Problem 27 - The Calculus with Analytic Geometry

Question

Suppose \(f\) is a differentiable function of \(x\) and \(y\) and \(u=f(x, y)\). Then if \(x=\cosh v \cos w\) and \(y=\sinh v \sin w\), express \(\partial u / \partial v\) and \(\partial u / \partial w\) in terms of \(\partial u / \partial x\) and \(\partial u / \partial y\).

Step-by-Step Solution

Answer

\( \frac{\partial u}{\partial v} = \frac{\partial u}{\partial x} \cos w \sinh v + \frac{\partial u}{\partial y} \sin w \cosh v \) and \( \frac{\partial u}{\partial w} = -\frac{\partial u}{\partial x} \cosh v \sin w + \frac{\partial u}{\partial y} \sinh v \cos w \).

Step 1: Identify the relationships

Given the transformations: \(x = \cosh v \cos w\) and \(y = \sinh v \sin w\), we need to express \partial u / \partial v\ and \partial u / \partial w\ in terms of \partial u / \partial x\ and \partial u / \partial y\.

Step 2: Apply the Chain Rule

Use the chain rule for partial derivatives: \[ \frac{\partial u}{\partial v} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial v} \] and \[ \frac{\partial u}{\partial w} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial w} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial w} \].

Step 3: Compute partial derivatives of x and y with respect to v

Calculate: \[ \frac{\partial x}{\partial v} = \cosh v \cos w \frac{\partial (\cosh v)}{\partial v} = \cosh v \cos w \{\sinh v\} = \cos w \sinh v \] and \[ \frac{\partial y}{\partial v} = \sinh v \sin w \frac{\partial (\sinh v)}{\partial v} = \sinh v \sin w \{\cosh v\} = \sin w \cosh v \].

Step 4: Compute partial derivatives of x and y with respect to w

Calculate: \[ \frac{\partial x}{\partial w} = \cosh v \cos w \frac{\partial (\cos w)}{\partial w} = \cosh v \cos w \{\-(\sin w)\} = -\cosh v \sin w \] and \[ \frac{\partial y}{\partial w} = \sinh v \sin w \frac{\partial (\sin w)}{\partial w} = \sinh v \sin w \{\cos w\} = \sinh v \cos w \].

Step 5: Substitute and simplify

Combine everything into the chain rule expressions: \[ \frac{\partial u}{\partial v} = \frac{\partial u}{\partial x} \cos w \sinh v + \frac{\partial u}{\partial y} \sin w \cosh v \] and \[ \frac{\partial u}{\partial w} = -\frac{\partial u}{\partial x} \cosh v \sin w + \frac{\partial u}{\partial y} \sinh v \cos w \].

Key Concepts

Chain Rule in Partial Derivatives

To solve problems involving transformations and partial derivatives, we often use the chain rule. The chain rule helps us express the derivative of a function in terms of other variables. Consider our problem where we have transformations:

• x = cosh v cos w
• y = sinh v sin w

We want to find \(\frac{\partial u}{\partial v}\) and \(\frac{\partial u}{\partial w}\) in terms of \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\).
The chain rule tells us that: \[ \frac{\partial u}{\partial v} = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial v} \] and \[ \frac{\partial u}{\partial w} = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial w} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial w} \] Using the chain rule makes it easier to handle functions depending on multiple variables by breaking down their derivatives into simpler pieces.

Understanding Hyperbolic Functions

Hyperbolic functions, such as sinh and cosh, often appear in problems involving transformations. They are similar to the trigonometric functions sine and cosine but are defined differently:

• sinh v = \frac{e^{v} - e^{-v}}{2}
• cosh v = \frac{e^{v} + e^{-v}}{2}

In our problem, we use hyperbolic functions to transform the variables x and y. Here are their derivatives, which we need:
\[ \frac{\partial}{\partial v} (\cosh v) = \sinh v \] and \[ \frac{\partial}{\partial v} (\sinh v) = \cosh v \]
By understanding these derivatives, we can substitute and simplify in the chain rule expressions correctly. These functions help model many natural phenomena, such as the shape of a hanging cable.

Introduction to Multivariable Calculus

In multivariable calculus, we handle functions with more than one input. Functions like u = f(x,y) depend on multiple variables. This adds complexity because we need to consider partial derivatives, which represent how a function changes as we vary one variable while keeping the others constant.

For our problem, the given transformations are:

• x = cosh v cos w
• y = sinh v sin w

and we express \(\frac{\partial u}{\partial v}\) and \(\frac{\partial u}{\partial w}\) in terms of other partial derivatives: \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\).
In multivariable calculus, the chain rule generalizes to handle partial derivatives. This allows us to transform coordinates and relate complex relationships between different variables systematically. This makes solving intricate problems, like the one at hand, feasible by breaking them down into manageable steps.