Question

Show that the following functions are independent of path in the \(x y\)-plane and evaluate them: a) \(\int_{(1,1)}^{(x, y)} 2 x y d x+\left(x^{2}-y^{2}\right) d y\) b) \(\int_{(0,0)}^{(x, y)} \sin y d x+x \cos y d y\).

Short answer

- The differential form \(M dx + N dy\) is exact if and only if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). 2. What are the two given functions for which we need to check if they are independent of path in the \(xy\)-plane? - The two given functions are: a. \(\int_{(1,1)}^{(x, y)} 2 x y d x+\left(x^{2}-y^{2}\right) d y\) b. \(\int_{(0,0)}^{(x, y)} \sin y d x+x \cos y d y\) 3. What are the potential functions obtained for each integrand? - For the first integrand: \(f(x,y) = x^2y - \frac{1}{3}y^3\) - For the second integrand: \(f(x,y) = x\sin y\) 4. Determine the values of the definite integrals for both functions. - For the first function: \(\int_{(1,1)}^{(x, y)} 2 x y d x+\left(x^{2}-y^{2}\right) d y = x^2y - \frac{1}{3}y^3- \frac{2}{3}\) - For the second function: \(\int_{(0,0)}^{(x, y)} \sin y d x+x \cos y d y = x\sin y\)

Step-by-step solution

Check if the integrand is an exact differential form

We need to check whether: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), where \(M = 2xy\) and \(N = x^2 - y^2\). Compute the partial derivatives: \(\frac{\partial M}{\partial y} = 2x\) \(\frac{\partial N}{\partial x} = 2x\) Since both partial derivatives are equal, the integrand is an exact differential form.

Find the potential function

Since the integrand is an exact differential form, a potential function \(f(x,y)\) exists. We can find it by integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\): \(\int M dx = \int 2xy dx = x^2y + C_1(y)\) \(\int N dy = \int (x^2 - y^2) dy = x^2y - \frac{1}{3}y^3 + C_2(x)\) Now, we need to find the constants \(C_1(y)\) and \(C_2(x)\). Since the integrals should be the same, we can equate the expressions: \(x^2y + C_1(y) = x^2y - \frac{1}{3}y^3 + C_2(x)\) Comparing both sides, we have: \(C_1(y) = -\frac{1}{3}y^3\) \(C_2(x) = 0\) So, the potential function is: \(f(x,y) = x^2y - \frac{1}{3}y^3\)

Evaluate the integral

Let's evaluate the definite integral: \(\int_{(1,1)}^{(x, y)} 2 x y d x+\left(x^{2}-y^{2}\right) d y = f(x,y) - f(1,1)\) Substitute the values of \((x, y)\) and \((1, 1)\) in the potential function: \(f(x, y) - f(1,1) = (x^2y - \frac{1}{3}y^3) - (1^2\cdot 1 - \frac{1}{3}1^3) = x^2y - \frac{1}{3}y^3 - \frac{2}{3}\) So, the integral is: \(\int_{(1,1)}^{(x, y)} 2 x y d x+\left(x^{2}-y^{2}\right) d y = x^2y - \frac{1}{3}y^3- \frac{2}{3}\) ## Part b ##

Check if the integrand is an exact differential form

We need to check whether: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), where \(M = \sin y\) and \(N = x\cos y\). Compute the partial derivatives: \(\frac{\partial M}{\partial y} = \cos y\) \(\frac{\partial N}{\partial x} = \cos y\) Since both partial derivatives are equal, the integrand is an exact differential form.

Find the potential function

Since the integrand is an exact differential form, a potential function \(f(x,y)\) exists. We can find it by integrating \(M\) with respect to \(x\) and \(N\) with respect to \(y\): \(\int M dx = \int \sin y dx = x \sin y + C_1(y)\) \(\int N dy = \int x\cos y dy = x \sin y + C_2(x)\) Now, we need to find the constants \(C_1(y)\) and \(C_2(x)\). Since the integrals should be the same, we can equate the expressions: \(x\sin y + C_1(y) = x\sin y + C_2(x)\) Comparing both sides, we have: \(C_1(y) = 0\) \(C_2(x) = 0\) So, the potential function is: \(f(x,y) = x\sin y\)

Evaluate the integral

Let's evaluate the definite integral: \(\int_{(0,0)}^{(x, y)} \sin y d x+x \cos y d y = f(x,y) - f(0,0)\) Substitute the values of \((x, y)\) and \((0, 0)\) in the potential function: \(f(x, y) - f(0,0) = (x\sin y) - (0\sin 0) = x\sin y\) So, the integral is: \(\int_{(0,0)}^{(x, y)} \sin y d x+x \cos y d y = x\sin y\)

Key concepts

Exact Differential Forms

When we discuss exact differential forms in mathematics and especially in calculus, we refer to a specific type of differential equation that can be expressed in a manner where its integrability conditions are satisfied. In simpler terms, a differential form \( P(x,y)dx + Q(x,y)dy \) is considered exact if there is a potential function, \( f(x, y) \), such that the differential form is exactly the derivative of this potential function.

For a form to be exact in two dimensions, we use a particular condition: the mixed partial derivatives should be equal. This means \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). Meeting this condition implies path independence. Therefore, the integral of the differential form between two points does not depend on the path taken but only on the endpoints.

In the problem provided, we demonstrated exactness by showing the equality of the partial derivatives. This concept helps in simplifying complex integrals by transforming them into easier tasks of finding potential functions.

Potential Functions

A potential function is a scalar function associated with a vector field, where the vector field can be expressed as the gradient of this potential function. In simpler terms, for a differential form \( P(x,y)dx + Q(x,y)dy \), if it is exact, there exists a function \( f(x, y) \) such that \( df = Pdx + Qdy \). This function \( f(x, y) \) is the potential function.

Finding a potential function involves integrating the components of the differential form. For example, integrating \( M \) with respect to \( x \) and \( N \) with respect to \( y \), ensures that the resulting function is consistent and satisfies the original differential form. In the exercise, we found the potential function \( f(x,y) \) for both parts a and b by carefully integrating and comparing consistency to identify the constants involved, such as \( C_1(y) \) and \( C_2(x) \).

Once identified, the potential function enables us to easily evaluate definite integrals via evaluating the difference of the function at the endpoints, reflecting how potential energy in physics only depends on initial and final states, not the transition path.

Partial Derivatives

Partial derivatives are at the heart of understanding multivariable functions and are crucial in verifying exactness. They represent the rate of change of a function with respect to one variable while holding the other variables constant. For a function \( f(x, y) \), the partial derivative \( \frac{\partial f}{\partial x} \) is the derivative relative to \( x \) while \( y \) is held constant, and vice versa for \( \frac{\partial f}{\partial y} \).

In the task of checking the exactness of a differential form, we compute and compare the partial derivatives. Specifically, we calculate \( \frac{\partial P}{\partial y} \) and \( \frac{\partial Q}{\partial x} \). If they equal each other, it indicates that the differential form is exact, providing assurance of path independence.

Understanding and applying the concept of partial derivatives allows evaluation of complex problems, ensuring that multi-variable functions are analyzed accurately. It’s through the analysis of partial derivatives that one verifies whether a potential function exists, thus simplifying further calculations.