Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=x^{2} y ; \mathbf{p}=(1,2) ; \mathbf{a}=3 \mathbf{i}-4 \mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative of \(f\) at \(\mathbf{p}\) in the direction of \(\mathbf{a}\) is \(\frac{8}{5}\).

Step 1: Find the gradient of the function

The gradient of the function \(f(x, y) = x^2 y\) is given by \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). First, compute the partial derivatives: \(\frac{\partial f}{\partial x} = 2xy \) and \(\frac{\partial f}{\partial y} = x^2\). Thus, the gradient vector is \(abla f(x, y) = (2xy, x^2)\).

Step 2: Evaluate the gradient at the point \(\mathbf{p}\)

Substitute the point \(\mathbf{p} = (1, 2)\) into the gradient vector \(abla f(x, y)\). This gives \(abla f(1, 2) = (2 \times 1 \times 2, 1^2) = (4, 1)\).

Step 3: Normalize the direction vector

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

Step 4: Compute the directional derivative

The directional derivative of \(f\) at \(\mathbf{p}\) in the direction of \(\mathbf{u}\) is given by \(abla f(1, 2) \cdot \mathbf{u}\). Substitute the gradient \((4, 1)\) and unit vector \(\mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right)\) into the dot product: \(abla f(1, 2) \cdot \mathbf{u} = 4 \cdot \frac{3}{5} + 1 \cdot \frac{-4}{5} = \frac{12}{5} - \frac{4}{5} = \frac{8}{5}\).

Key Concepts

Gradient Vector

In the world of multivariable calculus, the gradient vector plays a critical role. It is a vector composed of all the partial derivatives of a function. For a function of two variables, like \(f(x, y)\), the gradient vector is represented as \(abla f(x, y)\). This vector points in the direction of the greatest rate of increase of the function. Using our function \(f(x, y) = x^2 y\), the gradient vector is computed as follows:

• First, find the partial derivative of \(f\) with respect to \(x\), which is \(\frac{\partial f}{\partial x} = 2xy\).
• Next, find the partial derivative of \(f\) with respect to \(y\), which is \(\frac{\partial f}{\partial y} = x^2\).

Thus, the gradient vector is \(abla f(x, y) = (2xy, x^2)\), encapsulating how the function changes at any point \((x, y)\). This tool is invaluable for navigating the landscape of a function's possible values.

Partial Derivatives

Partial derivatives are used to understand how a multivariable function changes with respect to each individual variable. It is akin to taking the derivative of the function while keeping all other variables constant.
For example, with the function \(f(x, y) = x^2 y\), when we take the partial derivative with respect to \(x\), all \(y\) values remain constant, resulting in \(\frac{\partial f}{\partial x} = 2xy\). Similarly, the partial derivative with respect to \(y\), holding \(x\) constant, gives us \(\frac{\partial f}{\partial y} = x^2\).
This method provides insight into how changes in each specific variable affect the overall function's output, assisting in visualizing changes from a slice of the entire surface.

Unit Vector

A unit vector is a vector with a magnitude of one. It serves to indicate direction without affecting the scale. To find a unit vector, you simply divide a given vector by its magnitude. In our given exercise, the direction vector is \(\mathbf{a} = (3, -4)\).
First, calculate its magnitude:

• \(\|\mathbf{a}\| = \sqrt{3^2 + (-4)^2} = \sqrt{25} = 5\).

Then, the unit vector \(\mathbf{u}\) is determined as:

• \(\mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right)\).

This vector now effectively helps to describe the direction of \(\mathbf{a}\) without altering the properties of length or magnitude, ensuring calculations remain consistent in context for directional derivatives.

Dot Product

The dot product is a fundamental operation that takes two vectors and returns a scalar. It signifies the magnitude of one vector in the direction of another. In simpler terms, it's like asking "how much of one vector lies in the direction of the other".
The calculation involves multiplying corresponding components of two vectors and summing the results. For instance, using the final step of our example, where we need to find the directional derivative, the vectors involved are the gradient vector \((4, 1)\) and the unit direction vector \(\mathbf{u} = \left(\frac{3}{5}, \frac{-4}{5}\right)\).
The dot product \(abla f(1, 2) \cdot \mathbf{u}\) is calculated as:

• \(4 \times \frac{3}{5} + 1 \times \frac{-4}{5} = \frac{12}{5} - \frac{4}{5} = \frac{8}{5}\)

The result tells us how rapidly the function \(f\) changes in the direction of \(\mathbf{a}\), with context to its orientation described by unit vector \(\mathbf{u}\). The concept of dot product is critical when calculating directional derivatives.