Problem 12 - Calculus Early Transcendentals

Question

II-17 Find the directional derivative of the function at the given point in the direction of the vector \(\mathbf{v}\) . $$f(x, y)=\ln \left(x^{2}+y^{2}\right), \quad(2,1), \quad \mathbf{v}=\langle- 1,2\rangle$$

Step-by-Step Solution

Answer

The directional derivative is 0.

Step 1: Compute the Gradient of the Function

The function is given as \(f(x, y) = \ln(x^2 + y^2)\). First, we need to find the gradient \(abla f(x, y)\). The gradient is a vector consisting of partial derivatives: \(abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). Calculate \(\frac{\partial f}{\partial x}\) using the chain rule:\[\frac{\partial f}{\partial x} = \frac{1}{x^2 + y^2} \cdot 2x = \frac{2x}{x^2 + y^2}\]Calculate \(\frac{\partial f}{\partial y}\) using the chain rule:\[\frac{\partial f}{\partial y} = \frac{1}{x^2 + y^2} \cdot 2y = \frac{2y}{x^2 + y^2}\]Thus, \[abla f(x, y) = \left(\frac{2x}{x^2 + y^2}, \frac{2y}{x^2 + y^2}\right)\].

Step 2: Evaluate the Gradient at the Given Point

Substitute the given point \((2, 1)\) into the gradient formula. We need \(abla f(2, 1)\).Calculate: \[abla f(2, 1) = \left(\frac{2 \times 2}{2^2 + 1^2}, \frac{2 \times 1}{2^2 + 1^2}\right) = \left(\frac{4}{5}, \frac{2}{5}\right)\].

Step 3: Normalize the Direction Vector

Given vector \(\mathbf{v} = \langle -1, 2 \rangle\), we need to find its unit vector \(\mathbf{u}\). This is achieved by dividing each component by the magnitude of \(\mathbf{v}\).Calculate the magnitude of \(\mathbf{v}\): \[\| \mathbf{v} \| = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}\]Thus, the unit vector \(\mathbf{u}\) is:\[\mathbf{u} = \left(\frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\].

Step 4: Compute the Directional Derivative

The directional derivative of a function \(f\) at a point \((x, y)\) in the direction of a unit vector \(\mathbf{u} = \langle u_1, u_2 \rangle \) is given by the dot product:\[D_{\mathbf{u}} f(x, y) = abla f(x, y) \cdot \mathbf{u}\]Substituting the values:\[D_{\mathbf{u}} f(2, 1) = \left(\frac{4}{5}, \frac{2}{5}\right) \cdot \left(\frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\]Calculate the dot product:\[D_{\mathbf{u}} f(2, 1) = \frac{4}{5} \times \frac{-1}{\sqrt{5}} + \frac{2}{5} \times \frac{2}{\sqrt{5}} = \frac{-4}{5\sqrt{5}} + \frac{4}{5\sqrt{5}} = 0\].

Step 5: Conclusion

The directional derivative of the function at point \((2, 1)\) in the direction of vector \(\mathbf{v}\) is \(0\).

Key Concepts

Gradient Vector

In the world of multivariable calculus, the gradient vector is an essential tool. Think of it as a multi-dimensional generalization of the derivative. For a function of two variables, like our function \[f(x, y) = \ln(x^2 + y^2),\]the gradient vector \(abla f(x, y)\) is composed of partial derivatives: the derivative of the function with respect to each variable.
The gradient vector points in the direction of the greatest rate of increase of the function. The components of the gradient vector are:

• \(\frac{\partial f}{\partial x}\): How much \(f\) changes as \(x\) changes, with \(y\) held constant.
• \(\frac{\partial f}{\partial y}\): How much \(f\) changes as \(y\) changes, with \(x\) held constant.

At any point, evaluating the gradient gives you a snapshot of where the function is "steepest".

Partial Derivatives

Partial derivatives are a core concept when analyzing functions of several variables. They represent the rate of change of a function concerning one of its variables, while all other variables are fixed.
In our function, the partial derivatives are:

• \(\frac{\partial f}{\partial x} = \frac{2x}{x^2+y^2}\), which shows how the function \(f\) changes as \(x\) changes.
• \(\frac{\partial f}{\partial y} = \frac{2y}{x^2+y^2}\), which indicates how \(f\) changes with a change in \(y\).

Why are they important? Because they allow us to build the gradient vector. Each partial derivative contributes one component to this vector, defining its direction and magnitude.

Vector Normalization

Normalizing a vector is a process that scales the vector to have a magnitude (or length) of one, while maintaining its direction. This new vector is referred to as a unit vector.
To normalize a vector \( \mathbf{v} = \langle -1, 2 \rangle \), the steps are:

• Calculate the magnitude \(\| \mathbf{v} \|\) using: \(\sqrt{(-1)^2 + 2^2} = \sqrt{5}\).
• Divide each component of the vector by \(\| \mathbf{v} \|\) to get the unit vector: \(\mathbf{u} = \left(\frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\).

Normalizing vectors is crucial for directional derivatives. This is because directional derivatives require unit vectors to ensure the rate of change is measured consistently.

Unit Vector

A unit vector is simply a vector with a magnitude of one. It's important because it indicates direction without altering the scale. In directional derivatives, using a unit vector enables us to measure the rate of change of the function along a specific direction in a consistent manner.
For example, if we have a vector \( \mathbf{v} = \langle -1, 2 \rangle \), its corresponding unit vector is:\[\mathbf{u} = \left(\frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\]This unit vector maintains the direction of \(\mathbf{v}\) but ensures the measurement scales are aligned with the unit system used in calculus. Thus, they are pivotal when calculating directional derivatives, ensuring the change we compute reflects the true rate of change in the specified direction.