Problem 13 - Advanced Engineering Mathematics

Question

Find the points on the given surface at which the gradient is parallel to the indicated vector. $$ z=x^{2}+y^{2} ; 4 \mathbf{i}+\mathbf{j}+\frac{1}{2} \mathbf{k} $$

Step-by-Step Solution

Answer

The point is \((-4, -1, 17)\).

Step 1: Identify the Function and Vector

We're given the function for the surface: \( z = x^2 + y^2 \), and the vector: \( \mathbf{v} = 4\mathbf{i} + \mathbf{j} + \frac{1}{2}\mathbf{k} \). We need to find points on the surface where the gradient of the function is parallel to this vector.

Step 2: Find the Gradient of the Function

The gradient of the function \( z = x^2 + y^2 \) is given by \( abla z = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z} \right) \). Calculate each partial derivative: \[ \frac{\partial z}{\partial x} = 2x, \quad \frac{\partial z}{\partial y} = 2y, \quad \frac{\partial z}{\partial z} = -1. \]Thus, the gradient is \( abla z = (2x, 2y, -1) \).

Step 3: Set the Gradient Parallel to the Vector

For \( abla z = (2x, 2y, -1) \) to be parallel to \( \mathbf{v} = (4, 1, \frac{1}{2}) \), we need \( abla z = k\mathbf{v} \) for some scalar \( k \). This results in the equations:\[ 2x = 4k, \quad 2y = k, \quad -1 = \frac{1}{2}k. \]

Step 4: Solve for the Scalar and Coordinates

Solve \(-1 = \frac{1}{2}k \) to find \( k = -2 \). Using this in the other equations:- \( 2x = 4(-2) \Rightarrow x = -4. \)- \( 2y = -2 \Rightarrow y = -1. \)Insert \( x \) and \( y \) back into the equation for the surface \( z = x^2 + y^2 \) to find \( z = (-4)^2 + (-1)^2 = 17. \)

Step 5: Find Points on Surface

The point on the surface with the desired property is \((-4, -1, 17)\) where the gradient of the surface is parallel to the given vector.

Key Concepts

Partial Derivatives

When dealing with mathematical functions, the concept of partial derivatives plays a crucial role. Imagine a function that depends on multiple variables, such as the surface function given: \( z = x^2 + y^2 \).
Partial derivatives help us understand how this function changes when we vary only one variable while keeping the others constant. For example, the partial derivative of \( z \) with respect to \( x \) is noted as \( \frac{\partial z}{\partial x} \), which represents how \( z \) changes as \( x \) changes, while \( y \) remains fixed.
Here, the partial derivatives are:

• \( \frac{\partial z}{\partial x} = 2x \)
• \( \frac{\partial z}{\partial y} = 2y \)
• \( \frac{\partial z}{\partial z} = -1 \)

These derivatives provide the building blocks for understanding the more comprehensive concept of the gradient.

Gradient of a Function

The gradient of a function provides a way to understand the slope or rate of change at any given point on a surface. For a function like \( z = x^2 + y^2 \), its gradient \( abla z \) is a vector that consists of all the partial derivatives of the function.
In this case, the gradient is expressed as:

• \( abla z = (\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z}) = (2x, 2y, -1) \)

The elements of the gradient are the rates of change of the function with respect to each variable. This vector points in the direction of the steepest ascent on the surface, and its magnitude tells us how steep that ascent is at the point.
Understanding the gradient is essential for finding points where it is parallel to a given vector, as this involves analyzing the directions and magnitudes of these vectors.

Scalar Multiplication in Vectors

Vector parallelism can be understood through the concept of scalar multiplication. If two vectors are parallel, one is simply a scalar multiple of the other. In mathematical terms, if \( abla z \) is parallel to \( \mathbf{v} \), it means there exists a scalar \( k \) such that:

• \( abla z = k \mathbf{v} \)

For our example, we aimed to find a \( k \) for which:

• \( (2x, 2y, -1) = k(4, 1, \frac{1}{2}) \)

Solving for \( k \) involves equating corresponding elements of these vectors:

• \( 2x = 4k \)
• \( 2y = k \)
• \( -1 = \frac{1}{2}k \)

With \( k \) found as \(-2\), scalar multiplication helps us ensure precise and parallel alignment of the vectors.

Points on a Surface

Determining points on a surface where a vector condition is satisfied involves strategic calculations. With the function \( z = x^2 + y^2 \) and the relevance to the gradient \( abla z \), we then explore where this gradient becomes parallel to a specified vector.
We found \( k = -2 \), leading to solutions of the coordinates \( x \) and \( y \):

• \( x = -4 \)
• \( y = -1 \)

Finally, put these values into the surface function to find the specific \( z \) value:

• \( z = (-4)^2 + (-1)^2 = 17 \)

This provides the point \((-4, -1, 17)\) on the given surface where the gradient is parallel to the vector.
Finding such points is crucial in fields like physics and engineering, where understanding spatial relationships and vector behaviors on surfaces are often needed.