Problem 9 - 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 \(\left(\frac{1}{x^{2}}+\frac{1}{y^{2}}\right) \mathrm{i}+\left(\frac{1-2 x}{y^{3}}\right) \mathrm{j}\)

Step-by-Step Solution

Answer

The vector is a gradient; the potential function is \(\phi(x,y) = -\frac{1}{x} + \frac{x}{y^{2}} + \frac{1}{2y^{2}} + C\).

Step 1: Identify Gradient Conditions

A vector field \(F = M(x,y)\bf{i} + N(x,y)\bf{j}\) is a gradient if \(\frac{\partial N}{\partial x} = \frac{\partial M}{\partial y}\). Here, \(M(x,y) = \frac{1}{x^{2}} + \frac{1}{y^{2}}\) and \(N(x,y) = \frac{1-2x}{y^{3}}\). Calculate these partial derivatives.

Step 2: Compute Partial Derivatives

Calculate the partial derivative of \(M(x,y) = \frac{1}{x^{2}} + \frac{1}{y^{2}}\) with respect to \(y\): \(\frac{\partial M}{\partial y} = \frac{\partial }{\partial y}\left(\frac{1}{x^{2}} + \frac{1}{y^{2}}\right) = -2y^{-3} = -\frac{2}{y^{3}}\). Compute the partial derivative of \(N(x,y) = \frac{1 - 2x}{y^{3}}\) with respect to \(x\): \(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( \frac{1 - 2x}{y^{3}} \right) = -\frac{2}{y^{3}}\).

Step 3: Verify Gradient Condition

Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} = -\frac{2}{y^{3}}\), the vector field is a gradient.

Step 4: Integrate to Find Potential Function

Integrate \(\frac{1}{x^{2}} + \frac{1}{y^{2}}\) with respect to \(x\): \(\phi(x,y) = \int \left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) \, dx = -\frac{1}{x} + \frac{x}{y^{2}} + h(y)\).

Step 5: Determine \(h(y)\) by Differentiation

Differentiate \(\phi(x,y) = -\frac{1}{x} + \frac{x}{y^{2}} + h(y)\) with respect to \(y\) and set equal to \(\frac{1-2x}{y^{3}}\). \(\frac{\partial \phi}{\partial y} = -\frac{2x}{y^{3}} + h'(y) = \frac{1-2x}{y^{3}}\). Thus, \(\frac{2}{y^{3}} + h'(y) = \frac{1}{y^{3}}\), so \(h'(y) = -\frac{1}{y^{3}}\). Integrate to find \(h(y)\): \(h(y) = \int -\frac{1}{y^{3}} \, dy = \frac{1}{2y^{2}} + C\).

Step 6: Write Final Potential Function

The function \(\phi(x,y)\) is \(\phi(x,y) = -\frac{1}{x} + \frac{x}{y^{2}} + \frac{1}{2y^{2}} + C\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They represent the rate of change of a function with respect to one variable while keeping the other variables constant. In the given exercise, we have two partial derivatives to compute.

Consider the function components: \(M(x, y) = \frac{1}{x^{2}} + \frac{1}{y^{2}}\) and \(N(x, y) = \frac{1 - 2x}{y^{3}}\).

To find the partial derivative of \(M(x, y)\) with respect to \(y\), keep \(x\) constant and differentiate:

\(\frac{\partial M}{\partial y} = \frac{\partial }{\partial y}\left(\frac{1}{x^{2}} + \frac{1}{y^{2}}\right) = -2y^{-3} = -\frac{2}{y^{3}}\).

For \(N(x, y)\), differentiate with respect to \(x\), keeping \(y\) constant:

\(\frac{\partial N}{\partial x} = \frac{\partial}{\partial x} \left( \frac{1 - 2x}{y^{3}} \right) = -\frac{2}{y^{3}}\).

Understanding partial derivatives is essential for analyzing how functions change with multiple variables. It helps in identifying gradients and solving problems in fields like physics and engineering.

Potential Function

A potential function, often denoted by \(\phi\), is a scalar function from which a vector field can be derived. If a vector field \(F\) is a gradient field, then there exists a potential function such that the gradient of \(\phi\) equals \(F\).

In the given exercise, we determined that the vector field is a gradient field. To find the potential function, we integrated the components of the vector field with respect to the corresponding variables.

Start with integrating \(\frac{1}{x^{2}} + \frac{1}{y^{2}}\) with respect to \(x\):

\(\begin{aligned} \phi(x,y) &= \int \left( \frac{1}{x^{2}} + \frac{1}{y^{2}} \right) \, dx \ &= -\frac{1}{x} + \frac{x}{y^{2}} + h(y) \end{aligned}\)

Here, \(h(y)\) is an unknown function of \(y\). To determine \(h(y)\), we differentiate \(\phi(x,y)\) with respect to \(y\) and set it equal to \(\frac{1-2x}{y^{3}}\). This process reveals \(h(y) = \frac{1}{2y^{2}} + C\).

The final potential function is:

\(\phi(x,y) = -\frac{1}{x} + \frac{x}{y^{2}} + \frac{1}{2y^{2}} + C\).

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of more than one variable. It involves techniques like partial derivatives, multiple integrals, and vector calculus.

In the given exercise, we handle functions of two variables, \(x\) and \(y\). We compute partial derivatives to check if the given vector field is a gradient field. This involves:

* Differentiating with respect to one variable while keeping the others constant.
* Ensuring the gradient condition is met by comparing mixed partial derivatives.

Multivariable calculus also includes finding potential functions for gradient fields. This requires integration of the vector field components.

Mastering these techniques is crucial for applications in various scientific domains like physics, engineering, and economics.

Gradient Condition

The gradient condition ensures that a vector field is the gradient of some potential function. For a vector field \(F = M(x,y)\mathbf{i} + N(x,y)\mathbf{j}\), the condition is:

\( \frac{\partial N}{\partial x} = \frac{\partial M}{\partial y} \)

In the exercise, we verified this condition by computing the partial derivatives:

\( \frac{\partial M}{\partial y} = -\frac{2}{y^{3}} \) and \( \frac{\partial N}{\partial x} = -\frac{2}{y^{3}} \)

Since these are equal, the vector field is confirmed to be a gradient field.

This allows us to proceed with finding a potential function by integrating the components of the vector field.

Understanding the gradient condition is key to identifying gradient fields and solving related problems in multivariable calculus.