Problem 26 - APEX Calculus

Question

A function \(w=F(x, y, z),\) a vector \(\vec{v}\) and a point \(P\) are given. (a) Find \(\nabla F(x, y, z)\). (b) Find \(D_{\vec{u}} F\) at \(P,\) where \(\vec{u}\) is the unit vector in the direction of \(\vec{v}\). $$ F(x, y, z)=\sin (x) \cos (y) e^{z}, \vec{v}=\langle 2,2,1\rangle, P=(0,0,0) $$

Step-by-Step Solution

Answer

\(\nabla F(0,0,0) = \langle 1,0,0 \rangle; D_{\vec{u}} F(0,0,0) = \frac{2}{3}\).

Step 1: Find the Gradient ∇F

The gradient of a function \(F(x, y, z)\) is given as \(abla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right)\). For \(F(x, y, z) = \sin(x) \cos(y) e^{z}\), calculate each partial derivative separately:1. \(\frac{\partial F}{\partial x} = \cos(x) \cos(y) e^{z}\)2. \(\frac{\partial F}{\partial y} = -\sin(x) \sin(y) e^{z}\)3. \(\frac{\partial F}{\partial z} = \sin(x) \cos(y) e^{z}\)Thus, \(abla F = \langle \cos(x) \cos(y) e^{z}, -\sin(x) \sin(y) e^{z}, \sin(x) \cos(y) e^{z} \rangle\).

Step 2: Evaluate ∇F at Point P

Substitute the point \(P = (0, 0, 0)\) into the gradient \(abla F\):1. \(\frac{\partial F}{\partial x}(0,0,0) = \cos(0) \cos(0) e^{0} = 1\)2. \(\frac{\partial F}{\partial y}(0,0,0) = -\sin(0) \sin(0) e^{0} = 0\)3. \(\frac{\partial F}{\partial z}(0,0,0) = \sin(0) \cos(0) e^{0} = 0\)Thus, the gradient at \(P\) is \(abla F (0, 0, 0) = \langle 1, 0, 0 \rangle\).

Step 3: Find the Unit Vector \(\vec{u}\) of \(\vec{v}\)

The unit vector \(\vec{u}\) in the direction of \(\vec{v} = \langle 2, 2, 1 \rangle\) is given by \(\vec{u} = \frac{\vec{v}}{\|\vec{v}\|} \). First, calculate the magnitude of \(\vec{v}\):\[ \|\vec{v}\| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3 \]Then, the unit vector is:\[ \vec{u} = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) \]

Step 4: Find the Directional Derivative \(D_{\vec{u}} F\) at P

The directional derivative of \(F\) in the direction of \(\vec{u}\) at point \(P\) is calculated using:\[ D_{\vec{u}} F = abla F \cdot \vec{u} \]We have \(abla F (0, 0, 0) = \langle 1, 0, 0 \rangle\) and \(\vec{u} = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\):\[ D_{\vec{u}} F = \langle 1, 0, 0 \rangle \cdot \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right) = 1 \cdot \frac{2}{3} + 0 \cdot \frac{2}{3} + 0 \cdot \frac{1}{3} = \frac{2}{3} \]

Step 5: Final Answer

The gradient \(abla F\) at \(P\) is \(\langle 1, 0, 0 \rangle\), and the directional derivative \(D_{\vec{u}} F\) at \(P\) is \(\frac{2}{3}\).

Key Concepts

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. It is an essential concept in multivariable calculus. To compute the gradient of a function of several variables, such as \( F(x, y, z) = \sin(x) \cos(y) e^{z} \), we need to find the partial derivatives of the function with respect to each of its variables.
The gradient \( abla F \) is expressed as a vector:

• \( \frac{\partial F}{\partial x} \): the rate of change of the function concerning \( x \).
• \( \frac{\partial F}{\partial y} \): the rate of change of the function concerning \( y \).
• \( \frac{\partial F}{\partial z} \): the rate of change of the function concerning \( z \).

Each component of the gradient vector corresponds to one of these partial derivatives. Therefore, the gradient \( abla F \) is \(\langle \cos(x) \cos(y) e^{z}, -\sin(x) \sin(y) e^{z}, \sin(x) \cos(y) e^{z} \rangle\).
The gradient provides us with a complete picture of how the function behaves in different directions around a point. At a given point, substituting the point's coordinates into the gradient yields a vector that indicates the direction and rate of steepest ascent from that point.

Partial Derivatives

Partial derivatives measure how a function changes as one of its input variables changes, while all others are held constant. They are the building blocks for understanding how multivariable functions behave.
When you take a partial derivative with respect to one variable, you treat all other variables as constants. This means that if you find the partial derivative \( \frac{\partial F}{\partial x} \), you consider \( y \) and \( z \) as constants. For the function \( F(x, y, z) = \sin(x) \cos(y) e^{z} \), the partial derivatives are computed as follows:

• \( \frac{\partial F}{\partial x} = \cos(x) \cos(y) e^{z} \)
• \( \frac{\partial F}{\partial y} = -\sin(x) \sin(y) e^{z} \)
• \( \frac{\partial F}{\partial z} = \sin(x) \cos(y) e^{z} \)

Each of these derivatives tells us how the function's value changes when we make a small change in one of the variables while keeping the others fixed.
Understanding partial derivatives is crucial for analyzing functions involving several variables, as they help build the gradient and, ultimately the directional derivative.

Unit Vector

A unit vector is a vector of length one and is used to indicate direction without regard to magnitude. In the context of directional derivatives, it gives direction to measure the rate of change of a function.
To find a unit vector \( \vec{u} \) in the direction of a given vector \( \vec{v} = \langle 2, 2, 1 \rangle \), you first calculate the magnitude (or length) of \( \vec{v} \):\[\|\vec{v}\| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{9} = 3\]Then, divide each component of \( \vec{v} \) by its magnitude to get the unit vector:\[\vec{u} = \left(\frac{2}{3}, \frac{2}{3}, \frac{1}{3}\right)\]The unit vector tells us the direction in which we want to measure the change in the function when calculating the directional derivative.
Unit vectors are fundamental in many areas of mathematics and physics, offering a simple way to work with directions and make vector calculations simpler.