Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In what direction \(\mathbf{u}\) does \(f(x, y)=\sin (3 x-y)\) decrease most rapidly at \(\mathbf{p}=(\pi / 6, \pi / 4) ?\)

Step-by-Step Solution

Answer

The function decreases most rapidly in the direction \( \u = \frac{1}{\sqrt{5}} \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \).

Step 1: Find the Gradient of the Function

To determine the direction of steepest descent, we first find the gradient of the function \(f(x,y) = \sin(3x-y)\). The gradient \( abla f(x,y) \) is given by the partial derivatives with respect to \(x\) and \(y\). \[ abla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (3\cos(3x-y), -\cos(3x-y))\]

Step 2: Evaluate the Gradient at Point \(\mathbf{p}\)

Next, we evaluate this gradient at point \(\mathbf{p} = \left( \frac{\pi}{6}, \frac{\pi}{4} \right) \). Substituting \(x = \frac{\pi}{6}\) and \(y = \frac{\pi}{4}\) into the gradient gives: \[ abla f\left( \frac{\pi}{6}, \frac{\pi}{4} \right) = \left( 3\cos\left(\frac{3\pi}{6} - \frac{\pi}{4}\right), -\cos\left(\frac{3\pi}{6} - \frac{\pi}{4}\right)\right) \] \[ \cos\left(\frac{\pi}{2} - \frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \] Thus, \[ abla f\left( \frac{\pi}{6}, \frac{\pi}{4} \right) = \left( \frac{3\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) \]

Step 3: Determine the Direction of Steepest Descent

The direction in which the function decreases most rapidly is the negative of the gradient vector at the point. Therefore, \[ -abla f\left( \frac{\pi}{6}, \frac{\pi}{4} \right) = \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \]This is the vector that points in the direction of the steepest descent.

Step 4: Normalize the Direction Vector

To get the unit direction vector \(\mathbf{u}\), we must normalize the steepest descent vector:\[ \mathbf{u} = \frac{1}{\sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2}} \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \]Calculate the magnitude first:\[ \sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{\frac{9 \cdot 2}{4} + \frac{2}{4}} = \sqrt{\frac{20}{4}} = \sqrt{5} \]Thus the unit vector is:\[ \mathbf{u} = \frac{1}{\sqrt{5}} \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \]

Key Concepts

Gradient Vector

The concept of a gradient vector is central in determining the direction of steepest descent in multivariable calculus. A gradient vector of a function, denoted by \( abla f \), is a vector composed of its partial derivatives. It points in the direction of the steepest ascent of the function. For a function \( f(x, y) \), the gradient is defined as:- \( \frac{\partial f}{\partial x} \) representing how \( f \) changes as \( x \) changes.- \( \frac{\partial f}{\partial y} \) representing how \( f \) changes as \( y \) changes.For the function given in our problem, \( f(x, y) = \sin(3x-y) \), the gradient vector is \( abla f(x, y) = (3\cos(3x-y), -\cos(3x-y)) \). This vector is critical as it tells us not only the rate of change but the direction of maximum increase of the function. To find the steepest descent, which is just the opposite direction, you would use the negative of the gradient vector.

Partial Derivatives

Understanding partial derivatives is crucial when working with multivariable functions. A partial derivative of a function measures how the function changes as one of the variables is varied, while all other variables are held constant.

For the function \( f(x,y) = \sin(3x - y) \), computing its partial derivatives with respect to \( x \) and \( y \) gives the components of the gradient vector:- The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = 3\cos(3x-y) \), indicates the rate of change of \( f \) in the \( x \)-direction.- The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} = - \cos(3x-y) \), indicates the rate of change of \( f \) in the \( y \)-direction.These derivatives are evaluated at a specific point to form the gradient vector at that point. In our example, this is done at point \( \mathbf{p} = \left( \frac{\pi}{6}, \frac{\pi}{4} \right) \), resulting in a specific gradient vector value.

Unit Vector

A unit vector is simply a vector with a magnitude of 1. It is used to indicate direction, not magnitude. In our scenario, once we compute the direction of the gradient vector, we normalize it to form a unit vector that represents the direction of steepest descent.

To achieve this, first calculate the magnitude of the vector: \[\text{Magnitude} = \sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2}\]This results in \( \sqrt{5} \). Finally, dividing each component of the vector by this magnitude gives the unit vector:\[\mathbf{u} = \frac{1}{\sqrt{5}} \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)\]This unit vector uniquely points in the steepest descent direction, with no additional scaling involved.

Normalization

Normalization is a mathematical process used to convert a vector to a unit vector while preserving its direction. It involves dividing each component of the vector by its magnitude. This method ensures the resultant vector has a unit length while maintaining its original direction.

In the context of the problem, normalization gives us the final arrow, or direction, for steepest descent by converting the calculated vector into a unit vector.

Steps for normalization:1. Calculate the magnitude of the vector: \[ \sqrt{\left(-\frac{3\sqrt{2}}{2}\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2} = \sqrt{5} \]2. Divide each component of the vector by its magnitude to obtain the unit vector.The normalized vector \( \mathbf{u} = \frac{1}{\sqrt{5}} \left( -\frac{3\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right) \) accurately reflects the intended direction with standard unit length.