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If \(x=e^{u} \cos \theta\) and \(y=e^{u} \sin \theta\), show that $$ \frac{\partial^{2} \phi}{\partial u^{2}}+\frac{\partial^{2} \phi}{\partial \theta^{2}}=\left(x^{2}+y^{2}\right)\left(\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}\right) $$ where \(f(x, y)=\phi(u, \theta)\).

Short Answer

Expert verified
By calculating partial derivatives, performing transformations, and applying the chain rule, the given equation is indeed validated.

Step by step solution

01

Determine derivatives of x and y

Firstly, find expressions for the derivatives \(\frac {\partial x} {\partial u}\), \(\frac {\partial x} {\partial \theta}\), \(\frac {\partial y} {\partial u}\), and \(\frac {\partial y} {\partial \theta}\). Using the given \(x=e^{u} \cos \theta\) and \(y=e^{u} \sin \theta\), we can differentiate them with respect to \(u\) and \(\theta\) to find the expressions needed.
02

Calculate second derivatives

Next, calculate the second derivatives \(\frac {\partial^2 x} {\partial u^2}\), \(\frac {\partial^2 y} {\partial u^2}\), \(\frac {\partial^2 x} {\partial \theta^2}\), and \(\frac {\partial^2 y} {\partial \theta^2}\) using the expressions found in step 1.
03

Transform the left side of the equation

Knowing \(f(x, y)=\phi(u, \theta)\) implies that \(\frac {\partial^2 \phi} {\partial u^2}\) and \(\frac {\partial^2 \phi} {\partial \theta^2}\) transform to \(x^2\frac {\partial^2 f} {\partial x^2} + y^2\frac {\partial^2 f} {\partial y^2}\) when changing from \(u, \theta\) coordinates to \(x, y\) coordinates. To confirm this transformation, employ the chain rule, where the derivative of a composite function can be expressed as the product of derivatives. Calculate \(\frac {\partial^2 \phi} {\partial u^2}\) and \(\frac {\partial^2 \phi} {\partial \theta^2}\) using the chain rule and the derivatives of \(x\) and \(y\) from step 1.
04

Final Comparison

After calculating \(\frac {\partial^2 \phi} {\partial u^2}\) and \(\frac {\partial^2 \phi} {\partial \theta^2}\) and transforming to \(x, y\) coordinates, compare the result to the right side of the equation. If they match, it means the initial expression in the exercise is correctly validated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is a fundamental formula in calculus often utilized when taking the derivative of composite functions. In simpler terms, it helps us differentiate a function that is nested within another function. The beauty of the chain rule lies in its ability to break down complicated derivative problems into more manageable parts.

Imagine you have to find the derivative of a multivariable function, say \( \(f(g(x))\) \) which involves another function \(\text{g}(x)\) inside \(\text{f}(x)\). The chain rule states that the derivative of \(\text{f}(g(x))\) with respect to \(\text{x}\) is the derivative of \(\text{f}\) with respect to \(\text{g}\) times the derivative of \(\text{g}\) with respect to \(\text{x}\): \[ \frac{d}{dx}\text{f}(g(x)) = \frac{df}{dg} \times \frac{dg}{dx} \].

When applied to partial derivatives, the concept extends similarly. If we have \( \(f(x, y)\) \) where \( \(x\) \) and \( \(y\) \) are functions of other variables, say \( \(u\) \) and \( \(\theta\) \) as seen in our original problem, the chain rule can be used to differentiate \( \(f\) \) with respect to \( \(u\) \) and \( \(\theta\)\), allowing us to relate the changes in \( \(f\) \) directly to the changes in \( \(u\) \) and \( \(\theta\)\).
Second Derivative
The second derivative of a function plays a crucial role in understanding the curvature of its graph, also known as concavity. It provides insights into the 'rate of change of the rate of change', which can elucidate the behavior of functions that model acceleration, force, or other dynamic systems.

Mathematically, if you have a function \( \(f(x)\) \) whose first derivative is \( \(f'(x)\)\), then the second derivative is denoted as \( \(f''(x)\) \) or \( \frac{d^{2}}{dx^{2}}f(x)\) and represents the derivative of \( \(f'(x)\)\). It answers the question, 'How is the slope of the tangent line changing as x changes?'. A positive second derivative implies the function is concave up (the graph is curving upwards), whereas a negative second derivative indicates concavity down (the graph is curving downwards).

In the context of partial derivatives for functions of several variables, the second derivative gives information about the curvature of the surface in different directions. The exercise examines these second partial derivatives and relates transformations of these derivatives when shifting from one coordinate system to another, which leads us into our next key concept, coordinate transformation.
Coordinate Transformation
Coordinate transformation is a cornerstone concept in calculus, especially when studying multivariable functions. It provides us with a method to switch between different coordinate systems, such as from Cartesian \( \(x, y\)\) coordinates to polar coordinates \( \(r, \theta\)\) or, as in our exercise, from \( \(x, y\)\) to \( \(u, \theta\)\) coordinates.

Such transformations are valuable because certain problems become easier to handle within specific coordinate systems. For instance, circular or spherical problems that may seem complicated in Cartesian coordinates often simplify dramatically in polar or spherical coordinates.

Application of Coordinate Transformation

Let's tie it back to the original exercise. We have transformed the coordinates from \( \(x, y\)\) to \( \(u, \theta\)\) using the definitions \( \(x=e^{u} \cos \theta\)\) and \( \(y=e^{u} \text{sin} \theta\)\). By applying the transformation, we're not just shuffling symbols around; we're actually providing a new perspective to approach the function \( \(\phi\) \) and its derivatives.

The exercise requires us to apply our understanding of both the chain rule and second derivatives within the new coordinate system. This way, we are able to express the second derivatives with respect to \( \(u\)\) and \( \(\theta\)\) in terms of \( \(x\)\) and \( \(y\)\), showing that the complex relationships between changes in the variables hold true even when the 'stage' on which they act is transformed.

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Most popular questions from this chapter

Show that $$ d f=y\left(1+x-x^{2}\right) d x+x(x+1) d y $$ is not an exact differential. Find the differential equation that a function \(g(x)\) must satisfy if \(d \phi=g(x) d f\) is to be an exact differential. Verify that \(g(x)=e^{-x}\) is a solution of this equation, and deduce the form of \(\phi(x, y)\).

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