Problem 1 - The Calculus with Analytic Geometry

Question

In the following exercises, determine if the vector is a gradient. If it is, find a function having the given gradient \(4 x \mathbf{i}-3 y \mathbf{j}\)

Step-by-Step Solution

Answer

Yes, the vector is a gradient. The potential function is \[f(x,y) = 2x^2 - \frac{3}{2} y^2 + C\].

Step 1 - Identify the components of the vector

The given vector is \[4x \, \mathbf{i} - 3y \, \mathbf{j}\]. The components are:\[F_1 = 4x\] and \[F_2 = -3y\]

Step 2 - Integrate the first component

To check if the vector is a gradient, integrate \[F_1 = 4x\] with respect to \[x\]:\[\frac{\partial f}{\partial x} = 4x\]\[f(x,y) = \int 4x \, dx = 2x^2 + g(y)\]where \[g(y)\] is an unknown function of \[y\].

Step 3 - Determine \[g(y)\] using the second component

Next, use \[F_2 = -3y\] to find \[g(y)\]. We have:\[\frac{\partial f}{\partial y} = -3y\]Taking the partial derivative of \[f(x,y) = 2x^2 + g(y)\] with respect to \[y\], we get:\[\frac{\partial f}{\partial y} = g'(y) = -3y\]Integrate \[g'(y) = -3y\] with respect to \[y\]:\[g(y) = \int -3y \, dy = -\frac{3}{2} y^2 + C\]where \[C\] is a constant.

Step 4 - Form the potential function

Combine the results to form the potential function \[f(x,y)\]:\[f(x,y) = 2x^2 - \frac{3}{2} y^2 + C\]

Key Concepts

Vector Calculus

In vector calculus, we study vector fields and how vectors change in space. Let's talk about one crucial idea: the gradient. A vector field represents how a vector quantity varies in space. In the exercise above, we have the vector \(4x \mathbf{i} -3y \mathbf{j}\). The question is whether this vector can be written as the gradient of some function.The gradient of a scalar function \(f(x,y)\) is a vector made up of its partial derivatives. Mathematically, the gradient \(abla f\) in two-dimensional space is written as:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\]In other words, we differentiate the function \(f\) with respect to each variable to get the components of the gradient vector. In vector calculus, finding the gradient tells us which way the function \(f\) increases the fastest.

Partial Derivatives

Partial derivatives help us understand how a multivariable function changes when we slightly adjust one of its variables. In the given problem, we calculated the partial derivatives of a function \(f(x,y)\) to check if it matches the given vector \(4x \mathbf{i} -3y \mathbf{j}\).Let's break it down:

• First, we found \(F_1 = 4x\) which represents the partial derivative of \(f\) with respect to \(x\).
• Then, we integrated \(F_1\) with respect to \(x\) to find a part of \(f\).
• Next, we looked at \(F_2 = -3y\) which represents the partial derivative of \(f\) with respect to \(y\).
• We then took the derivative of our function with respect to \(y\) to find the unknown part \(g(y)\) and integrated it.

By integrating these partial derivatives, we formed a potential function for the given vector.

Potential Function

A potential function \(f(x,y)\) is a scalar function whose gradient matches the given vector field. In this exercise, we wanted to find a function \(f\) such that \(abla f\) equals \(4x\mathbf{i} -3y \mathbf{j}\).Here’s a recap:

• We integrated \(4x\) with respect to \(x\), creating part of the function \(f(x,y) = 2x^2 + g(y)\).
• To get \(g(y)\), we differentiated this partial result with respect to \(y\), then set it equal to \(F_2 = -3y\).
• By integrating \(g'(y) = -3y\), we got \(g(y) = -\frac{3}{2} y^2 + C\).

Combining these results, our potential function became:\[f(x,y) = 2x^2 - \frac{3}{2} y^2 + C\]This function tells us the original vector field is indeed a gradient, as the function's gradient matches the vector components.