Problem 48 - Calculus Early Transcendentals

Question

\(45-48\) Assume that all the given functions are differentiable. If \(z=f(x, y),\) where \(x=s+t\) and \(y=s-t,\) show that $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial s} \frac{\partial z}{\partial t}$$

Step-by-Step Solution

Answer

The identity is proven by expressing \(z\) using \(x\) and \(y\) and applying the chain rule for partial derivatives. The manipulations lead to the verification of the given condition.

Step 1: Express z as a function of s and t

Given that \(z=f(x, y)\), where \(x=s+t\) and \(y=s-t\), we can express \(z\) in terms of \(s\) and \(t\). Thus, \(z=f(s+t, s-t)\).

Step 2: Calculate partial derivatives of z with respect to x and y

Using the chain rule, we find the partial derivatives of \(z\):\(\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x} \) and \( \frac{\partial z}{\partial y} = \frac{\partial f}{\partial y}\), where \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) are the respective partial derivatives of \(f\).

Step 3: Express partial derivatives in terms of s and t

Using the expressions of \(x\) and \(y\) in terms of \(s\) and \(t\), apply the chain rule to express the derivatives:\[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial s} \cdot \frac{\partial s}{\partial x} + \frac{\partial z}{\partial t} \cdot \frac{\partial t}{\partial x} \]\[ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial s} \cdot \frac{\partial s}{\partial y} + \frac{\partial z}{\partial t} \cdot \frac{\partial t}{\partial y} \]

Step 4: Determine partial derivatives with respect to s and t

Calculate necessary partial derivatives from the equations \(x=s+t\) and \(y=s-t\):- From \(x=s+t\) and \(y=s-t\), we know: \[\frac{\partial s}{\partial x} = \frac{1}{2}, \quad \frac{\partial t}{\partial x} = \frac{1}{2} \] \[\frac{\partial s}{\partial y} = \frac{1}{2}, \quad \frac{\partial t}{\partial y} = -\frac{1}{2} \]Thus,\[\frac{\partial z}{\partial x} = \frac{1}{2}\left( \frac{\partial z}{\partial s} + \frac{\partial z}{\partial t}\right)\]\[\frac{\partial z}{\partial y} = \frac{1}{2}\left( \frac{\partial z}{\partial s} - \frac{\partial z}{\partial t}\right)\]

Step 5: Confirm the given condition

Now we have:\[ \left(\frac{\partial z}{\partial x}\right)^2 - \left(\frac{\partial z}{\partial y}\right)^2 \]\[= \left\{\frac{1}{2}( \frac{\partial z}{\partial s} + \frac{\partial z}{\partial t} )\right\}^2 - \left\{\frac{1}{2}( \frac{\partial z}{\partial s} - \frac{\partial z}{\partial t} )\right\}^2 \]Simplifying using the identity for the difference of squares:\[ = \frac{1}{4} \left( ( \frac{\partial z}{\partial s} + \frac{\partial z}{\partial t} )^2 - ( \frac{\partial z}{\partial s} - \frac{\partial z}{\partial t} )^2 \right) \]This simplifies to:\[ = \frac{1}{4} \left( 4 \frac{\partial z}{\partial s} \frac{\partial z}{\partial t} \right) \]\[ = \frac{\partial z}{\partial s} \frac{\partial z}{\partial t} \]Thus proving the given condition.

Key Concepts

Partial Derivatives

Partial derivatives are essential tools in multivariable calculus. They provide us the rate of change of a function with respect to one variable while keeping others constant.
For a function like \(z = f(x, y)\), the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) indicate how \(z\) changes when we vary \(x\) or \(y\) alone.
This concept is especially pivotal in fields where multiple variables influence outcomes, such as physics or economics.

• Partial derivatives are notated with \(\frac{\partial}{\partial}\) to distinguish them from regular derivatives.
• They are computed by treating all other variables as constants.

Understanding these derivatives gives insight into the behavior of a function in a specific direction while leaving other variables unchanged.
Mastering partial derivatives involves practicing with different functions to comprehend how each variable affects the outcome.

Differentiable Functions

Differentiability in the context of multivariable calculus means a function has defined partial derivatives at all points in its domain.
If a function \(z = f(x, y)\) is differentiable, it guarantees a certain smoothness akin to having a tangent plane at each point.
This smoothness helps in approximating the function close to any given point using its linearization.

• A function being differentiable implies it can be 'tamed' nicely with calculus techniques.
• Differentiability is a stronger condition than simply having partial derivatives; it ensures no abrupt changes in any direction.

This property is crucial for applying calculus tools like the chain rule, helping in expansions like Taylor series, and in predicting behavior in small neighborhoods around a point.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of two or more variables. This involves studying changes and behaviors in higher dimensions.
Within this realm, partial derivatives, gradient vectors, and multiple integrals become vital components.
In the exercise, multivariable calculus allows us to explore how a function \(z\) behaves concerning transformations of \(x = s + t\) and \(y = s - t\).

• Concepts like curves and surfaces explore the geometric interpretation of functions of several variables.
• Integration and differentiation in multiple dimensions solve problems ranging from physics to engineering and computer graphics.

This broader perspective prepares students to tackle complex, real-world problems where multiple factors interact continuously.

Function Transformation

Function transformation in calculus involves changing variables to simplify a problem.
In the given exercise, we see how \(z=f(x, y)\) can be redefined using \(x = s + t\) and \(y = s - t\), leading to a different way to analyze the function's behavior.

• Transformations help break down complex functions into simpler components.
• They can reveal hidden properties by recasting the problem in a more informative context.

This technique is pivotal in many applications, including simplifying integrals, solving differential equations, and more abstractly, in coordinate changes across geometry and physics.
Understanding function transformations equips students with powerful strategies to simplify and solve intricate mathematical problems efficiently.