Problem 272 - Calculus 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: Understand the Problem

We need to find the directional derivative of the function \(f(x, y) = x^2 y\) at a given point \(P(-5, 5)\) in the direction of the vector \(\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}\).

Step 2: Calculate the Gradient

The gradient of \(f\), denoted as \(abla f\), is calculated by taking the partial derivatives with respect to \(x\) and \(y\).The partial derivative with respect to \(x\) is \(f_x = 2xy\).The partial derivative with respect to \(y\) is \(f_y = x^2\).Thus, \(abla f = \langle 2xy, x^2 \rangle\).

Step 3: Evaluate the Gradient at Point P

Substitute \(x = -5\) and \(y = 5\) into \(abla f = \langle 2xy, x^2 \rangle\).\(f_x(-5,5) = 2(-5)(5) = -50\).\(f_y(-5,5) = (-5)^2 = 25\).Therefore, \(abla f(-5, 5) = \langle -50, 25 \rangle\).

Step 4: Normalize the Direction Vector

The direction vector \(\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}\) needs to be a unit vector.Calculate the magnitude: \(|\mathbf{v}| = \sqrt{3^2 + (-4)^2} = 5\).The unit vector \(\mathbf{u}\) in the direction of \(\mathbf{v}\) is \(\langle \frac{3}{5}, \frac{-4}{5} \rangle\).

Step 5: Calculate the Directional Derivative

The directional derivative \(D_{\mathbf{u}}f\) is the dot product of \(abla f\) evaluated at the point \(P\) and the unit vector \(\mathbf{u}\).Compute \(D_{\mathbf{u}}f = \langle -50, 25 \rangle \cdot \langle \frac{3}{5}, \frac{-4}{5} \rangle\).\(D_{\mathbf{u}}f = (-50)(\frac{3}{5}) + 25(\frac{-4}{5})\).\(D_{\mathbf{u}}f = -30 - 20 = -50\).

Step 6: Interpret the Result

The directional derivative at the point \((-5, 5)\) in the direction of the vector \(\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}\) is \(-50\). This indicates the rate of change of the function in that specific direction.

Key Concepts

Gradient

The concept of the Gradient is central when dealing with directional derivatives. Essentially, the gradient of a function provides us with a vector that describes the direction and rate of the steepest ascent of that function. If you have a function, say like our example function, the gradient is notated as \( abla f \). This vector is calculated using partial derivatives of the function with respect to each of its variables.
For the function \(f(x, y)=x^{2}y\), we compute the gradient by finding these partial derivatives. Specifically, the partial derivative with respect to \(x\) is \(2xy\), and with respect to \(y\) is \(x^2\). Thus, the gradient at any point \((x, y)\) can be described by the vector \(\langle 2xy, x^2 \rangle\).
When evaluated at a specific point, such as \((x, y) = (-5, 5)\), these partial derivatives give us a vector \(\langle -50, 25 \rangle\). This tells us in which direction the function \(f\) increases most rapidly and how fast it does so.

Partial Derivatives

Partial Derivatives are crucial for understanding how a function changes. They measure the rate of change of a function with respect to one of its variables, keeping other variables constant. Imagine the full function as a landscape, and each partial derivative gives you a different slice of this landscape.
For the function \(f(x, y) = x^2 y\), the partial derivative with respect to \(x\) means treating \(y\) as a constant. This gives us \(f_x = 2xy\). Similarly, the partial derivative with respect to \(y\) means treating \(x\) as constant, resulting in \(f_y = x^2\). Each partial derivative provides a piece of the puzzle, and together they form the gradient.
By evaluating these at a particular point, such as \((x, y) = (-5, 5)\), you can determine specific directions and rates of change for that function. It is like getting insight on how the landscape changes as we move along these two directions.

Unit Vector

The Unit Vector plays an important role in finding directional derivatives. It ensures that we are moving in a specified direction without changing how far we move in terms of magnitude. Essentially, a unit vector has a magnitude of 1, making it perfect for direction purposes without scaling any distances.
To find a unit vector in the direction of any given vector \(\mathbf{v}\), you divide each component of \(\mathbf{v}\) by the magnitude of \(\mathbf{v}\). For the vector \(\mathbf{v} = 3\mathbf{i} - 4\mathbf{j}\), the magnitude \(|\mathbf{v}|\) is \(\sqrt{3^2 + (-4)^2} = 5\). Thus, the unit vector \(\mathbf{u}\) in the direction of \(\mathbf{v}\) is derived as \(\langle \frac{3}{5}, \frac{-4}{5} \rangle\).
Using a unit vector, we can assess how a function changes in a specific direction accurately, which is crucial for determining the directional derivative.

Dot Product

The Dot Product is an essential concept that links the gradient and unit vector to find the directional derivative. It provides a measure of the magnitude of one vector in the direction of another. Mathematically, the dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + \dots\).
In the context of directional derivatives, you'll often compute the dot product of the gradient \(abla f\) and the direction's unit vector \(\mathbf{u}\). This dot product returns the directional derivative, providing the rate of change of the function in the specified direction. For our problem, it is calculated as \(D_\mathbf{u}f = \langle -50, 25 \rangle \cdot \langle \frac{3}{5}, \frac{-4}{5} \rangle\).
Thus, the dot product combines both vectors to provide a scalar, revealing how much the function changes as you move in the precise direction of interest.