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Chapter 2: Problem 7

If \(u=f(x, y)\) and \(x=r \cos \theta, y=r \sin \theta\), then show that $$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} . $$ [Hint: Use the chain rules to evaluate the derivatives on the right-hand side.]

Short Answer

Expert verified
Question: Show that the following identity holds for any differentiable function u(x, y): $$\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} $$, where r and 胃 are polar coordinates related to x and y by x = rcos胃 and y = rsin胃. Answer: We can verify the given identity by following these steps: 1. Find the partial derivatives of u with respect to x and y. 2. Replace x and y as polar coordinates (r, 胃). 3. Apply the chain rule to find the partial derivatives with respect to r and 胃, and square them. 4. Substitute the obtained expressions into the given equation and simplify. By following these steps, we arrive at the desired identity, thus verifying it: $$\left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} $$

Step by step solution

01

Find partial derivatives of u with respect to x and y

Let's find the partial derivatives of the function u(x, y) with respect to x and y: $$ \frac{\partial u}{\partial x} = \frac{\partial f(x,y)}{\partial x} $$ $$ \frac{\partial u}{\partial y} = \frac{\partial f(x,y)}{\partial y} $$
02

Replace x and y as polar coordinates (r, 胃)

Now replace x and y with their polar coordinate representations: $$ x = r \cos \theta $$ $$ y = r \sin \theta $$
03

Apply the chain rule to partial derivative with respect to r

Now we'll apply the chain rule to find the partial derivative of u with respect to r: $$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial r} $$ Using the polar coordinate representations for x and y, we can calculate their derivatives with respect to r: $$ \frac{\partial x}{\partial r} = \cos \theta $$ $$ \frac{\partial y}{\partial r} = \sin \theta $$ Substitute these into the expression for the partial derivative of u with respect to r: $$ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta $$
04

Apply the chain rule to partial derivative with respect to 胃

Now we'll apply the chain rule to find the partial derivative of u with respect to 胃: $$ \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial \theta}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial \theta} $$ Using the polar coordinate representations for x and y, we can calculate their derivatives with respect to 胃: $$ \frac{\partial x}{\partial \theta} = -r \sin \theta $$ $$ \frac{\partial y}{\partial \theta} = r \cos \theta $$ Substitute these into the expression for the partial derivative of u with respect to 胃: $$ \frac{\partial u}{\partial \theta} = -\frac{\partial u}{\partial x}r \sin \theta + \frac{\partial u}{\partial y}r \cos \theta $$
05

Substitute the expressions obtained into the given equation

Now we'll square both the partial derivatives with respect to r and 胃, and substitute them into the given equation: $$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2} = \left(\frac{\partial u}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} $$ Squaring both partial derivatives, we get: $$ \left(\frac{\partial u}{\partial r}\right)^{2} = \left(\frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta\right)^2 $$ $$ \frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} = \frac{1}{r^2}\left(-\frac{\partial u}{\partial x}r \sin \theta + \frac{\partial u}{\partial y}r \cos \theta\right)^2 $$ Plugging these expressions into the given equation, we obtain: $$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2} = \left(\frac{\partial u}{\partial x} \cos \theta + \frac{\partial u}{\partial y} \sin \theta\right)^2 + \frac{1}{r^2}\left(-\frac{\partial u}{\partial x}r \sin \theta + \frac{\partial u}{\partial y}r \cos \theta\right)^2 $$ This equation simplifies to the desired identity: $$ \left(\frac{\partial u}{\partial x}\right)^{2}+\left(\frac{\partial u}{\partial y}\right)^{2}=\left(\frac{\partial u}{\partial r}\right)^{2}+\frac{1}{r^{2}}\left(\frac{\partial u}{\partial \theta}\right)^{2} $$ This verifies the given equation using the chain rule and polar coordinate transformations.

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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 an essential tool in calculus that allows us to find the derivative of composite functions. When dealing with functions of several variables, the Chain Rule helps us differentiate a function regarding its various inputs, even when those inputs themselves are functions of other variables.
Imagine you have a function \( u = f(x, y) \) and another set of functions relating \( x \) and \( y \) to different variables like in polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). The Chain Rule acts as a bridge. It links the changes in \( u \) to changes in \( r \) and \( \theta \) by accounting for how \( u \) depends on both \( x \) and \( y \).
When applying the Chain Rule to functions of multiple variables, you need to multiply the partial derivative of the main function by the partial derivatives of its components. For example:
  • \( \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} \)
  • \( \frac{\partial u}{\partial \theta} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial \theta} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial \theta} \)

This process allows us to express derivatives in terms of \( r \) and \( \theta \), making it extremely useful for problems involving coordinate transformations.
Partial Derivatives
Partial Derivatives are a way to understand how a function changes as each individual variable changes, while keeping other variables constant. If you have a function \( u = f(x, y) \), its partial derivatives convey how \( u \) reacts to changes in \( x \) and \( y \) separately.
When working with variables in rectangular coordinates (\( x, y \)), the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \) tell us the rate of change of \( u \) with respect to \( x \) or \( y \) individually, assuming the other variable is constant.
  • \( \frac{\partial u}{\partial x} \) involves differentiating \( f \) with respect to \( x \), treating \( y \) as a constant.
  • \( \frac{\partial u}{\partial y} \) involves differentiating \( f \) with respect to \( y \), treating \( x \) as a constant.

Partial derivatives form the crux of many more complex calculus operations, particularly in understanding multidimensional systems or functions that describe surfaces in space. They鈥檙e vital in getting a grasp on the gradient, divergence, and curl used in vector calculus.
Coordinate Transformation
Coordinate Transformation is the method of changing from one coordinate system to another. It's a fundamental concept in mathematics, especially useful in scenarios where changing the point of view simplifies the problem.
For instance, converting from Cartesian coordinates \( (x, y) \) to polar coordinates \( (r, \theta) \) involves replacing \( x \) and \( y \) through the equations \[ x = r \cos \theta \] and \[ y = r \sin \theta \]. This shift is not just a different way of writing coordinates. It can significantly simplify calculations, especially those involving symmetry and angles, where polar coordinates often make the math easier to handle.
In calculus and physics, transformations like these simplify the expression of physical laws or functions, making them more tractable. Once in a new coordinate system, applying calculus techniques like the Chain Rule often becomes smoother and more insightful.
Advanced Calculus
Advanced Calculus refers to the deeper exploration of calculus concepts beyond basic differentiation and integration. It covers a vast array of topics like multi-variable calculus, vector calculus, and the study of more complex functions.
In the realm of advanced calculus, understanding how derivatives apply in higher dimensions becomes crucial. It allows tackling complex phenomena where multiple variables and their intricate relationships are involved. Concepts such as Gradients, Jacobians, and Laplacians grow from this rich groundwork.
Advanced calculus provides the mathematical framework to navigate through fields like Physics, Engineering, and Economics, where systems being studied are often multi-dimensional. This is where polar coordinates, partial derivatives, and the chain rule all interlink to solve problems efficiently. Advanced calculus brings precision to these analyses, helping model and understand the world in its full complexity.

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Most popular questions from this chapter

Evaluate the indicated partial derivatives: a) \(\left(\frac{\partial u}{\partial x}\right)_{y}\) and \(\left(\frac{\partial y}{\partial y}\right)_{x}\) if \(u=x^{2}-y^{2}, v=x-2 y\) b) \(\left(\frac{\partial x}{\partial u}\right)_{v}\) and \(\left(\frac{\partial y}{\partial v}\right)_{u}\) if \(x=e^{u} \cos v, y=e^{u} \sin v\) c) \(\left(\frac{\partial x}{\partial u}\right)_{y}\) and \(\left(\frac{\partial y}{\partial v}\right)_{u}\) if \(u=x-2 y, v=u-2 y\) d) \(\left(\frac{\partial r}{\partial x}\right)_{y}\) and \(\left(\frac{\partial r}{\partial \theta}\right)_{x}\) if \(r=\sqrt{x^{2}+y^{2}}, x=r \cos \theta\)

Let \(u_{1}=x_{1}-3 x_{2}+2 x_{1} x_{2}, u_{2}=2 x_{1}+5 x_{2}-3 x_{1} x_{2}\). Let \(\mathbf{w}=\left(w_{1}, w_{2}\right)\) be a vector function of \(\mathbf{u}=\left(u_{1}, u_{2}\right)\) such that \(\mathbf{w}_{\mathbf{u}}=\left[\begin{array}{cc}2 & 1 \\ 7 & 5\end{array}\right]\) for \(\mathbf{u}=(3,3)\). Find the Jacobian matrix at \(\mathbf{x}=(2,1)\) for the composite function \(\mathbf{w}[\mathbf{u}(\mathbf{x})]\).

. An (infinite) sequence of points \(P_{1} \ldots P_{n} \ldots\) in the plane is said to converge and have limit \(P_{0}\) : $$ \lim _{n \rightarrow \infty} P_{n}=P_{0} \quad \text { or } \quad P_{n} \rightarrow P_{0} $$ if for each \(\epsilon>0\) there is an integer \(N\) such that \(d\left(P_{n}, P_{0}\right)<\epsilon\) for \(n \geq N\). Show that the limit \(P_{0}\) is unique. [Hint: If \(P_{n} \rightarrow P_{0}\) and \(P_{n} \rightarrow P_{0}^{\prime}, P_{0}^{\prime} \neq P_{0}\), then take \(\epsilon=\frac{1}{3} d\left(P_{00} . P_{0}^{\prime}\right)\) and obtain a contradiction.]

Two functions \(f(x, y)\) and \(g(x, y)\) are such that $$ \nabla f \equiv \nabla g $$ in a domain \(D\). Show that $$ f=g+c $$ for some constant \(c\).

If \(z=e^{x} \cos y\), while \(x\) and \(y\) are implicit functions of \(t\) defined by the equations $$ x^{3}+e^{x}-t^{2}-t=1, \quad y t^{2}+y^{2} t-t+y=0 . $$ then find \(d z / d t\) for \(t=0\). [Note that \(x=0\) and \(y=0\) for \(t=0\).]
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