Problem 17 - Advanced Engineering Mathematics

Question

In Problems, find the directional derivative of the given function at the given point in the indicated direction. $$ F(x, y, z)=x^{2} y^{2}(2 z+1)^{2} ;(1,-1,1),\langle 0,3,3\rangle $$

Step-by-Step Solution

Answer

Directional derivative is \(-3\sqrt{2}\).

Step 1: Calculate the Gradient of the Function

The gradient of the function \( F(x, y, z) = x^2 y^2 (2z+1)^2 \) is given by the vector of partial derivatives \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).First, find \( \frac{\partial F}{\partial x} = 2xy^2(2z+1)^2 \).Then, \( \frac{\partial F}{\partial y} = 2x^2y(2z+1)^2 \).Finally, \( \frac{\partial F}{\partial z} = 4x^2y^2(2z+1) \).Thus, the gradient vector is \( abla F = \left( 2xy^2(2z+1)^2, 2x^2y(2z+1)^2, 4x^2y^2(2z+1) \right) \).

Step 2: Evaluate the Gradient at the Given Point

Substitute the point \((1, -1, 1)\) into the gradient \( abla F \).The x-component is \( 2 \cdot 1 \cdot (-1)^2 \cdot (2 \cdot 1 + 1)^2 = 18 \).The y-component is \( 2 \cdot 1^2 \cdot (-1) \cdot (2 \cdot 1 + 1)^2 = -18 \).The z-component is \( 4 \cdot 1^2 \cdot (-1)^2 \cdot (2 \cdot 1 + 1) = 12 \).So, \( abla F(1, -1, 1) = \langle 18, -18, 12 \rangle \).

Step 3: Find the Unit Vector in the Direction

The given direction vector is \( \langle 0, 3, 3 \rangle \).Calculate its magnitude: \( \sqrt{0^2 + 3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \).The unit vector \( \mathbf{u} \) is obtained by dividing the direction vector \( \langle 0, 3, 3 \rangle \) by its magnitude: \( \mathbf{u} = \left\langle 0, \frac{3}{3\sqrt{2}}, \frac{3}{3\sqrt{2}} \right\rangle = \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \).

Step 4: Compute the Directional Derivative

The directional derivative \( D_{\mathbf{u}}F \) at the point in the direction of the unit vector \( \mathbf{u} \) is given by the dot product \( abla F(1, -1, 1) \cdot \mathbf{u} \).Calculate \( D_{\mathbf{u}}F = \langle 18, -18, 12 \rangle \cdot \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle = 18 \cdot 0 + (-18) \cdot \frac{1}{\sqrt{2}} + 12 \cdot \frac{1}{\sqrt{2}} = \frac{-18}{\sqrt{2}} + \frac{12}{\sqrt{2}} = \frac{-6}{\sqrt{2}} = -3\sqrt{2}. \)Thus, the directional derivative is \( -3\sqrt{2} \).

Key Concepts

Gradient

In calculus, the gradient provides a multi-variable function with direction and rate of change information. It is essentially a vector of partial derivatives. For a function like \( F(x, y, z) = x^2 y^2 (2z+1)^2 \), the gradient is represented as \( abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \).
This vector tells us how the function changes as each variable is adjusted independently.

• Visualize it: Picture the gradient as a force pointing in the direction of the greatest increase of the function.
• Components matter: Each component reveals how the function changes concerning one particular variable.

For our specific function, the gradient becomes \( abla F = \left( 2xy^2(2z+1)^2, 2x^2y(2z+1)^2, 4x^2y^2(2z+1) \right) \).
Calculating each component at a point like \((1, -1, 1)\) helps to understand how the function behaves exactly there. This interpretation of the gradient is crucial in various fields such as physics and engineering, where understanding the changes in multi-dimensional functions is pivotal.

Partial Derivatives

Partial derivatives allow us to understand how a multi-variable function changes concerning one variable while keeping others constant.
This is critical when dealing with functions like \( F(x, y, z) = x^2 y^2 (2z+1)^2 \), where we have multiple variables at play.

• Focused change: Think of partial derivatives as a zoom on how the function changes when only one specific variable changes.
• Building blocks: They form the components of the gradient vector, reflecting changes in each variable.

For instance, the partial derivative \( \frac{\partial F}{\partial x} = 2xy^2(2z+1)^2 \) tells us how \( F \) changes as \( x \) changes.
Similarly, \( \frac{\partial F}{\partial y} = 2x^2y(2z+1)^2 \) and \( \frac{\partial F}{\partial z} = 4x^2y^2(2z+1) \) provide specific information for changes in \( y \) and \( z \) respectively.
Understanding partial derivatives individually helps in constructing a comprehensive understanding of the function's overall behavior.

Unit Vector

Unit vectors are essential tools in directional derivatives as they help specify direction.
A unit vector has a magnitude of one and points in a given direction.

• Standardized direction: A unit vector allows us to measure direction without worrying about how long the vector is.
• Normalized vectors: Convert any vector to a unit vector by dividing it by its magnitude.

For example, given the direction vector \( \langle 0, 3, 3 \rangle \), its magnitude is \( 3\sqrt{2} \).
The corresponding unit vector \( \mathbf{u} \) becomes \( \left\langle 0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \).
This means we can use this unit vector to specify directions universally, without affecting our results negatively due to vector length.
Using unit vectors simplifies calculations, such as determining the directional derivative, ensuring precision and clarity in analysis.