Problem 54 - Calculus Early Transcendentals

Question

\(49-54\) Assume that all the given functions have continuous second-order partial derivatives. Suppose \(z=f(x, y),\) where \(x=g(s, t)\) and \(y=h(s, t) .\) (a) Show that $$\begin{aligned} \frac{\partial^{2} z}{\partial t^{2}}=\frac{\partial^{2} z}{\partial x^{2}} &\left(\frac{\partial x}{\partial t}\right)^{2}+2 \frac{\partial^{2} z}{\partial X \partial y} \frac{\partial x}{\partial t} \frac{\partial y}{\partial t}+\frac{\partial^{2} z}{\partial y^{2}}\left(\frac{\partial y}{\partial t}\right)^{2} \\ &+\frac{\partial z}{\partial x} \frac{\partial^{2} x}{\partial t^{2}}+\frac{\partial z}{\partial y} \frac{\partial^{2} y}{\partial t^{2}} \end{aligned}$$ (b) Find a similar formula for \(\partial^{2} z / \partial s \partial t\)

Step-by-Step Solution

Answer

For part (a), apply chain rule twice to show the expression. For part (b), use similar differentiation to derive a formula.

Step 1: Understanding the Problem

We need to show the given expression for \( \frac{\partial^{2} z}{\partial t^{2}} \) by using chain rule and to find a similar expression for \( \frac{\partial^{2} z}{\partial s \partial t} \). The functions and their derivatives must be manipulated appropriately using partial derivatives and chain rules.

Step 1: Differentiate \( z = f(x, y) \) once with respect to \( t \)

First apply the chain rule to differentiate \( z = f(x, y) \) with respect to \( t \). This gives us:\[\frac{\partial z}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}\]

Step 2: Differentiate \( \frac{\partial z}{\partial t} \) again with respect to \( t \)

Apply the product rule and chain rule:\[\frac{\partial^2 z}{\partial t^2} = \frac{\partial}{\partial t}\left( \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} \right) + \frac{\partial}{\partial t}\left( \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} \right)\]Calculate each derivative individually:\[= \frac{\partial^2 z}{\partial x^2} \left( \frac{\partial x}{\partial t} \right)^2 + \frac{\partial z}{\partial x} \frac{\partial^2 x}{\partial t^2} + \frac{\partial^2 z}{\partial x \partial y} \frac{\partial x}{\partial t} \frac{\partial y}{\partial t}\]\[+ \frac{\partial^2 z}{\partial y \partial x} \frac{\partial y}{\partial t} \frac{\partial x}{\partial t} + \frac{\partial^2 z}{\partial y^2} \left( \frac{\partial y}{\partial t} \right)^2 + \frac{\partial z}{\partial y} \frac{\partial^2 y}{\partial t^2}\]Rearrange terms to match the provided expression.

Step 3: Derive \( \frac{\partial^2 z}{\partial s \partial t} \) using similar principles

Start by differentiating \( \frac{\partial z}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s} \):\[\frac{\partial^2 z}{\partial s \partial t} = \frac{\partial}{\partial t}\left( \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \right)\]Compute each derivative:\[= \frac{\partial^2 z}{\partial x^2} \frac{\partial x}{\partial t} \frac{\partial x}{\partial s} + \frac{\partial^2 z}{\partial y \partial x} \left( \frac{\partial y}{\partial t} \frac{\partial x}{\partial s} + \frac{\partial x}{\partial t} \frac{\partial y}{\partial s} \right)\]\[+ \frac{\partial^2 z}{\partial y^2} \frac{\partial y}{\partial t} \frac{\partial y}{\partial s} + \frac{\partial z}{\partial x} \frac{\partial^2 x}{\partial s \partial t} + \frac{\partial z}{\partial y} \frac{\partial^2 y}{\partial s \partial t}\]This forms a similar expression for the mixed partial derivative.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used for differentiating composite functions. It allows us to find the derivative of a function with respect to an independent variable by considering the dependency of functions inside it. In our context, the function \( z = f(x, y) \) depends on \( x \) and \( y \), which are in turn functions of \( s \) and \( t \). The chain rule enables us to calculate partial derivatives of \( z \) with respect to \( t \). Here’s how we apply it:

• To find \( \frac{\partial z}{\partial t} \), use the chain rule for multivariable functions: \( \frac{\partial z}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} \).

The chain rule helps us develop expressions for derivatives where multiple variables interact. It provides a systematic method to differentiate complex expressions by breaking them down into simpler parts, specifically useful when variables are themselves functions of other variables.

Second-Order Derivatives

Second-order derivatives are derivatives of a derivative. They provide information about the curvature or the rate of change of the rate of change of a function. For functions of two variables, such as \( z = f(x, y) \), we are interested in second-order partial derivatives like \( \frac{\partial^2 z}{\partial t^2} \) and \( \frac{\partial^2 z}{\partial s \partial t} \).

• To compute \( \frac{\partial^2 z}{\partial t^2} \), we must differentiate \( \frac{\partial z}{\partial t} \) again with respect to \( t \). Apply the product and chain rules in this calculation.
• For mixed partial derivatives like \( \frac{\partial^2 z}{\partial s \partial t} \), differentiate \( \frac{\partial z}{\partial s} \) with respect to \( t \), again using the chain rule.

Second-order derivatives are crucial in understanding and predicting the behavior of physical systems described by functions of several variables, offering insights into concavity, convexity, and inflection points beyond what first-order derivatives provide.

Product Rule

The product rule is a useful tool when differentiating a product of two or more functions. It states that the derivative of a product \( u \cdot v \) is \( u'v + uv' \), where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. When dealing with partial derivatives, this concept is applied separately to each component of the product.

• In the expressions for \( \frac{\partial^2 z}{\partial t^2} \), use the product rule to manage derivatives like \( \frac{\partial}{\partial t}(\frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t}) \). Break it down into components that can be differentiated separately.

The product rule allows complexities in ensuring each part of a product is adequately differentiated, especially in multivariable calculus, making the process of finding higher-order derivatives more systematic and manageable.