Problem 272 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=x^{2} y, P(-5,5), \quad \mathbf{v}=3 \mathbf{i}-4 \mathbf{j}$$

Step-by-Step Solution

Answer

The directional derivative is -50.

Step 1: Find the gradient of the function

First, calculate the partial derivatives of the function \( f(x, y) = x^2 y \). The partial derivative with respect to \( x \) is \( f_x = 2xy \) and with respect to \( y \) is \( f_y = x^2 \). Thus, the gradient of the function \( abla f \) is given by \( abla f = (2xy, x^2) \).

Step 2: Evaluate the gradient at point P

Substitute the point \( P(-5, 5) \) into the gradient. This gives us \( abla f(-5, 5) = (2(-5)(5), (-5)^2) = (-50, 25) \).

Step 3: Normalize the direction vector \( \mathbf{v} \)

The direction vector is given as \( \mathbf{v} = 3 \mathbf{i} - 4 \mathbf{j} \). Find the magnitude of \( \mathbf{v} \) using \( ||\mathbf{v}|| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \). The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{v} \) is \( \mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right) \).

Step 4: Compute the directional derivative

The directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{v} \) is given by the dot product of the gradient at \( P \) and the unit vector \( \mathbf{u} \). This is \( abla f(-5, 5) \cdot \mathbf{u} = (-50, 25) \cdot \left(\frac{3}{5}, \frac{-4}{5}\right) \). Calculate this as \(-50 \cdot \frac{3}{5} + 25 \cdot \frac{-4}{5} = -30 - 20 = -50\).

Key Concepts

Gradient

The gradient is an essential concept in calculating directional derivatives. It is a vector that points in the direction of the greatest increase of a function. For a function of two variables, say, \( f(x, y) \), the gradient is represented as \( abla f \) and constitutes the partial derivatives of the function with respect to each variable.
This means the gradient vector will have components \( (f_x, f_y) \).
In this case, for the function \( f(x, y) = x^2y \), the partial derivatives are \( f_x = 2xy \) and \( f_y = x^2 \), leading to the gradient vector \( abla f = (2xy, x^2) \).

• The gradient vector is crucial as it aids in finding the directional derivative in any specified direction.
• Note that the gradient is not just direction-specific but rather tells us about the slope in all possible directions.

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one of multiple variables, keeping other variables constant.
This is like taking the derivative of a single-variable function while treating the other variables as constants.
In our example, we have the function \( f(x, y) = x^2y \), and we need to compute partial derivatives with respect to both \( x \) and \( y \).

• For \( x \), we treat \( y \) as constant, deriving \( 2xy \).
• For \( y \), we treat \( x \) as constant, deriving \( x^2 \).

These partial derivatives form the components of the gradient vector. Through partial derivatives, we can measure change one dimension or variable at a time, a fundamental aspect when dealing with functions of multiple variables.

Normalization

Normalization is a process in mathematics where we transform a vector so that it has a magnitude (length) of 1, making it a unit vector.
This is crucial when we want to analyze direction without concern for magnitude.
For a vector \( \mathbf{v} = 3 \mathbf{i} - 4 \mathbf{j} \), the magnitude is calculated by \( ||\mathbf{v}|| = \sqrt{3^2 + (-4)^2} = 5 \). The unit vector \( \mathbf{u} \) is then found by dividing each component of \( \mathbf{v} \) by its magnitude, i.e., \( \mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right) \).

• Normalization is useful as it allows us to focus on direction and scale it to a precedence of 1, simplifying calculations involving vectors.
• By normalizing, we remove the magnitude component, which ensures calculations linked to direction, such as dot products, give us meaningful directional results.

Dot Product

The dot product is a way of multiplying two vectors to obtain a scalar. In the context of directional derivatives, the dot product between the gradient and a unit vector gives the directional derivative itself. For two vectors \( \mathbf{a} = (a_1, a_2) \) and \( \mathbf{b} = (b_1, b_2) \), their dot product is calculated as \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \). In our example, we use the dot product to combine the gradient \( abla f(-5, 5) = (-50, 25) \) calculated at point \( P (-5,5) \), with the unit vector \( \mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right) \). The resulting dot product, \(-50 \cdot \frac{3}{5} + 25 \cdot \frac{-4}{5} = -50\), gives us the rate of change of the function in the direction of our vector.

• This scalar result can tell us whether the function is increasing or decreasing in that direction.
• A positive result indicates an increase, while a negative one represents a decrease in the function's value.

Understanding the dot product helps in comprehending how different directional influences combine mathematically.