Question

Find the gradient of \(\phi=z \sin y-x z\) 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 \(2 \mathbf{i}+3 \mathbf{j}\).

Short answer

The gradient at the point is (1, 0, -1). \(\phi\) decreases most rapidly in the direction (-1, 0, 1). The directional derivative along 2\mathbf{i}+3\mathbf{j} is \(\frac{2}{\sqrt{13}}\).

Step-by-step solution

- Define the function

Given the function \(\phi = z \sin y - xz\).This function will be used to find the gradient and evaluate it at a specific point.

- Compute partial derivatives

Find the partial derivatives with respect to each variable: 1. \(\frac{\partial \phi}{\partial x} = -z \) 2. \(\frac{\partial \phi}{\partial y} = z \cos y \) 3. \(\frac{\partial \phi}{\partial z} = \sin y - x \)

- Evaluate partial derivatives at point (2, \pi / 2, -1)

Substitute the values into the partial derivatives calculated in Step 2: 1. \(\frac{\partial \phi}{\partial x} \) at (2, \(\pi / 2\), -1) is -(-1) = 1 2. \(\frac{\partial \phi}{\partial y} \) at (2, \(\pi / 2\), -1) is (-1)\cos(\(\pi / 2\)) = 0 3. \(\frac{\partial \phi}{\partial z} \) at (2, \(\pi / 2\), -1) is \(\sin(\pi / 2\) - 2 = 1 - 2 = -1

- Write the gradient vector

The gradient at the point (2,\( \pi / 2\), -1) is the vector of the partial derivatives:\(abla \phi(2, \pi / 2, -1) = (1, 0, -1) \)

- Identify the direction of maximum decrease

The direction in which \(\phi\) decreases most rapidly is opposite to the gradient vector. Hence, the direction of maximum decrease is \(-(1, 0, -1) = (-1, 0, 1) \)

- Find the unit vector in direction 2\mathbf{i}+3\mathbf{j}

The given direction vector is 2\mathbf{i}+3\mathbf{j}. First, convert it to a unit vector:\(\mathbf{d} = \frac{1}{\sqrt{2^2+3^2}}(2\mathbf{i} + 3\mathbf{j}) = \frac{1}{\sqrt{13}}(2, 3, 0) \)

- Compute directional derivative

The directional derivative is the dot product of the gradient and the unit vector.\(\frac{\partial \phi}{\partial \mathbf{d}} = abla \phi \cdot \mathbf{d} = (1, 0, -1) \cdot \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) \)\( = \frac{1}{\sqrt{13}}(2 \times 1 + 3 \times 0 + (-1) \times 0) = \frac{2}{\sqrt{13}} \)

Key concepts

partial derivatives

To understand gradient in vector calculus, it's essential to get familiar with the concept of partial derivatives. Imagine you have a function that depends on multiple variables. In our case, we have \( \phi = z \sin y - xz \). Here, \( \phi \) depends on the variables \( x \), \( y \), and \( z \). When taking a partial derivative, we focus on how the function changes as we vary just one of these variables while holding the others constant.
So:

• For \( \frac{\partial \phi}{\partial x} \), we treat \( y \) and \( z \) as constants and only differentiate with respect to \( x \). This results in \( \frac{\partial \phi}{\partial x} = -z \).
  
• For \( \frac{\partial \phi}{\partial y} \), we treat \( x \) and \( z \) as constants. We get \( \frac{\partial \phi}{\partial y} = z \cos y \).
  
• For \( \frac{\partial \phi}{\partial z} \), we differentiate with respect to \( z \) while keeping \( x \) and \( y \) constant. This gives us \( \frac{\partial \phi}{\partial z} = \sin y - x \).
  

By calculating these partial derivatives, we now know how each variable influences the value of our function \( \phi \). This is crucial for finding the gradient, an essential concept in vector calculus.

directional derivative

Now, let's dive into the directional derivative. Imagine you are hiking up a hill. The directional derivative tells you how steep the hill is in the direction you are walking. To calculate this, we need two things: the gradient of the function and a unit vector in the desired direction.
In our exercise, the gradient was found to be \( \mathbf{abla \phi} (2, \pi / 2, -1) = (1, 0, -1) \). Think of the gradient as an arrow that points in the direction of the steepest ascent.
Next, we need to express our given direction \((2 \mathbf{i} + 3 \mathbf{j})\) as a unit vector—a vector with a length of one. We do this by dividing each component of the vector by the vector's length. The length, or magnitude \( \| \mathbf{d} \| \), is calculated as follows:
\ \ \| \mathbf{d} \| = \sqrt{2^2 + 3^2} = \sqrt{13} \
So, our unit vector becomes:\( \mathbf{d} = \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) \). The directional derivative is then the dot product of the gradient and this unit vector.
\ \ \frac{\partial \phi}{\partial \mathbf{d}} = \mathbf{abla \phi} \cdot \mathbf{d} = (1, 0, -1) \cdot \left(\frac{2}{\sqrt{13}}, \frac{3}{\sqrt{13}}, 0 \right) = \frac{2}{\sqrt{13}} \
This calculation reveals the rate at which \( \phi \) changes in the specified direction.

vector calculus

Understanding gradients, partial derivatives, and directional derivatives is all part of the broader field known as vector calculus. This field studies how functions change and how we can analyze these changes using vectors.
When dealing with functions of multiple variables, vectors become an invaluable tool for describing rates of change in various directions. The gradient vector \( \mathbf{abla \phi} \) gives the direction of steepest ascent and contains the partial derivatives of the function. The directional derivative then uses this gradient to find the rate of change in any specified direction.
Furthermore, vector calculus assists in solving real-world problems in physics, engineering, economics, and more. It helps us understand how fields such as temperature, velocity, or pressure change in space, leading to more informed decisions and better designs.
By mastering concepts like gradients, partial derivatives, and directional derivatives, you unlock a powerful toolkit for exploring and understanding the multidimensional world around us.