Problem 14 - Advanced Engineering Mathematics

Question

Find the points on the given surface at which the gradient is parallel to the indicated vector. $$ x^{3}+y^{2}+z=15 ; 27 \mathbf{i}+8 \mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

The points are (3, 4, -28) and (-3, 4, 26).

Step 1: Find the Gradient of the Surface

First, find the gradient of the surface defined by the function \( f(x, y, z) = x^3 + y^2 + z \). The gradient \( abla f(x, y, z) \) is given by the partial derivatives of \( f \) with respect to each variable. So we have: \[ abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (3x^2, 2y, 1) \].

Step 2: Set the Gradient Parallel to the Given Vector

We need to set the gradient \( abla f(x, y, z) = (3x^2, 2y, 1) \) parallel to the vector \( 27\mathbf{i} + 8\mathbf{j} + \mathbf{k} \). Two vectors are parallel if one is a scalar multiple of the other. Thus, equate: \( 3x^2 = 27k \), \( 2y = 8k \), and \( 1 = k \).

Step 3: Solve the System of Equations

From the third equation, we have \( k = 1 \). Substitute \( k = 1 \) into the first two equations: \( 3x^2 = 27 \times 1 \rightarrow 3x^2 = 27 \), \( 2y = 8 \times 1 \rightarrow 2y = 8 \). Solving these gives \( x^2 = 9 \Rightarrow x = \pm 3 \) and \( y = 4 \).

Step 4: Substitute and Find z-values

Substitute the values \( x = 3, -3 \) and \( y = 4 \) back into the original surface equation \( x^3 + y^2 + z = 15 \) to find the corresponding \( z \)-values. For \( x = 3 \): \[ (3)^3 + (4)^2 + z = 15 \Rightarrow 27 + 16 + z = 15 \Rightarrow z = 15 - 43 \Rightarrow z = -28 \]. For \( x = -3 \): \[ (-3)^3 + (4)^2 + z = 15 \Rightarrow -27 + 16 + z = 15 \Rightarrow z = 15 - (-11) \Rightarrow z = 26 \].

Step 5: Present the Solution Points

The points on the surface where the gradient is parallel to the vector \( 27\mathbf{i} + 8\mathbf{j} + \mathbf{k} \) are \( (3, 4, -28) \) and \( (-3, 4, 26) \).

Key Concepts

Partial Derivatives

Partial derivatives are a way to explore the change in a multi-variable function along one of its variables while keeping others constant. Think of them as a zoom-in tool for functions with several variables.
When we have a function like \( f(x, y, z) = x^3 + y^2 + z \), we can find how it changes in the \( x \), \( y \), and \( z \) directions individually using partial derivatives.

• \( \frac{\partial f}{\partial x} \) shows us the rate of change in the direction of \( x \), treating \( y \) and \( z \) as constant.
• \( \frac{\partial f}{\partial y} \) reveals how the function changes with respect to \( y \), with \( x \) and \( z \) fixed.
• \( \frac{\partial f}{\partial z} \) gives the change rate along \( z \), considering \( x \) and \( y \) as constants.

For this problem, the partial derivatives led us to the gradient: \( (3x^2, 2y, 1) \). Each part of this gradient helps understand the behavior of the surface in its respective direction.

Vector Calculus

Vector calculus is a field of mathematics that's crucial when dealing with vector fields and multidimensional functions. It combines vector algebra and calculus to allow for the study of curves, surfaces, and volume.
A central concept in vector calculus is the gradient of a function. The gradient, noted as \( abla f \), points in the direction of the greatest rate of increase of the function and its magnitude is the rate of that increase. This makes it invaluable when trying to find parallel vectors, like in our exercise.

• The gradient \( abla f \) is often represented by the vector of partial derivatives: \( \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).
• When the gradient of a surface is parallel to a vector, that means the directional slope matches that vector's direction.

This concept allows us to compare the behavior of the surface function with the vector \( 27\mathbf{i} + 8\mathbf{j} + \mathbf{k} \) and to set equations to find specific points on the surface.

Parallel Vectors

Parallel vectors have the same or exact opposite direction, and they are scalar multiples of each other. This means, given two vectors \( \mathbf{A} \) and \( \mathbf{B} \), they are parallel if \( \mathbf{A} = k \mathbf{B} \) for some scalar \( k \).
In our exercise, we had to find where the surface gradient is parallel to the vector \( 27\mathbf{i} + 8\mathbf{j} + \mathbf{k} \).

• To establish parallelism, the components of the gradient \( (3x^2, 2y, 1) \) had to satisfy: \( 3x^2 = 27k \), \( 2y = 8k \), and \( 1 = k \).
• From these equations, the value of \( k \) was found to be 1, thus simplifying the problem and allowing us to solve for \( x \) and \( y \).
• Parallel vectors are used in many applications, such as determining alignments of fields in physics and engineering.

Understanding when and how vectors can be parallel is crucial since it can drastically simplify complex problems by reducing them to more straightforward arithmetic relationships.