Problem 19 - Calculus Early Transcendentals

Question

Find the directional derivative of \(f(x, y)=\sqrt{x y}\) at \(P(2,8)\) in the direction of \(Q(5,4)\)

Step-by-Step Solution

Answer

The directional derivative is \( \frac{2}{5} \).

Step 1: Calculate the Gradient of f(x, y)

First, we identify the function's gradient. The gradient \( abla f(x, y) \) represents the vector of partial derivatives. For \( f(x, y) = \sqrt{xy} \), we have:- The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = \frac{y}{2\sqrt{xy}} \).- The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = \frac{x}{2\sqrt{xy}} \).Thus, the gradient is:\[ abla f(x, y) = \left( \frac{y}{2\sqrt{xy}}, \frac{x}{2\sqrt{xy}} \right) \]

Step 2: Evaluate the Gradient at P(2,8)

Substitute \( x = 2 \) and \( y = 8 \) into the gradient:- \( \frac{\partial f}{\partial x}(2, 8) = \frac{8}{2\sqrt{2 \times 8}} = \frac{8}{2\sqrt{16}} = \frac{8}{8} = 1 \)- \( \frac{\partial f}{\partial y}(2, 8) = \frac{2}{2\sqrt{2 \times 8}} = \frac{2}{8} = \frac{1}{4} \)So, \( abla f(2, 8) = (1, \frac{1}{4}) \).

Step 3: Find the Direction Vector from P(2,8) to Q(5,4)

The direction vector from \( P(2, 8) \) to \( Q(5, 4) \) is found by subtracting the coordinates:\[ \text{Direction vector} = (5 - 2, 4 - 8) = (3, -4) \]

Step 4: Normalize the Direction Vector

To calculate the directional derivative, we need a unit direction vector. Normalize the vector \((3, -4)\):1. Calculate the magnitude: \[ \text{Magnitude} = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]2. Divide each component by the magnitude: \[ \text{Unit vector} = \left( \frac{3}{5}, \frac{-4}{5} \right) \]

Step 5: Calculate the Directional Derivative

The directional derivative is the dot product of the gradient and the unit direction vector:\[ D_{\mathbf{u}}f(2, 8) = abla f(2, 8) \cdot \mathbf{u} = (1, \frac{1}{4}) \cdot \left( \frac{3}{5}, \frac{-4}{5} \right) \]Calculate the dot product:\[ = 1 \cdot \frac{3}{5} + \frac{1}{4} \cdot \frac{-4}{5} \]\[ = \frac{3}{5} + \frac{-4}{20} \]\[ = \frac{3}{5} - \frac{1}{5} \]\[ = \frac{2}{5} \]Thus, the directional derivative is \( \frac{2}{5} \).

Key Concepts

Gradient

The gradient is like a compass that shows us the direction of the steepest ascent of a function at a given point. It is a vector comprised of partial derivatives. For a function of two variables, such as \( f(x, y) = \sqrt{xy} \), the gradient \( abla f(x, y) \) is expressed as:

• The partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = \frac{y}{2\sqrt{xy}} \).
• The partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = \frac{x}{2\sqrt{xy}} \).

These partial derivatives are the components of the gradient vector \( abla f(x, y) \), and together, they point in the direction where the function increases most rapidly. Evaluating this gradient at a specific point, like \( P(2,8) \), gives a vector that describes the steepest slope at that particular spot.

Partial Derivatives

Partial derivatives help us understand how a function changes as one of the variables changes, while the other variables are held constant. In the context of the function \( f(x, y) = \sqrt{xy} \), we compute:

• The partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} \), which captures how the function changes when \( x \) changes, with \( y \) held constant.
• The partial derivative with respect to \( y \), \( \frac{\partial f}{\partial y} \), indicating the change in \( f \) when varying \( y \) while keeping \( x \) fixed.

These derivatives are fundamental building blocks in multivariable calculus as they allow us to break down complex problems into simpler, one-dimensional analyses.

Normalization

Normalization is the process of turning a vector into a unit vector. It means making the vector's length equal to 1. This is crucial when computing the directional derivative because we need to ensure that our direction vector doesn't include any scaling impacts caused by its length.
Take the vector from \( P(2,8) \) to \( Q(5,4) \):

• Find the magnitude: \( \sqrt{3^2 + (-4)^2} = 5 \).
• Normalize by dividing each component by its magnitude: the result is \( \left( \frac{3}{5}, \frac{-4}{5} \right) \).

This normalized vector is "direction-only" and is ready to be used in further calculations, like a directional derivative.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. It combines the vectors in a way that takes into account both magnitude and direction. To compute a directional derivative, we need the dot product of the gradient vector and the normalized direction vector.

• The dot product formula: \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \ldots \).
• For our gradient \( (1, \frac{1}{4}) \) and unit direction vector \( \left( \frac{3}{5}, \frac{-4}{5} \right) \), it becomes \( 1 \cdot \frac{3}{5} + \frac{1}{4} \cdot \frac{-4}{5} \).
• Calculating: \( \frac{3}{5} - \frac{1}{5} = \frac{2}{5} \).

Thus, we find how quickly the function \( f(x, y) = \sqrt{xy} \) changes in the direction of our vector at the point \( P(2,8) \).