Question

Find the gradient of \(\phi=2 \sin y-x z\) at the point \((2, \pi / 2,-1)\). Starting at this point, in what direction is \(\phi\) decreasing most rapidly? Find the derivative of \(\phi\) in the direction, \(2 \mathrm{i}+3 \mathrm{j}\)

Short answer

Gradient is (1, 0, -2). \ Direction of maximum decrease is (-1, 0, 2). \ Derivative in direction \( 2 \mathbf{i} + 3 \mathbf{j} \) is \frac{2}{\sqrt{13}}.

Step-by-step solution

Find Partial Derivatives

To find the gradient of \(\fxHint='fx-102Max1001';\phi = 2 \sin y - x z\), we need to calculate the partial derivatives with respect to \x\, \y\, and \z\.\[ \frac{\partial\phi}{\partial x} = -z \] \[ \frac{\partial\phi}{\partial y} = 2 \cos y \] \[ \frac{\partial\phi}{\partial z} = -x \]

Calculate the Gradient at the Given Point

Evaluate the partial derivatives at the point \(2, \frac{\pi}{2}, -1\).\[ \frac{\partial\phi}{\partial x} = -(-1) = 1 \] \[ \frac{\partial\phi}{\partial y} = 2 \cos \frac{\pi}{2} = 2 \times 0 = 0 \] \[ \frac{\partial\phi}{\partial z} = -2 \]So, the gradient at \(2, \frac{\pi}{2}, -1\) is \[ \abla\phi = (1, 0, -2) \]

Determine the Direction of Maximum Decrease

The direction of maximum decrease of \phi\ is given by the negative of the gradient.Thus, \[ -\abla\phi = (-1, 0, 2) \]

Find the Unit Vector in the Given Direction

First, convert the given direction vector \(2 \mathbf{i} + 3 \mathbf{j}\) to a unit vector.The magnitude of the vector \is:\[ \sqrt{2^2 + 3^2} = \sqrt{13} \]So the unit vector \in this direction\is:\[ \mathbf{u} = \left(\frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}, 0\right) \]

Compute the Directional Derivative

The directional derivative of \phi\ in the direction \[ 2 \mathbf{i} \+ 3 \mathbf{j}\] is the dot product of the gradient and the unit vector.\[ \abla \phi \cdot \mathbf{u} = (1, 0, -2) \cdot \left( \frac{2}{\sqrt{13}},\frac{3}{\sqrt{13}}, 0\right) = \frac{2}{\sqrt{13}} \]Hence, the derivative of \phi\ in the direction \[ 2 \mathbf{i} \+ 3 \mathbf{j} \] \is: \[ \frac{2}{\sqrt{13}} \]

Key concepts

partial derivatives

Partial derivatives are the derivatives of a function with respect to one of its variables, holding the other variables constant. This concept is crucial in multivariable calculus. It helps to understand the rate at which a function changes as one particular variable varies while keeping others fixed. For example, if we have a function \( \phi = 2 \sin y - xz \), we can compute its partial derivatives with respect to \(x\), \(y\), and \(z\).
\[ \frac{\partial \phi}{\partial x} = - z \]
\[ \frac{\partial \phi}{\partial y} = 2 \cos y \]
\[ \frac{\partial \phi}{\partial z} = - x \]

Evaluating these derivatives at the point \( (2, \pi / 2, -1) \), we get:
\[ \frac{\partial \phi}{\partial x} = 1 \]
\[ \frac{\partial \phi}{\partial y} = 0 \]
\[ \frac{\partial \phi}{\partial z} = -2 \]

These expressions show how the function changes as each variable changes individually at the given point, making them essential for understanding the local behavior of multivariable functions.

gradient vector

The gradient vector of a scalar field is a vector that points in the direction of the greatest rate of increase of the function. It is composed of partial derivatives with respect to its variables. For the function \(\phi = 2 \sin y - xz\), the gradient \(abla\phi\) is given by:
\[ abla \phi = \left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) \]
For our specific function and evaluated at point \((2, \pi / 2, -1)\), the gradient is:
\[ abla \phi = (1, 0, -2) \]

This vector tells us the direction and rate of the fastest increase of \phi\ at the given point. Conversely, the negative of this vector \((-1, 0, 2)\) points in the direction of the quickest decrease of \phi\.

directional derivative

The directional derivative measures how a function changes as you move in a specific direction. It is calculated as the dot product of the gradient of the function and the unit vector in the chosen direction. Given a direction vector \(2\mathbf{i} + 3\mathbf{j}\), we first convert it to a unit vector. The magnitude of the direction vector is:
\[ \sqrt{2^2 + 3^2} = \sqrt{13} \]
The unit vector \(\mathbf{u}\) in this direction, then, is:
\[ \mathbf{u} = \left( \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) \]

The directional derivative of \phi\ in this direction is the dot product of the gradient \( (1, 0, -2) \) and \mathbf{u}\:
\[ abla \phi \cdot \mathbf{u} = \left( 1, 0, -2 \right) \cdot \left( \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) = \frac{2}{\sqrt{13}} \]
This result indicates how much \phi\ changes at the given point when moving in the direction \( 2\mathbf{i} + 3\mathbf{j} \).

unit vector

A unit vector is a vector with a magnitude of 1. It is used to indicate direction without reference to the magnitude. To convert any given vector into a unit vector, you divide the vector by its magnitude. For instance, consider the vector \(2\mathbf{i} + 3\mathbf{j}\). Its magnitude is:
\[ \sqrt{2^2 + 3^2} = \sqrt{13} \]
Therefore, the unit vector \mathbf{u}\ in this direction is:
\[ \mathbf{u} = \left( \frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) \]

This unit vector retains the direction of the original vector but scales it down to have a magnitude of 1. Unit vectors are essential in computations involving vectors, such as when finding the directional derivative, as they ensure computations are directionally accurate but not influenced by the length of the vector.