Problem 7 - Calculus with Concepts in Calculus

Question

Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y)=\tan (x+2 y) ; P=(0, \pi / 6) ; \mathbf{a}=-4 \mathbf{i}+5 \mathbf{j} $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{24\sqrt{41}}{41}\).

Step 1: Calculate the Gradient of f

First, we need to calculate the gradient of the function \(f(x,y)\). The gradient is the vector of partial derivatives. We have:\[abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\]Calculate the partial derivative with respect to \(x\):\[\frac{\partial f}{\partial x} = \sec^2(x + 2y)\]Calculate the partial derivative with respect to \(y\):\[\frac{\partial f}{\partial y} = 2\sec^2(x + 2y)\]Thus, the gradient is:\[abla f(x, y) = (\sec^2(x + 2y), 2\sec^2(x + 2y))\]

Step 2: Evaluate the Gradient at P

Now, evaluate the gradient at the point \(P = (0, \pi/6)\):\[abla f(0, \pi/6) = (\sec^2(0 + 2\cdot\pi/6), 2\sec^2(0 + 2\cdot\pi/6))\]Simplify this:\[abla f(0, \pi/6) = (\sec^2(\pi/3), 2\sec^2(\pi/3))\]Recall that \(\sec(\pi/3) = 2\), so:\[\sec^2(\pi/3) = 4\]Thus, the gradient is:\[abla f(0, \pi/6) = (4, 8)\]

Step 3: Normalize the Direction Vector

The direction vector \(\mathbf{a} = -4\mathbf{i} + 5\mathbf{j}\) must be normalized. The magnitude of \(\mathbf{a}\) is:\[\|\mathbf{a}\| = \sqrt{(-4)^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41}\]Normalize \(\mathbf{a}\) by dividing each component by its magnitude:\[\mathbf{u} = \left(-\frac{4}{\sqrt{41}}, \frac{5}{\sqrt{41}}\right)\]

Step 4: Compute the Directional Derivative

The directional derivative \(D_{\mathbf{u}}f\) is given by the dot product of the gradient and the unit vector \(\mathbf{u}\):\[D_{\mathbf{u}}f = abla f(0, \pi/6) \cdot \mathbf{u}\]Plug in the values:\[D_{\mathbf{u}}f = (4, 8) \cdot \left(-\frac{4}{\sqrt{41}}, \frac{5}{\sqrt{41}}\right)\]This results in:\[D_{\mathbf{u}}f = 4\left(-\frac{4}{\sqrt{41}}\right) + 8\left(\frac{5}{\sqrt{41}}\right)\]\[D_{\mathbf{u}}f = -\frac{16}{\sqrt{41}} + \frac{40}{\sqrt{41}}\]\[D_{\mathbf{u}}f = \frac{24}{\sqrt{41}}\]

Step 5: Simplify the Final Result

To present the directional derivative in a simplified form, rationalize the denominator (optional in some cases):\[D_{\mathbf{u}}f = \frac{24\sqrt{41}}{41}\]This is the directional derivative of the function \(f\) at \(P\) in the direction of \(\mathbf{a}\).

Key Concepts

Gradient of a Function

The gradient of a function is a crucial concept in multivariable calculus. It serves as a vector showing the direction of the steepest ascent of a function at a particular point. For a function \( f(x, y) \), its gradient is represented by the symbol \( abla f \) and is found by taking the partial derivatives of \( f \) with respect to each of its variables.

The gradient is expressed as a vector:

• \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

This means you calculate the slope of the function along the \( x \) axis and the \( y \) axis. The gradient not only points in the direction of maximum increase but its magnitude gives the rate of increase as well. In the original solution, finding the gradient at a point like \( P=(0, \pi/6) \) involves substituting these coordinates into the gradient vector, revealing the local behavior of the function.

Partial Derivatives

Partial derivatives are essential tools in understanding how a function changes with respect to one of its variables while keeping others constant. When working with functions of multiple variables, like \( f(x, y) = \tan(x + 2y) \), you compute partial derivatives to examine how each variable individually influences the function's output.

In our exercise, the partial derivative of \( f \) with respect to \( x \) is computed as:

• \( \frac{\partial f}{\partial x} = \sec^2(x + 2y) \)

Similarly, with respect to \( y \), it is calculated as:

• \( \frac{\partial f}{\partial y} = 2\sec^2(x + 2y) \)

These derivatives tell us the rate of change of \( f \) in the direction of \( x \) or \( y \), providing critical insights when determining the gradient and directional derivatives.

Unit Vector

A unit vector is a vector with a magnitude of 1, making it particularly useful for defining a consistent direction. To compute a unit vector from any given vector, you divide each component of the vector by its magnitude. This process "normalizes" the vector.

For the vector \( \mathbf{a} = -4\mathbf{i} + 5\mathbf{j} \) in the problem, its unit vector \( \mathbf{u} \) is found by first calculating the magnitude of \( \mathbf{a} \):

• \( \|\mathbf{a}\| = \sqrt{(-4)^2 + 5^2} = \sqrt{41} \)

Then, each component of \( \mathbf{a} \) is divided by this magnitude, resulting in:

• \( \mathbf{u} = \left(-\frac{4}{\sqrt{41}}, \frac{5}{\sqrt{41}}\right) \)

This unit vector is critical for determining the directional derivative, as it ensures that the direction is purely directional and not influenced by the magnitude of \( \mathbf{a} \).

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. This operation is a cornerstone for determining angles between vectors and projecting vectors in directions defined by other vectors.

The dot product is calculated as the sum of the products of the corresponding entries of the two sequences of numbers. For vectors \( \mathbf{v} = (a, b) \) and \( \mathbf{w} = (c, d) \), the dot product \( \mathbf{v} \cdot \mathbf{w} \) is:

• \( ac + bd \)

This concept is applied when finding the directional derivative. Normally, the dot product between the gradient vector and a unit vector (which represents the direction of interest) is computed. As shown in the solution, in the form:

• \( D_{\mathbf{u}}f = abla f(0, \pi/6) \cdot \mathbf{u} \)
• The result provides a scalar value that reflects how sharply \( f \) changes in the specified direction.

Therefore, the dot product allows us to discern the effect of directions defined by different vectors and their interactions with function gradients.