Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a unit vector in the direction in which \(f\) increases most rapidly at \(\mathbf{p} .\) What is the rate of change in this direction? \(f(x, y)=e^{y} \sin x ; \mathbf{p}=(5 \pi / 6,0)\)

Step-by-Step Solution

Answer

The unit vector is \(\left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right)\), and the rate of change is 1.

Step 1: Find the Gradient

The gradient of a function \( f(x, y) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = e^y \sin x \), we calculate:\[ abla f = \left( \cos x \cdot e^y, \sin x \cdot e^y \right) \].

Step 2: Evaluate Gradient at Given Point

Now we need to evaluate the gradient at the given point \( \mathbf{p} = \left( \frac{5\pi}{6}, 0 \right) \). Substituting these values into \( abla f \), we find:\[ abla f(\frac{5\pi}{6}, 0) = \left( \cos \left( \frac{5\pi}{6} \right) \cdot e^0, \sin \left( \frac{5\pi}{6} \right) \cdot e^0 \right) = \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \].

Step 3: Compute the Magnitude of the Gradient

The magnitude of the gradient vector \( abla f(\frac{5\pi}{6}, 0) \) is calculated as:\[ \| abla f \| = \sqrt{\left( -\frac{\sqrt{3}}{2} \right)^2 + \left( \frac{1}{2} \right)^2} = \sqrt{\frac{3}{4} + \frac{1}{4}} = 1 \].

Step 4: Determine the Unit Vector in the Direction of the Gradient

A unit vector in the direction of the gradient is obtained by dividing the gradient vector by its magnitude:\[ \mathbf{u} = \frac{1}{1} \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) = \left( -\frac{\sqrt{3}}{2}, \frac{1}{2} \right) \].

Step 5: Rate of Change in the Direction of the Gradient

The rate of change of \( f \) in the direction of the gradient is given by the magnitude of the gradient \( \| abla f \| \), which we already found to be 1.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in calculus, particularly when dealing with functions of multiple variables. When a function is dependent on more than one variable, the partial derivative of the function with respect to one of those variables is the derivative of the function, considering all other variables constant.
For example, if you have a function \( f(x, y) \), which depends on \( x \) and \( y \), taking the partial derivative with respect to \( x \) means differentiating \( f \) as if \( y \) is just a constant, and vice versa. This allows us to understand how the function changes as we vary one of the input variables while the others are held steady.

• In our solution, the partial derivatives are \( \frac{\partial f}{\partial x} = \cos x \cdot e^y \) and \( \frac{\partial f}{\partial y} = \sin x \cdot e^y \).
• These partial derivatives are components of the gradient vector, which gives us the direction of steepest ascent for the function.

Unit Vector

A unit vector is a vector with a magnitude of 1. It is used to indicate direction without any magnitude. In mathematics, unit vectors are often used to build or decompose other vectors.
When working with gradients, the gradient vector itself indicates the direction of the greatest rate of increase of a function. But to express this direction with no consideration of magnitude, we use a unit vector. This is done by dividing the original vector by its magnitude.

• In the given exercise, our task requires us to find the unit vector in the direction of the gradient at a specific point.
• The gradient vector at \( \mathbf{p} = (\frac{5\pi}{6}, 0) \) is \( (-\frac{\sqrt{3}}{2}, \frac{1}{2}) \).
• We calculated the magnitude to be 1, so the unit vector remains \( (-\frac{\sqrt{3}}{2}, \frac{1}{2}) \).

Rate of Change

The rate of change of a function is a measure of how much the function's output changes as its inputs change. In calculus, for functions of multiple variables, the rate of change in a specific direction is indicated by the directional derivative.
The steepest possible rate of change at a given point is found in the direction of the gradient vector. This direction is significant because it tells us where the function is increasing most rapidly.

• In the exercise's context, the rate of change in the direction of the gradient gives us the magnitude of the gradient vector itself, which is 1.
• This indicates that the function \( f(x, y) = e^y \sin x \) increases most rapidly at the point \( \mathbf{p} \) in the direction of the gradient vector.