Problem 280 - Calculus 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

Gradient at P(1,2) is \( \left( \frac{-2}{3}, \frac{-4}{3} \right) \).

Step 1: Identify the Function

We start with the function given: \( f(x, y) = \frac{14 - x^2 - y^2}{3} \). This is the function whose gradient we need to calculate.

Step 2: Compute the Partial Derivative with Respect to x

To find the gradient, first find the partial derivative of \( f(x, y) \) with respect to \( x \).Use the power rule and constant multiple rule:\[ \frac{\partial}{\partial x}\left( \frac{14 - x^2 - y^2}{3} \right) = \frac{-2x}{3} \]

Step 3: Compute the Partial Derivative with Respect to y

Next, compute the partial derivative of \( f(x, y) \) with respect to \( y \).Use the power rule and constant multiple rule:\[ \frac{\partial}{\partial y}\left( \frac{14 - x^2 - y^2}{3} \right) = \frac{-2y}{3} \]

Step 4: Form the Gradient Vector

The gradient \( abla f(x, y) \) is a vector consisting of the partial derivatives we found:\[ abla f(x, y) = \left( \frac{-2x}{3}, \frac{-2y}{3} \right) \]

Step 5: Evaluate the Gradient at Point P(1, 2)

To find the gradient at the point \( P(1, 2) \), substitute \( x = 1 \) and \( y = 2 \) into the gradient vector:\[ abla f(1, 2) = \left( \frac{-2(1)}{3}, \frac{-2(2)}{3} \right) = \left( \frac{-2}{3}, \frac{-4}{3} \right) \]

Key Concepts

Partial Derivatives

Partial derivatives are like regular derivatives but for functions with more than one variable. In simple terms, they measure how a function changes as you tweak just one of the input variables, keeping the others constant.
For our function, which is defined as \( f(x, y) = \frac{14 - x^2 - y^2}{3} \), partial derivatives help us understand how changes in \( x \) or \( y \) alone affect \( f(x, y) \).

• Partial derivative with respect to \( x \): focuses on the impact of changes in \( x \) while keeping \( y \) constant. For this function, it is \( \frac{-2x}{3} \).
• Partial derivative with respect to \( y \): looks at the influence of changes in \( y \) while keeping \( x \) constant. For this function, it is \( \frac{-2y}{3} \).

Partial derivatives are foundational in multivariable calculus and crucial for forming other important concepts like the gradient.

Gradient Vector

The gradient vector is a vital concept in calculus involving functions with several variables. Essentially, it is a multi-directional derivative, combining the partial derivatives of a function into a single vector.
For the function \( f(x, y) = \frac{14 - x^2 - y^2}{3} \), the gradient vector \( abla f(x, y) \) is built from its partial derivatives. It looks like this: \( \left( \frac{-2x}{3}, \frac{-2y}{3} \right) \).

The gradient points in the direction of the steepest ascent of the function—think of it like a compass showing where the hill gets steeper fastest.

• The first component \( \frac{-2x}{3} \) represents how \( f \) changes with variations in \( x \).
• The second component \( \frac{-2y}{3} \) shows how \( f \) changes with changes in \( y \).

By finding these vector components, you gather all the information about how the function changes with respect to both variables. Evaluating the gradient at a specific point, like \( P(1, 2) \), provides the precise direction of maximum increase at that location.

Function of Two Variables

A function of two variables, such as \( f(x, y) = \frac{14 - x^2 - y^2}{3} \), assigns a single output value to each pair of input values \( (x, y) \). Imagine it as a surface over a 2D plane where every point on the surface corresponds to values \( x \) and \( y \).

These functions can depict real-world relationships where outcomes depend on two changing inputs.

• In this case, as \( x \) and \( y \) vary, \( f(x, y) \) describes how the surface height changes.
• Functions of two variables are essential in fields like physics, economics, and engineering to model systems with interconnected influencing factors.

Understanding such functions allows you to analyze situations where multiple variables interact, making it possible to predict outcomes under different circumstances. With the partial derivatives and gradient, you gain insight into how the function behaves locally around each point in its domain.