Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=x^{2}-3 x y+2 y^{2} ; \mathbf{p}=(-1,2) ; \mathbf{a}=2 \mathbf{i}-\mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative is \(-\frac{27}{\sqrt{5}}\).

Step 1: Normalize the Direction Vector

First, we need to ensure that the direction vector \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} \) is a unit vector. To do this, we calculate its magnitude: \( ||\mathbf{a}|| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \). Thus, the unit vector in the direction of \( \mathbf{a} \) is \( \hat{\mathbf{a}} = \frac{1}{\sqrt{5}}(2\mathbf{i} - \mathbf{j}) \).

Step 2: Compute the Gradient of the Function

The gradient of the function \( f(x, y) = x^2 - 3xy + 2y^2 \) is given by the vector of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute each of the partial derivatives: \( \frac{\partial f}{\partial x} = 2x - 3y \) and \( \frac{\partial f}{\partial y} = -3x + 4y \). So, \( abla f = (2x - 3y, -3x + 4y) \).

Step 3: Evaluate the Gradient at the Point \( \mathbf{p} \)

Substitute the point \( \mathbf{p} = (-1,2) \) into the gradient: \(abla f(-1, 2) = (2(-1) - 3(2), -3(-1) + 4(2)) = (-2 - 6, 3 + 8)\)This results in \( abla f(-1, 2) = (-8, 11) \).

Step 4: Calculate the Directional Derivative

The directional derivative of \( f \) at \( \mathbf{p} \) in the direction of \( \hat{\mathbf{a}} \) is given by \(D_{\hat{\mathbf{a}}} f = abla f \cdot \hat{\mathbf{a}} = (-8, 11) \cdot \frac{1}{\sqrt{5}}(2, -1).\)Compute the dot product and simplify:\(D_{\hat{\mathbf{a}}} f = \frac{1}{\sqrt{5}}((-8)(2) + (11)(-1)) = \frac{1}{\sqrt{5}}(-16 - 11) = \frac{-27}{\sqrt{5}}.\)

Key Concepts

Gradient Vector

The gradient vector is an essential concept in calculus, especially when dealing with functions of several variables. Its chief role is to point in the direction of the steepest ascent of a function. For a function like the given one, \( f(x, y) = x^2 - 3xy + 2y^2 \), the gradient vector \( abla f \) is a vector composed of the function's partial derivatives. This vector, \( abla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \), highlights how the function \( f \) changes in response to small changes in \( x \) and \( y \).
In the context of the exercise, calculating the gradient vector involves determining each partial derivative and evaluating them at a specific point. The resulting vector provides direction, giving insight into how the function elevates at that point. Remember, gradients point towards the increasing values of \( f \).

Partial Derivatives

Partial derivatives represent the rate of change of a multivariable function with respect to one variable, while keeping others constant. They form the building blocks of the gradient vector.
In this exercise, we determined the partial derivatives of \( f(x, y) = x^2 - 3xy + 2y^2 \) to find \( abla f \).

• \( \frac{\partial f}{\partial x} = 2x - 3y \)
• \( \frac{\partial f}{\partial y} = -3x + 4y \)

These calculations reveal how \( f \) changes if we nudge \( x \) or \( y \) a little. Evaluating these derivatives at the point \( \mathbf{p} = (-1, 2) \) gives a concrete gradient vector, making it possible to compute the directional derivative.

Unit Vector

Direction is crucial when computing directional derivatives, which is why vectors must often be normalized into unit vectors. Normalization shrinks the vector to a unit of 1 without changing its direction. For instance, in this problem, the direction vector \( \mathbf{a} = 2\mathbf{i} - \mathbf{j} \) was transformed into a unit vector \( \hat{\mathbf{a}} \).
The process involves calculating the vector's magnitude: \( ||\mathbf{a}|| = \sqrt{2^2 + (-1)^2} = \sqrt{5} \). The normalized vector is then \( \hat{\mathbf{a}} = \frac{1}{\sqrt{5}}(2\mathbf{i} - \mathbf{j}) \). With a unit length, this allows for accurate computation of the directional derivative without exaggerating the effect of vector scaling.

Dot Product

The dot product is pivotal for determining directional derivatives as it allows for the combination of the gradient and the direction. It presents a way to measure how much one vector "projects" onto another, like projecting \( abla f \) onto \( \hat{\mathbf{a}} \) to find the directional derivative.
In the calculation, the dot product \( abla f \cdot \hat{\mathbf{a}} \) provides a scalar that conveys the extent of the influence of \( \hat{\mathbf{a}} \) on the rate of change of \( f \.\) For \( abla f(-1, 2) = (-8, 11) \) and \( \hat{\mathbf{a}} = \frac{1}{\sqrt{5}}(2, -1) \), this resulted in \( D_{\hat{\mathbf{a}}} f = \frac{1}{\sqrt{5}}(-16 - 11) = \frac{-27}{\sqrt{5}} \).
Dot products thus make directional derivatives go beyond simple direction, reflecting collective contribution of both gradient inclination and direction stability.