Problem 27 - 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 \(\vec{v}\). $$ F(x, y, z)=x^{2} y^{2}-y^{2} z^{2}, \vec{v}=\langle-1,7,3\rangle, P=(1,0,-1) $$

Step-by-Step Solution

Answer

(a) \( \nabla F(x, y, z) = \left\langle 2xy^2, 2x^2y - 2yz^2, -2y^2z \right\rangle \), (b) \( D_{\vec{u}} F = 0 \).

Step 1: Find Partial Derivatives of F

To find the gradient \( abla F(x, y, z) \), we need to compute the partial derivatives of the function \( F(x, y, z) = x^2 y^2 - y^2 z^2 \) with respect to \( x \), \( y \), and \( z \). - The partial derivative with respect to \( x \) is: \( \frac{\partial F}{\partial x} = 2xy^2 \).- The partial derivative with respect to \( y \) is: \( \frac{\partial F}{\partial y} = 2x^2y - 2yz^2 \).- The partial derivative with respect to \( z \) is: \( \frac{\partial F}{\partial z} = -2y^2z \).

Step 2: Form the Gradient Vector

The gradient of \( F \) is denoted by \( abla F \) and is given by the vector of the partial derivatives:\[ abla F(x, y, z) = \left\langle 2xy^2, 2x^2y - 2yz^2, -2y^2z \right\rangle \].

Step 3: Evaluate \( \nabla F \) at Point P

Substitute the point \( P = (1, 0, -1) \) into \( abla F(x, y, z) \):\[ abla F(1, 0, -1) = \left\langle 2(1)(0)^2, 2(1)^2(0) - 2(0)(-1)^2, -2(0)^2(-1) \right\rangle = \left\langle 0, 0, 0 \right\rangle \].

Step 4: Determine the Unit Vector \( \vec{u} \)

The unit vector \( \vec{u} \) in the direction of \( \vec{v} = \langle -1, 7, 3 \rangle \) is found by dividing \( \vec{v} \) by its magnitude:1. Calculate the magnitude of \( \vec{v} \): \[ ||\vec{v}|| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59} \].2. The unit vector \( \vec{u} \) is: \[ \vec{u} = \frac{1}{\sqrt{59}} \langle -1, 7, 3 \rangle \].

Step 5: Compute Directional Derivative at Point P

The directional derivative \( D_{\vec{u}} F \) at point \( P \) is the dot product of \( abla F(P) \) and \( \vec{u} \):\[ D_{\vec{u}} F = abla F(P) \cdot \vec{u} = \left\langle 0, 0, 0 \right\rangle \cdot \left\langle \frac{-1}{\sqrt{59}}, \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}} \right\rangle = 0 \].

Key Concepts

Partial Derivatives

Partial derivatives are foundational concepts in calculus for functions of multiple variables. They represent the rate of change of the function with respect to one variable, while keeping the other variables constant. Imagine you have a surface, like a hilly landscape described by a function, and you want to understand how steep the hill is if you walk straight along the x-axis or y-axis.
To find the partial derivative of a function with respect to a particular variable, you differentiate the function treating all other variables as constants. For example, given a function like \( F(x, y, z) = x^2 y^2 - y^2 z^2 \), the partial derivatives are:

• With respect to \( x \): \( \frac{\partial F}{\partial x} = 2xy^2 \)
• With respect to \( y \): \( \frac{\partial F}{\partial y} = 2x^2y - 2yz^2 \)
• With respect to \( z \): \( \frac{\partial F}{\partial z} = -2y^2z \)

These derivatives help us form the gradient vector, which is crucial for understanding how our function changes in space.

Directional Derivative

The directional derivative is an extension of the concept of partial derivatives. It measures how a function changes as you move in any arbitrary direction, not just along the axes. This is especially useful in real-world scenarios where movements are rarely aligned with coordinate axes.
To calculate the directional derivative of a function at a given point in a specific direction, you need:

• The gradient vector, \( abla F \), at the point of interest, which provides the direction and rate of steepest ascent.
• A unit vector, \( \vec{u} \), which specifies the direction in which you want to assess the rate of change.

The directional derivative \( D_{\vec{u}} F \) is then computed as the dot product of the gradient vector and the unit vector. In our example, if the gradient \( abla F(P) = \langle 0, 0, 0 \rangle \), then regardless of the direction, the directional derivative will be zero.
This tells us that at point \( P \), the function isn't increasing or decreasing, indicating a flat or critical point.

Unit Vector

A unit vector is a vector that has a length or magnitude of exactly one. It is used to specify a direction without scaling up the vector's effect. This is essential when calculating directional derivatives because we want to measure the function's rate of change purely due to direction, not the magnitude.
To find the unit vector in the direction of a non-zero vector \( \vec{v} \), you divide the vector by its magnitude:

• Calculate the magnitude: \( ||\vec{v}|| = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{59} \).
• The unit vector \( \vec{u} \) is then: \( \vec{u} = \frac{1}{\sqrt{59}} \langle -1, 7, 3 \rangle \).

Using unit vectors allows us to normalize the direction, enabling us to accurately compute things like the directional derivative no matter the original vector's magnitude. This versatility makes unit vectors indispensable in vector calculus.