Problem 7 If \(z=f(x, y), x=\rho \cos \the... [FREE SOLUTION] (2024)

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Chapter 22: Problem 7

If \(z=f(x, y), x=\rho \cos \theta\), and \(y=\rho \sin \theta\), show that for\(z=F(\rho, \theta)\) $$ z_{\rho}^{2}+\frac{1}{\rho^{2}} z_{\theta}^{2}=z_{x}^{2}+z_{y}^{2} $$

Short Answer

Expert verified

Using partial derivatives and the chain rule, expressions for \(z_{\rho}\) and \(z_{\theta}\) were found in terms of \(z_x\) and \(z_y\), and after applying trigonometric identities, the desired identity \(z_{\rho}^{2} + \frac{1}{\rho^{2}} z_{\theta}^{2} = z_{x}^{2} + z_{y}^{2}\) was established.

Step by step solution

01

Calculate Partial Derivatives with respect to Cartesian Coordinates

Given that \(z = f(x, y)\), we first find the partial derivatives of \(z\) with respect to \(x\) and \(y\), denoted as \(z_x\) and \(z_y\) respectively. We will use these partial derivatives to express the left side of the equation in terms of Cartesian coordinates. The derivatives are given by:\[ z_x = \frac{\partial z}{\partial x} \]and\[ z_y = \frac{\partial z}{\partial y} \].

02

Express the Partial Derivatives in Polar Coordinates

To relate the Cartesian coordinates to the polar coordinates, the chain rule must be applied using \(x = \rho \cos \theta\) and \(y = \rho \sin \theta\). The chain rule gives us the following:\[ z_{\rho} = z_x \frac{\partial x}{\partial \rho} + z_y \frac{\partial y}{\partial \rho} \]\[ z_{\theta} = z_x \frac{\partial x}{\partial \theta} + z_y \frac{\partial y}{\partial \theta} \].The partial derivatives of \(x\) and \(y\) with respect to \(\rho\) and \(\theta\) can be computed as follows:\[ \frac{\partial x}{\partial \rho} = \cos \theta \]\[ \frac{\partial y}{\partial \rho} = \sin \theta \]\[ \frac{\partial x}{\partial \theta} = -\rho \sin \theta \]\[ \frac{\partial y}{\partial \theta} = \rho \cos \theta \].

03

Substitute and Simplify the Expressions

Substituting the derived partial derivatives of \(x\) and \(y\) with respect to \(\rho\) and \(\theta\) into the equations for \(z_{\rho}\) and \(z_{\theta}\), we get:\[ z_{\rho} = z_x \cos \theta + z_y \sin \theta \]\[ z_{\theta} = -z_x \rho \sin \theta + z_y \rho \cos \theta \].Now square both \(z_{\rho}\) and \(z_{\theta}\) to get:\[ z_{\rho}^2 = (z_x \cos \theta + z_y \sin \theta)^2 \]and\[ z_{\theta}^2 = (-z_x \rho \sin \theta + z_y \rho \cos \theta)^2 \].Applying the trigonometric identities and distributing the squares will allow us to relate these expressions to \(z_x^2\) and \(z_y^2\).

04

Apply Trigonometric Identities and Expand the Squares

By expanding the squares obtained in Step 3, and using trigonometric identities, we get:\[ z_{\rho}^2 = z_x^2 \cos^2 \theta + 2z_x z_y \cos \theta \sin \theta + z_y^2 \sin^2 \theta \]and\[ z_{\theta}^2 = z_x^2 \rho^2 \sin^2 \theta - 2z_x z_y \rho^2 \sin \theta \cos \theta + z_y^2 \rho^2 \cos^2 \theta \].Summing both equations and factoring out the common terms \(z_x^2\) and \(z_y^2\) yields:\[ z_{\rho}^2 + \frac{1}{\rho^2} z_{\theta}^2 = z_x^2 ( \cos^2 \theta + \sin^2 \theta) + z_y^2 (\sin^2 \theta + \cos^2 \theta) \].

05

Apply the Pythagorean Identity to Simplify

Utilize the Pythagorean identity \(\cos^2 \theta + \sin^2 \theta = 1\) to simplify the equation further:\[ z_{\rho}^2 + \frac{1}{\rho^2} z_{\theta}^2 = z_x^2 \cdot 1 + z_y^2 \cdot 1 \]\[ z_{\rho}^2 + \frac{1}{\rho^2} z_{\theta}^2 = z_x^2 + z_y^2 \].This proves the required identity.

Key Concepts

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

Chain Rule Calculus

The chain rule in calculus is a fundamental theorem used to take derivatives of composite functions. When we have a function that is a combination of two or more functions, the chain rule enables us to differentiate it with respect to one of its variables. In the context of changing between coordinate systems, like Cartesian and polar coordinates, the chain rule becomes particularly useful.

For example, if we have a function in terms of polar coordinates, say, \( F(\rho, \theta) \), and we wish to find how \( F \) changes with respect to the variables \( x \) and \( y \) of Cartesian coordinates, we would apply the chain rule. It allows us to link the rate of change of \( F \) with respect to \( x \) and \( y \) to its rate of change with respect to \( \rho \) and \( \theta \).

By using the relationships between the coordinates, \( x = \rho \cos \theta \) and \( y = \rho \sin \theta \), we can express the partial derivatives with respect to \( \rho \) and \( \theta \) in terms of \( x \) and \( y \). This is done by differentiating the function firstly with respect to \( x \) and \( y \), and then considering the derivatives of \( x \) and \( y \) with respect to \( \rho \) and \( \theta \).

The application of the chain rule in polar coordinates, therefore, links the partial derivatives of a function in polar coordinates to its partial derivatives in Cartesian coordinates, bridging the gap between the two systems and facilitating the process of transitioning from one to another when solving problems in multi-variable calculus.

Trigonometric Identities

Trigonometric identities are equalities that involve trigonometric functions and are true for every value where both sides of the equality are defined. These identities are very useful when simplifying expressions and solving equations involving trigonometric functions.

Key Trigonometric Identities

Some of the most commonly used trigonometric identities include the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \), which was used in the step-by-step solution to the exercise. This identity comes from the Pythagorean theorem and the unit circle definition of sine and cosine functions.
Another set of useful identities is the angle sum and difference identities, such as \( \sin(\alpha \pm \beta) = \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin(\beta) \) and \( \cos(\alpha \pm \beta) = \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta) \).

When dealing with polar coordinates, these identities allow us to connect the trigonometric functions of \( \theta \) with the Cartesian coordinates, leading to simplifications that can be vital for solving calculus problems. Understanding and utilizing these identities are crucial for students working with trigonometric functions and is especially important in the study of calculus, where they frequently appear in integration and differentiation problems.

Polar to Cartesian Coordinate Transformation

Transformation between coordinate systems is a common task in mathematics, physics, and engineering. The polar to Cartesian coordinate transformation is one such example, wherein we switch from the polar coordinate system (\( \rho, \theta \)) to the Cartesian coordinate system (\( x, y \)). This process often utilizes the relationships \( x = \rho \cos \theta \) and \( y = \rho \sin \theta \), derived from the definition of polar coordinates.

Converting Between Systems

The conversion formulas instantiate how a point's position is viewed in different systems. In the polar system, a point's location is identified by the distance from the origin, \( \rho \), and the angle, \( \theta \), from the positive x-axis. In the Cartesian system, the point's position is given by its horizontal (x) and vertical (y) distances from the origin.

These transformations are not only critical for changing the coordinate representation of points but also play a significant role in problems involving calculus, as seen in the exercise provided. The ability to convert between polar and Cartesian coordinates allows one to select the most convenient framework for tackling a problem, which can lead to simpler computations or more intuitive solutions. For calculus students, mastering these transformations is essential when working with functions represented in different coordinates and when integrating or differentiating in multiple dimensions.

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Problem 7 If \(z=f(x, y), x=\rho \cos \the... [FREE SOLUTION] (3)

Most popular questions from this chapter

Show that the line joining the fixed point \((a, b, c)\) to the point on thesurface \(\Phi(x, y, z)=0\) for which the distance is least is normal to \(\Phi=\)0 .If \(z=\log \frac{\sqrt{(x-1)^{2}+y^{2}}}{\sqrt{(x+1)^{2}+y^{2}}}\), show that\(\frac{\partial^{2} z}{\partial x^{2}}+\frac{\partial^{2} z}{\partialy^{2}}=0\).Suppose that \(u=f(x, y, z)\) while \(x=g(r, s, t), y=h(r, s, t)\), and \(z=k(r\),\(s, t)\). Write the chain rule for \(\partial f / \partial r\). Ans. \(\partial f/ \partial r=f_{x} g_{r}+f_{y} h_{r}+f_{z} k_{r}\).What is the domain of \(x\) and \(y\) for which the function\(z=\sqrt{1-x^{2}-y^{2}}\) has real values?Show that the tangent plane to the surface of \(z=f(x, y)\) at a relativemaximum must be parallel to the \(x y\)-plane.
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Problem 7 If \(z=f(x, y), x=\rho \cos \the... [FREE SOLUTION] (2024)

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