Problem 280 - Calculus Volume 3

Question

For the following exercises, find the gradient. Find the gradient of \(f(x, y)=\frac{14-x^{2}-y^{2}}{3}\). Then, find the gradient at point \(P(1,2)\).

Step-by-Step Solution

Answer

The gradient of \(f(x, y)\) at point \(P(1,2)\) is \((-\frac{2}{3}, -\frac{4}{3})\).

Step 1: Understanding the Gradient

The gradient of a function is a vector of its partial derivatives. It points in the direction of the greatest rate of increase of the function.

Step 2: Find the Partial Derivative with Respect to x

Differentiate the function with respect to the variable \( x \). For \( f(x, y)=\frac{14-x^{2}-y^{2}}{3} \), calculate \( \frac{\partial f}{\partial x} = \frac{-2x}{3} \).

Step 3: Find the Partial Derivative with Respect to y

Differentiate the function with respect to the variable \( y \). For \( f(x, y)=\frac{14-x^{2}-y^{2}}{3} \), calculate \( \frac{\partial f}{\partial y} = \frac{-2y}{3} \).

Step 4: Construct the Gradient Vector

Combine the partial derivatives into a gradient vector: \( abla f(x, y) = \left( \frac{-2x}{3}, \frac{-2y}{3} \right) \).

Step 5: Evaluate the Gradient at the Given Point

Substitute \( x = 1 \) and \( y = 2 \) into the gradient \( abla f(x, y) \) to find the gradient at point \( P(1, 2) \). \( abla f(1, 2) = \left( \frac{-2(1)}{3}, \frac{-2(2)}{3} \right) = \left( -\frac{2}{3}, -\frac{4}{3} \right) \).

Key Concepts

Partial Derivative

Partial derivatives are an essential concept in calculus, particularly when dealing with functions of several variables. Imagine you have a surface in three-dimensional space, defined by a function like \(f(x, y)\). To understand how the function behaves as we change just one variable at a time, we use partial derivatives.
For the function \(f(x, y) = \frac{14-x^{2}-y^{2}}{3}\), the partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), measures the rate at which \(f\) changes as \(x\) changes while keeping \(y\) constant. Similarly, the partial derivative with respect to \(y\), \(\frac{\partial f}{\partial y}\), measures the rate of change with \(y\) while \(x\) stays constant.

• For \(\frac{\partial f}{\partial x}\), differentiate the function and get \(\frac{-2x}{3}\).
• For \(\frac{\partial f}{\partial y}\), differentiate to find \(\frac{-2y}{3}\).

These partial derivatives are then combined into a vector known as the gradient.

Vector Calculus

Vector calculus extends all the basics of calculus to vector fields, allowing us to explore functions with multiple inputs and outputs.
The gradient, in particular, is a vector that shows the direction and rate of fastest increase of a function. It's composed of the partial derivatives, making it a fundamental concept in vector calculus. For \(f(x, y) = \frac{14-x^{2}-y^{2}}{3}\), the gradient \(abla f\) is \(\left( \frac{-2x}{3}, \frac{-2y}{3} \right)\).

• The gradient vector gives both magnitude and direction.
• It's essential for understanding multidimensional surfaces and fields.

When the gradient is calculated for a specific point, it reveals how the function behaves locally at that point.

Directional Derivative

The directional derivative is another vector calculus concept that supplements our understanding of gradients. While the gradient points to the maximum rate of increase, the directional derivative measures how a function changes in any given direction.
Suppose you want to know the rate of change of \(f(x, y)\) in a direction specified by a vector \(\mathbf{u}\). You would compute the dot product of the gradient \(abla f\) and your unit direction vector \(\mathbf{u}\).
Some key points about directional derivatives include:

• They provide a flexible tool to analyze changes in various directions.
• The result tells us how steep or flat the surface is in the chosen direction.

Thus, understanding the directional derivative can provide more detailed insights beyond what the gradient alone can offer.

Multivariable Calculus

Multivariable Calculus deals with functions that have more than one input variable. It's a natural extension of single-variable calculus, opening the door to understanding complex systems and surfaces.
Here, a central challenge is to generalize concepts from basic calculus, such as derivatives and integrals, into a world where variables operate simultaneously.

• Functions like \(f(x, y) = \frac{14-x^{2}-y^{2}}{3}\) are commonplace in this field.
• Partial derivatives help clarify how each individual variable affects the overall function.

With multivariable calculus, you can tackle real-world applications in physics, engineering, and economics by examining the interactions between variables and their collective effects.