Problem 283 - Calculus Volume 3

Question

For the following exercises, find the gradient. \(f(x, y, z)=4 x^{5} y^{2} z^{3}, P(2,-1,1), \quad \mathbf{u}=\frac{1}{3} \mathbf{i}+\frac{2}{3} \mathbf{j}-\frac{2}{3} \mathbf{k}\).

Step-by-Step Solution

Answer

The gradient is \((320, -256, 384)\) and the directional derivative is \(-320\).

Step 1: Understand the Gradient Concept

The gradient of a function \( f(x, y, z) \) of several variables is a vector that points in the direction of the greatest rate of increase of the function, and is composed of the partial derivatives with respect to each variable. It is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).

Step 2: Compute Partial Derivatives

To find \( abla f \), we need to calculate the partial derivatives of \( f(x, y, z) = 4x^5y^2z^3 \).- \( \frac{\partial f}{\partial x} = 20x^4y^2z^3 \)- \( \frac{\partial f}{\partial y} = 8x^5yz^3 \)- \( \frac{\partial f}{\partial z} = 12x^5y^2z^2 \)

Step 3: Evaluate the Gradient at Point P

Substitute \( P(2,-1,1) \) into the partial derivatives to find the gradient at this point:- \( \frac{\partial f}{\partial x} \bigg|_{P} = 20(2)^4(-1)^2(1)^3 = 320 \)- \( \frac{\partial f}{\partial y} \bigg|_{P} = 8(2)^5(-1)(1)^3 = -256 \)- \( \frac{\partial f}{\partial z} \bigg|_{P} = 12(2)^5(-1)^2(1)^2 = 384 \)So, \( abla f(P) = (320, -256, 384) \).

Step 4: Compute Directional Derivative

To find the directional derivative of \( f \) at \( P \) in the direction of a unit vector \( \mathbf{u} = (\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}) \), calculate \( abla f \cdot \mathbf{u} \):\[D_{\mathbf{u}} f = (320, -256, 384) \cdot \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right) \]Calculate the dot product:\(D_{\mathbf{u}} f = 320 \times \frac{1}{3} + (-256) \times \frac{2}{3} + 384 \times (-\frac{2}{3}) = \frac{320}{3} - \frac{512}{3} - \frac{768}{3} = -\frac{960}{3} = -320.\)

Key Concepts

Partial Derivatives

In the realm of multivariable calculus, partial derivatives are essential tools. When working with a function of several variables, like \( f(x, y, z) = 4x^5y^2z^3 \), partial derivatives help determine how the function changes as one of the variables changes, with the others held constant. Here, the partial derivative with respect to \( x \) is represented as \( \frac{\partial f}{\partial x} \), while keeping \( y \) and \( z \) constant.
For the given function, the process is straightforward:

• The partial derivative with respect to \( x \) is \( 20x^4y^2z^3 \), indicating change along the axis of \( x \).
• With respect to \( y \), it becomes \( 8x^5yz^3 \), measuring variations along \( y \).
• Finally, for \( z \), it results in \( 12x^5y^2z^2 \), showing influence along the \( z \)-axis.

Understanding these derivatives forms the base for further calculation of gradients and other vector calculus concepts.

Directional Derivative

Once the gradient is known, the next step is to translate this information to a specific direction using the directional derivative. This value provides insight into how swiftly the function \( f \) changes in a particular direction given by a unit vector \( \mathbf{u} \).
In this exercise, the unit vector \( \mathbf{u} = \left( \frac{1}{3}, \frac{2}{3}, -\frac{2}{3} \right) \) signifies the intended direction. The directional derivative \( D_{\mathbf{u}} f \) is calculated by performing the dot product of the gradient vector and \( \mathbf{u} \):

• Firstly, compute the gradient at the point \( P(2,-1,1) \) yielding \( abla f(P) = (320, -256, 384) \).
• Then, calculate the dot product: \( D_{\mathbf{u}} f = 320 \times \frac{1}{3} + (-256) \times \frac{2}{3} + 384 \times \left(-\frac{2}{3}\right) \).
• Upon simplifying, we find \( D_{\mathbf{u}} f = -320 \).

This negative value suggests that the function \( f \) decreases as we move in the direction of \( \mathbf{u} \).

Vector Calculus

Vector calculus is a branch of mathematics dealing with vector fields and operations such as differentiation and integration over these fields. In the context of function \( f(x, y, z) = 4x^5y^2z^3 \), this involves computing gradients and directional derivatives.
Key concepts include:

• **Gradient:** A vector in vector calculus that has numerous practical applications. It points in the direction of the steepest increase of the function and its magnitude represents the rate of ascent.
• **Dot Product:** Central to finding directional derivatives, the dot product measures how aligned two vectors are. A higher positive value indicates more significant alignment in terms of direction.

These foundations enable solving complex physical problems, such as determining how temperature changes in a region or the velocity of fluid in 3D space. Mastery of vector calculus allows for detailed analysis and understanding of multidimensional functions and systems.