Problem 27 - Advanced Engineering Mathematics

Question

Find a vector that gives the direction in which the given function decreases most rapidly at the indicated point. Find the minimum rate. $$ f(x, y)=\tan \left(x^{2}+y^{2}\right) ;(\sqrt{\pi / 6}, \sqrt{\pi / 6}) $$

Step-by-Step Solution

Answer

The vector is \( \left( -\frac{4\sqrt{\pi}}{3\sqrt{6}}, -\frac{4\sqrt{\pi}}{3\sqrt{6}} \right) \) with a rate of \( \frac{4\sqrt{2\pi}}{3\sqrt{3}} \).

Step 1: Compute the Gradient of the Function

The gradient of a function \(f(x, y)\) is the vector of partial derivatives \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). For \(f(x, y) = \tan(x^2 + y^2)\), we find:\[\frac{\partial f}{\partial x} = \sec^2(x^2 + y^2) \cdot 2x\]\[\frac{\partial f}{\partial y} = \sec^2(x^2 + y^2) \cdot 2y\]So, the gradient is:\[abla f = \left( 2x \sec^2(x^2 + y^2), 2y \sec^2(x^2 + y^2) \right)\]

Step 2: Evaluate the Gradient at the Given Point

Substitute \( (x, y) = (\sqrt{\pi/6}, \sqrt{\pi/6}) \) into \( abla f\):\[abla f\bigg|_{(x, y) = (\sqrt{\pi/6}, \sqrt{\pi/6})} = \left( 2\sqrt{\pi/6} \sec^2(\pi/3), 2\sqrt{\pi/6} \sec^2(\pi/3) \right)\]Since \( \sec^2(\pi/3) = 4/3 \), the gradient becomes:\[abla f = \left( \frac{4\sqrt{\pi}}{3\sqrt{6}}, \frac{4\sqrt{\pi}}{3\sqrt{6}} \right)\]

Step 3: Determine the Direction of Steepest Descent

The direction of the steepest descent is the negative gradient vector. Therefore, the vector pointing in the direction of the steepest decrease is:\[\mathbf{v} = -abla f = \left( -\frac{4\sqrt{\pi}}{3\sqrt{6}}, -\frac{4\sqrt{\pi}}{3\sqrt{6}} \right)\]

Step 4: Calculate the Minimum Rate of Decrease

The rate of most rapid decrease is the magnitude of the negative gradient vector. Calculate the magnitude:\[\|\mathbf{v}\| = \left\| -abla f \right\| = \sqrt{\left(-\frac{4\sqrt{\pi}}{3\sqrt{6}}\right)^2 + \left(-\frac{4\sqrt{\pi}}{3\sqrt{6}}\right)^2}\]Simplifying, we get:\[\|\mathbf{v}\| = \frac{8\sqrt{\pi}}{3\sqrt{6}} \times \frac{1}{\sqrt{2}} = \frac{4\sqrt{2\pi}}{3\sqrt{3}}\]

Step 5: Conclusion: State the Vector and Minimum Rate

The vector giving the direction of steepest descent is \( \left( -\frac{4\sqrt{\pi}}{3\sqrt{6}}, -\frac{4\sqrt{\pi}}{3\sqrt{6}} \right) \) and the minimum rate of decrease is \( \frac{4\sqrt{2\pi}}{3\sqrt{3}} \).

Key Concepts

Understanding Partial Derivatives

Partial derivatives are like specialized derivatives that focus on how a function changes with respect to one variable, while keeping the others constant. If you have a multivariable function like \( f(x, y) = \tan(x^2 + y^2) \), taking partial derivatives helps you to see its behavior along each axis.

Here's how it works:

• \( \frac{\partial f}{\partial x} \) means we see how the function changes as \( x \) changes, ignoring \( y \).
• Similarly, \( \frac{\partial f}{\partial y} \) examines changes in \( y \) ignoring \( x \).

In our example, we calculate both of these to find the gradient. For \( f(x, y) = \tan(x^2 + y^2) \):

• \( \frac{\partial f}{\partial x} = \sec^2(x^2 + y^2) \cdot 2x \)
• \( \frac{\partial f}{\partial y} = \sec^2(x^2 + y^2) \cdot 2y \)

The partial derivatives reveal how sharply \( f \) increases or decreases along each direction (\( x \) and \( y \)), a crucial step toward understanding the gradient—and subsequently how the function behaves in multidimensional space.

The Concept of Steepest Descent

Steepest descent is all about finding the fastest way down a hill—mathematically speaking. In optimization, the direction of steepest descent is guided by the negative of the gradient vector. Why negative? Because it points us downhill, where the function decreases the fastest.

Let's break this down:

• The gradient vector \( abla f \) points towards the steepest increase of a function.
• Negating it, or \( -abla f \), points downward where the function drops steeply.

For the function \( f(x, y) = \tan(x^2 + y^2) \) evaluated at \( (\sqrt{\pi / 6}, \sqrt{\pi / 6}) \), the gradient is \( \left( \frac{4\sqrt{\pi}}{3\sqrt{6}}, \frac{4\sqrt{\pi}}{3\sqrt{6}} \right) \). The steepest descent vector is then:

• \( \mathbf{v} = -abla f = \left( -\frac{4\sqrt{\pi}}{3\sqrt{6}}, -\frac{4\sqrt{\pi}}{3\sqrt{6}} \right) \)

This shows how to navigate directly to where the slope allows for the quickest reduction in \( f \). In practical terms, you follow this vector in iterative optimization algorithms to find local minima efficiently.

Understanding Minimum Rate of Decrease

The minimum rate of decrease represents how fast a function can decrease if we proceed in the direction of steepest descent, using the gradient. It is the magnitude of the negative gradient, and here's why it's important.

In practical terms:

• The steeper the slope, the faster you can reach the lowest point or the local minimum.
• To calculate this rate, we measure the length or magnitude of the negative gradient vector.

In our function example, the magnitude of the steepest descent vector \( \mathbf{v} \) is calculated as follows:

• Calculate the magnitude: \( \|\mathbf{v}\| = \sqrt{\left(-\frac{4\sqrt{\pi}}{3\sqrt{6}}\right)^2 + \left(-\frac{4\sqrt{\pi}}{3\sqrt{6}}\right)^2} \)
• The result simplifies to \( \frac{4\sqrt{2\pi}}{3\sqrt{3}} \).

This value, \( \frac{4\sqrt{2\pi}}{3\sqrt{3}} \), communicates just how quickly the function's value will drop as you follow that path of steepest descent per unit of distance you travel.