Problem 14 - APEX Calculus

Question

A function \(z=f(x, y)\) and a point \(P\) are given. Find the directional derivative of \(f\) in the indicated directions. Note: these are the same functions as in Exercises 7 through 12. \(f(x, y)=\sin x \cos y, P=\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\) (a) In the direction of \(\vec{v}=\langle 1,1\rangle .\) (b) In the direction toward the point \(Q=(0,0)\).

Step-by-Step Solution

Answer

(a) \( \frac{1 - \sqrt{3}}{4} \). (b) \( \frac{4\sqrt{6} - 3\sqrt{2}}{20} \).

Step 1: Find the Gradient of f

The gradient of a function, represented as \( abla f(x, y) \), is the vector of its partial derivatives. For the function \( f(x, y) = \sin x \cos y \), the partial derivatives are:\( f_x = \frac{\partial}{\partial x}(\sin x \cos y) = \cos x \cos y \) \( f_y = \frac{\partial}{\partial y}(\sin x \cos y) = -\sin x \sin y \) Thus, \( abla f(x, y) = \langle \cos x \cos y, -\sin x \sin y \rangle \).

Step 2: Evaluate Gradient at P

Evaluate the gradient at the point \( P = \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \):\( abla f\left( \frac{\pi}{4}, \frac{\pi}{3} \right) = \langle \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{3}\right), -\sin\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{3}\right) \rangle \)Using trigonometric values,\( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}, \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)Thus, the gradient at P is \( abla f\left( \frac{\pi}{4}, \frac{\pi}{3} \right) = \langle \frac{\sqrt{2}}{2} \cdot \frac{1}{2}, -\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \rangle = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \).

Step 3: Directional Derivative in Direction of \(\vec{v}\)

The directional derivative of \( f \) in the direction of a vector \( \vec{v} \) is defined as:\[ D_{\vec{v}}f = abla f \cdot \frac{\vec{v}}{\|\vec{v}\|} \]First, normalize \( \vec{v} = \langle 1, 1 \rangle \):\[ \|\vec{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2} \]\( \frac{\vec{v}}{\|\vec{v}\|} = \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \)Now, compute the dot product:\( D_{\vec{v}}f = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \cdot \left\langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = \frac{\sqrt{2}}{4} \cdot \frac{1}{\sqrt{2}} + (-\frac{\sqrt{6}}{4}) \cdot \frac{1}{\sqrt{2}} \)\[ D_{\vec{v}}f = \frac{1}{4} - \frac{\sqrt{3}}{4} = \frac{1 - \sqrt{3}}{4} \]

Step 4: Directional Derivative Toward Point Q

For the direction toward point \( Q = (0,0) \) from \( P = \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \), the direction vector is \( \vec{u} = \left\langle 0 - \frac{\pi}{4}, 0 - \frac{\pi}{3} \right\rangle = \left\langle -\frac{\pi}{4}, -\frac{\pi}{3} \right\rangle \).Normalize \( \vec{u} \):\[ \|\vec{u}\| = \sqrt{\left(-\frac{\pi}{4}\right)^2 + \left(-\frac{\pi}{3}\right)^2} = \sqrt{\frac{\pi^2}{16} + \frac{\pi^2}{9}} = \sqrt{\frac{25\pi^2}{144}} = \frac{5\pi}{12} \]\( \frac{\vec{u}}{\|\vec{u}\|} = \left\langle -\frac{3}{5}, -\frac{4}{5} \right\rangle \) (from simplifying \( \frac{\left(-\frac{\pi}{4}\right)}{\frac{5\pi}{12}} \) and \( \frac{\left(-\frac{\pi}{3}\right)}{\frac{5\pi}{12}} \))Now compute the dot product with the gradient:\( D_{\vec{u}}f = \langle \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \rangle \cdot \langle -\frac{3}{5}, -\frac{4}{5} \rangle \)\[ D_{\vec{u}}f = \frac{\sqrt{2}}{4} \cdot -\frac{3}{5} + \left(-\frac{\sqrt{6}}{4}\right) \cdot -\frac{4}{5} = -\frac{3\sqrt{2}}{20} + \frac{4\sqrt{6}}{20} \]\[ D_{\vec{u}}f = \frac{4\sqrt{6} - 3\sqrt{2}}{20} \]

Key Concepts

Gradient

The concept of a gradient is central in understanding directional derivatives. The gradient of a function, denoted as \( abla f(x, y) \), is essentially a vector that combines the partial derivatives of a function. For a function like \( f(x, y) = \sin x \cos y \), the gradient gives the direction of the steepest increase.

To find the gradient, you take the derivative of \( f \) with respect to each variable separately:

• First, find the partial derivative with respect to \( x \): \( f_x = \frac{\partial}{\partial x}(\sin x \cos y) = \cos x \cos y \).
• Next, find the partial derivative with respect to \( y \): \( f_y = \frac{\partial}{\partial y}(\sin x \cos y) = -\sin x \sin y \).

Therefore, the gradient is \( abla f(x, y) = \langle \cos x \cos y, -\sin x \sin y \rangle \). It points in the direction where \( f \) increases most rapidly. The magnitude of the gradient vector acts as a measure of how fast \( f \) is increasing.

Partial Derivatives

Partial derivatives are an essential part of working with functions of several variables. When calculating a partial derivative, you are focusing on how the function changes as you slightly vary only one of its variables, keeping the others constant.

For example, with the function \( f(x, y) = \sin x \cos y \):

• The partial derivative with respect to \( x \), denoted as \( f_x \), measures the rate of change of \( f \) along the \( x \)-axis. In this case, we treat \( y \) as a constant, resulting in \( f_x = \cos x \cos y \).
• Similarly, the partial derivative with respect to \( y \), denoted as \( f_y \), measures the rate of change of \( f \) along the \( y \)-axis, giving us \( f_y = -\sin x \sin y \) when \( x \) is treated as constant.

Understanding partial derivatives helps in finding the gradient of a function, which is crucial for finding directional derivatives.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are fundamental in mathematics. They often appear in problems relating to angles and periodic phenomena. In the context of this problem, they're used to express the function \( f(x, y) = \sin x \cos y \), where both \( \sin x \) and \( \cos y \) contribute to the behavior of \( f \).

These functions have specific values at commonly used angles, such as \( \frac{\pi}{4} \) and \( \frac{\pi}{3} \), which can simplify calculations:

• \( \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• \( \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \)
• \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \)

These values simplify the evaluation of derivatives at specific points, which is crucial for calculating the directional derivative effectively.