Problem 37 - Advanced Engineering Mathematics

Question

If \(f(x, y)=x^{3}-12 x+y^{2}-10 y\), find all points at which \(\|\nabla f\|=0 .\)

Step-by-Step Solution

Answer

The points are \((2, 5)\) and \((-2, 5)\).

Step 1: Find the Gradient

To find the critical points where the magnitude of the gradient is zero, first calculate the gradient of the function \(f(x, y)\). The gradient \(abla f\) in two variables is given by the vector of partial derivatives: \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). Calculate \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\): - \(\frac{\partial f}{\partial x} = \frac{d}{dx}(x^3 - 12x + y^2 - 10y) = 3x^2 - 12\) - \(\frac{\partial f}{\partial y} = \frac{d}{dy}(x^3 - 12x + y^2 - 10y) = 2y - 10\). Thus, \(abla f = (3x^2 - 12, 2y - 10)\).

Step 2: Set Gradient to Zero

To find where \(\|abla f\| = 0\), set each component of \(abla f\) to zero: - \(3x^2 - 12 = 0\) - \(2y - 10 = 0\). Solve these equations to find the critical points.

Step 3: Solve for x

Solve the equation \(3x^2 - 12 = 0\): - Divide the equation by 3, \(x^2 - 4 = 0\). - Factor as \((x - 2)(x + 2) = 0\). - The solutions are \(x = 2\) and \(x = -2\).

Step 4: Solve for y

Solve the equation \(2y - 10 = 0\): - Add 10 to both sides giving \(2y = 10\). - Divide by 2 to find \(y = 5\).

Step 5: Determine Critical Points

Combine the solutions for \(x\) and \(y\) to list out all critical points. The potential points where \(\|abla f\| = 0\) are: - For \(x = 2, y = 5\), the point is \((2, 5)\). - For \(x = -2, y = 5\), the point is \((-2, 5)\). Thus, the points at which \(\|abla f\| = 0\) are \((2, 5)\) and \((-2, 5)\).

Key Concepts

Partial Derivatives

When we talk about functions with multiple variables, partial derivatives help us understand how the function changes with respect to one variable at a time, while keeping others constant. This is crucial for exploring the behavior of these functions in a localized fashion. To compute a partial derivative, select the variable you want to differentiate with respect to, and treat all remaining variables as constants.

For example, let's consider the function given in the exercise, which is \(f(x, y) = x^3 - 12x + y^2 - 10y\).

• To find \(\frac{\partial f}{\partial x}\), differentiate \(f\) with respect to \(x\) as if \(y\) were a constant: thus, \(\frac{\partial f}{\partial x} = 3x^2 - 12\).
• To find \(\frac{\partial f}{\partial y}\), differentiate \(f\) with respect to \(y\) and treat \(x\) as a constant: therefore, \(\frac{\partial f}{\partial y} = 2y - 10\).

These partial derivatives provide the components of the gradient vector which describes the direction of the steepest ascent for the function.

Critical Points

Critical points are locations on a graph where changes occur in the behavior of the function. In the context of functions of several variables, these are the points where the gradient vector equals zero. It means that at this point, there is no slope or direction favored for increasing or decreasing in the function's value, implying potential maxima, minima, or saddle points.

To find these critical points for the function \(f(x, y) = x^3 - 12x + y^2 - 10y\), we need to examine where each component of the gradient vector equals zero.

• Set \(3x^2 - 12 = 0\) to find correct \(x\) values, as the x-component: solve to get \(x = 2\) and \(x = -2\).
• Set \(2y - 10 = 0\) to determine the correct \(y\) value, giving \(y = 5\).

Match these solutions to pinpoint the specific coordinates that are critical points. For this function, these coordinates are \((2, 5)\) and \((-2, 5)\). These points will undergo further analysis to establish if they're actually local mins, maxs, or saddles.

Magnitude of Gradient

The magnitude of the gradient, denoted as \(\|abla f\|\), indicates how steeply inclined or how flat the surface of the function is at any given point. When the magnitude is zero, the function has leveled out completely at that point, suggesting a potential peak, trough, or flat area. This concept serves as a navigational tool for identifying where optimal points or stationary points occur.

For the function \(f(x, y)\), once the components \(3x^2 - 12\) and \(2y - 10\) are zero, it means \(\|abla f\| = 0\).

• These components being zero aligns with the vector's length being zero, hence no direction for the steepest ascent is preferred at these points.
• This makes the specific coordinates \((2, 5)\) and \((-2, 5)\) critical as there is no further increase or decrease in the function's height in the neighborhood surrounding these points.

Understanding the magnitude and direction of the gradient offers a comprehensive view of the function's topology, allowing us to interpret its behavior, identify critical points, and understand the implications of these flat regions.