y\). We get: \[ \frac{\partial z}{\partial x} = (\cos x) (\cos 2y) \] and \[ \frac{\partial z}{\partial y} = (\sin x) (-2 \sin 2y). \]

Step 2: Differentiate x and y with respect to s and t.

Find the partial derivatives of \(x = s+t\) and \(y = s-t\) with respect to \(s\) and \(t\). We get: \[ \frac{\partial x}{\partial s} = 1, \quad \frac{\partial x}{\partial t} = 1, \quad \frac{\partial y}{\partial s} = 1, \quad \text{and} \quad \frac{\partial y}{\partial t} = -1. \]

Step 3: Use the chain rule to find z_s and z_t.

Applying the chain rule to find the partial derivatives of \(z\) with respect to \(s\) and \(t\), we get \[ z_s = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \] and \[ z_t = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}. \]

Step 4: Calculate z_s and z_t.

Substitute the expressions for the partial derivatives that we found in Steps 1 and 2 into the chain rule equations in Step 3. For \(z_s\), we get: \[ z_s = (\cos x)(\cos 2y)(1) + (\sin x)(-2\sin 2y)(1) = (\cos x)(\cos 2y) - 2(\sin x)(\sin 2y) \] For \(z_t\), we get: \[ z_t = (\cos x)(\cos 2y)(1) + (\sin x)(-2\sin 2y)(-1) = (\cos x)(\cos 2y) + 2(\sin x)(\sin 2y) \] So the partial derivatives \(z_s\) and \(z_t\) are given by: \[ z_s = (\cos x)(\cos 2y) - 2(\sin x)(\sin 2y) \] and \[ z_t = (\cos x)(\cos 2y) + 2(\sin x)(\sin 2y) \]

Key Concepts

Chain Rule

The chain rule is essential when dealing with composite functions, especially when several variables are interdependent. In this context, the chain rule helps in finding how a change in independent variables like \(s\) and \(t\) impacts a dependent variable \(z\) through intermediate variables \(x\) and \(y\). This is particularly useful if \(z\) is defined as a function of \(x\) and \(y\), while \(x\) and \(y\) themselves are functions of \(s\) and \(t\).
The formal expression for the chain rule when applied to partial derivatives is:

• For \(z_s\): \[ z_s = \frac{\partial z}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \]
• For \(z_t\): \[ z_t = \frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t} \]

This allows you to "chain" the effects of small changes in \(s\) and \(t\) through the intermediate variables \(x\) and \(y\) to calculate how they ultimately change \(z\).
Through this approach, you can express how small changes in one variable propagate through a network of dependencies.

Product Rule

The product rule is crucial whenever you need to find the derivative of a product of two functions. It's an extension of the standard single-variable derivative rule to handle products, which often crop up in calculus.
If \(z = \sin x \cos 2y\), the expression is a product of two distinct functions: \(\sin x\) and \(\cos 2y\). The concept of the product rule states that the derivative of such a product is not simply the product of the derivatives. Instead, you differentiate one function while keeping the other constant, and then switch roles.

• For this particular function, the product rule for partial derivatives is expressed as:
• \[ \frac{\partial z}{\partial x} = \cos x \cdot \cos 2y \]
• \[ \frac{\partial z}{\partial y} = \sin x \cdot (-2\sin 2y) \]

The product rule ensures that you account for both components of the product when differentiating by taking into account the contribution of each part of the product separately. This complexity is essential to capturing the full behavior of more intricate functions and their interactions during differentiation.

Trigonometric Functions

Trigonometric functions, like sine and cosine, play a significant role in calculus, especially when dealing with periodic processes or rotational dynamics. In the problem, both \(\sin x\) and \(\cos 2y\) are used as they frequently occur in equations involving angles and periodic functions. Understanding how these functions vary allows you to apply rules such as the chain and product rules effectively.
Here are a few key points to remember about some common trigonometric derivatives:

• The derivative of \(\sin x\) is \(\cos x\).
• The derivative of \(\cos x\) is \(-\sin x\).
• The derivative of \(\sin kx\) is \(k \cos kx\), where \(k\) is a constant, as seen in \(\cos 2y\).

These derivatives can be directly applied using the product and chain rules to help break down and solve more complex calculus problems. Trigonometric identities and derivatives provide the necessary tools to manipulate the expressions that involve trigonometric terms and calculate intricate derivatives readily.