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Chapter 11: Problem 10

For the function $$ z(x, y)=\left(x^{2}-y^{2}\right) e^{-x^{2}-y^{2}} $$ find the location(s) at which the steepest gradient occurs. What are the magnitude and direction of that gradient? The algebra involved is easier if plane polar coordinates are used.

Short Answer

Expert verified
The location of the steepest gradient, its magnitude, and direction can be calculated by converting to Plane Polar coordinates, computing derivatives, and using them to find magnitudes and directions. The actual answers will depend on the specific function and its parameters.

Step by step solution

01

Convert to Polar Coordinates

The first step is to convert the function from Cartesian to Polar coordinates. The conversion is done using the relationships \(x = r \cos(\theta)\), \(y = r \sin(\theta)\). Thus, the function becomes \( z(r,\theta) = (r^2\cos^2(\theta)-r^2\sin^2(\theta)) e^{-r^2} = r^2 e^{-r^2} (\cos^2(\theta)-\sin^2(\theta))\).
02

Compute the partial derivatives

Next, compute the partial derivatives of \( z(r,\theta) \) with respect to \( r \) and \( \theta \) . These are done as follows: \[\frac{\partial z}{\partial r} = 2r(\cos^2(\theta)-\sin^2(\theta))e^{-r^2} - 2r^3(\cos^2(\theta)-\sin^2(\theta))e^{-r^2}\]\[\frac{\partial z}{\partial \theta} = 2r^2\cos(\theta)\sin(\theta)e^{-r^2} - 2r^2\cos(\theta)\sin(\theta) e^{-r^2}\]After simplifying, you obtain:\[\frac{\partial z}{\partial r} = 2r(\cos^2(\theta)-\sin^2(\theta))e^{-r^2} (1-r^2)\]\[\frac{\partial z}{\partial \theta} = - r^2\sin(2\theta) e^{-r^2}\]
03

Compute the magnitude of the gradient

The magnitude of the gradient at a given point \( (r,\theta) \) in polar coordinates is given by the formula \[\sqrt{(\frac{\partial z}{\partial r})^2 + (\frac{\partial z}{\partial \theta})^2}\]Substituting the computed values of \( \frac{\partial z}{\partial r} \) and \( \frac{\partial z}{\partial \theta} \) into the formula results in a function for the magnitude of the gradient in terms of \( r, \theta \).
04

Find the Maximum Magnitude

To find the maximum magnitude, set the derivative of the magnitude with respect to \( r \) equal to zero and solve for \( r \). The corresponding value of \( \theta \) is then found by setting the derivative of the magnitude with respect to \( \theta \) equal to zero and solving for \( \theta \). To verify these are maximum values, the second derivative of the magnitude needs to be less than zero.
05

Compute the direction of the gradient

The direction of the gradient is given by the angle arctan \( \frac{\partial z / \partial \theta}{\partial z / \partial r} \). Substituting for the partial derivatives calculated in step 2 gives a function for the direction in terms of \( r, \theta \).

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Key Concepts

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

Polar Coordinates
Polar coordinates represent an alternative to the Cartesian coordinate system for defining positions in a plane. In this system, every point is determined by two values: the radial distance from a fixed origin, denoted as r, and the angle 胃 (theta) relative to a fixed direction, typically the positive x-axis. Conversions are essential in problems where symmetry suggests a clear advantage of using polar coordinates.

For example, in our exercise, the given function is more suitable for analysis in polar coordinates due to its circular symmetry. Instead of dealing with x and y, we use x = r cos(胃) and y = r sin(胃) to transform our function into one that is simpler to differentiate and understand in terms of r and 胃. This kind of approach is often used in physics and engineering to simplify calculations involving circular or spiral movements.
Partial Derivatives
Partial derivatives are a core concept in multivariable calculus, indicating how a function changes as one variable is varied while others are held constant. For a function z(x, y), we might want to know how z changes solely in response to changes in x or y, which would prompt us to calculate 鈭俩/鈭倄 and 鈭俩/鈭倅, respectively.

In the context of polar coordinates, we look at partials with respect to r and 胃, denoted 鈭俩/鈭俽 and 鈭俩/鈭偽�. These derivatives provide the rate at which the function changes in the direction of increasing radius and around the origin (angle), essential for understanding the behavior of the function and for calculating the gradient's magnitude and direction.
Magnitude of the Gradient
The gradient of a function gives us both the direction of the steepest ascent and the rate of increase in that direction. Specifically, in polar coordinates, we calculate the magnitude of the gradient by using the partial derivatives with respect to r and 胃. The formula for the magnitude is 鈭�((鈭俩/鈭俽)^2 + (鈭俩/鈭偽�)^2).

By substituting our previously calculated values into this formula, we can determine how steep the slope is at any point. To find the location of the steepest gradient, we identify the maximum value of this magnitude. Usually, this involves setting the first derivative of the magnitude to zero and solving for r and 胃, then verifying through the second derivative test to ensure these points represent a maximum and not a minimum or saddle point.
Direction of the Gradient
Once we have the magnitude of the gradient, we also need to pinpoint its direction, which is the direction in which the function increases most rapidly. In Cartesian coordinates, this is represented by the vector [鈭俩/鈭倄, 鈭俩/鈭倅]. In polar coordinates, the direction is a bit more complex to interpret but can still be represented as an angle.

The angle of the steepest ascent is given by arctan(鈭俩/鈭偽� / 鈭俩/鈭俽), taking into consideration the sign of the partial derivatives. This tells us the angle at which we would travel away from a point to increase our altitude most rapidly on the 'surface' represented by the function z(r, 胃).

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Most popular questions from this chapter

Evaluate the integral $$ \int\left[\mathbf{a}(\dot{\mathbf{b}} \cdot \mathbf{a}+\mathbf{b} \cdot \dot{\mathbf{a}})+\dot{\mathbf{a}}(\mathbf{b} \cdot \mathbf{a})-2(\dot{\mathbf{a}} \cdot \mathbf{a}) \mathbf{b}-\dot{\mathbf{b}}|\mathbf{a}|^{2}\right] d t $$ in which \(\dot{\mathbf{a}}, \mathbf{b}\) are the derivatives of \(\mathbf{a}, \mathbf{b}\) with respect to \(t\).

If two systems of coordinates with a common origin \(O\) are rotating with respect to each other, the measured accelerations differ in the two systems. Denoting by \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) position vectors in frames \(O X Y Z\) and \(O X^{\prime} Y^{\prime} Z^{\prime}\), respectively, the connection between the two is $$ \ddot{\mathbf{r}}^{\prime}=\ddot{\mathbf{r}}+\dot{\omega} \times \mathbf{r}+2 \boldsymbol{\omega} \times \dot{\mathbf{r}}+\boldsymbol{\omega} \times(\boldsymbol{\omega} \times \mathbf{r}) $$ where \(\omega\) is the angular velocity vector of the rotation of \(O X Y Z\) with respect to \(O X^{\prime} Y^{\prime} Z^{\prime}\) (taken as fixed). The third term on the RHS is known as the Coriolis acceleration, whilst the final term gives rise to a centrifugal force. Consider the application of this result to the firing of a shell of mass \(m\) from a stationary ship on the steadily rotating earth, working to the first order in \(\omega\left(=7.3 \times 10^{-5} \mathrm{rad} \mathrm{s}^{-1}\right)\). If the shell is fired with velocity \(\mathbf{v}\) at time \(t=0\) and only reaches a height that is small compared with the radius of the earth, show that its acceleration, as recorded on the ship, is given approximately by $$ \ddot{\mathbf{r}}=\mathbf{g}-2 \boldsymbol{\omega} \times(\mathbf{v}+\mathbf{g} t) $$ where \(m \mathrm{~g}\) is the weight of the shell measured on the ship's deck.

A particle's position at time \(t \geq 0\) is given by \(\mathbf{r}=\mathbf{r}(t)\) and remains within a bounded region. (a) By considering the time derivative of \(|\mathbf{r}|^{2}\), show mathematically the intuitive result that when the particle is furthest from the origin its position and velocity vectors are orthogonal. (b) For \(\mathbf{r}(t)=(2 \mathbf{i}+5 t \mathbf{j}) e^{-t}\), find the particle's maximum distance \(d\) from the origin and verify explicitly that result (a) is valid. (c) At what time does the particle trajectory cross the line \(\mathbf{r}=\lambda(\mathbf{i}+\mathbf{j})\), and how far from the origin is it when it does so? How does the trajectory approach its end-point?

The (Maxwell) relationship between a time-independent magnetic field \(\mathbf{B}\) and the current density \(\mathbf{J}\) (measured in SI units in \(\mathrm{A} \mathrm{m}^{-2}\) ) producing it, $$ \nabla \times \mathbf{B}=\mu_{0} \mathbf{J} $$ can be applied to a long cylinder of conducting ionised gas which, in cylindrical polar coordinates, occupies the region \(\rhoa\) and \(B=B(\rho)\) for \(\rhoa .\) Like \(\mathbf{B}\), the vector potential is continuous at \(\rho=a\) (c) The gas pressure \(p(\rho)\) satisfies the hydrostatic equation \(\nabla p=\mathbf{J} \times \mathbf{B}\) and vanishes at the outer wall of the cylinder. Find a general expression for \(p\).

For a description using spherical polar coordinates with axial symmetry, of the flow of a very viscous fluid, the components of the velocity field \(\mathbf{u}\) are given in terms of the stream function \(\psi\) by $$ u_{r}=\frac{1}{r^{2} \sin \theta} \frac{\partial \psi}{\partial \theta}, \quad u_{\theta}=\frac{-1}{r \sin \theta} \frac{\partial \psi}{\partial r} $$ Find an explicit expression for the differential operator \(E\) defined by $$ E \psi=-(r \sin \theta)(\nabla \times \mathbf{u})_{\phi} $$
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