Question

Which of the following are exact differentials? (a) \(6 x^{5} d x+d y\) (b) \(x^{2} y^{2} d x+3 x^{2} y^{3} d y\) (c) \((1 / y) d x-\left(x / y^{2}\right) d y\) (d) \(y d x+2 x d y\) (e) \(\cos x d x-\sin y d y\) (f) \(\left(x^{2}+y\right) d x+\left(x+y^{2}\right) d y\) (g) \(x d x+\sin y d y\)

Short answer

(a), (c), (e), (f), and (g) are exact differentials.

Step-by-step solution

Evaluate condition for exact differential

An expression of the form M(x,y)dx + N(x,y)dy is an exact differential if \ \ \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\). We need to check this condition for each given differential.

Check (a) \(6 x^{5} d x+d y\)

Here, M(x,y) = 6x^5 and N(x,y) = 1. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 0\) and \(\frac{\partial N}{\partial x} = 0\). These are equal, so it is an exact differential.

Check (b) \(x^{2} y^{2} d x+3 x^{2} y^{3} d y\)

Here, M(x,y) = x^2 y^2 and N(x,y) = 3x^2 y^3. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 2x^2 y\) and \(\frac{\partial N}{\partial x} = 6x y^3\). These are not equal, so it is not an exact differential.

Check (c) \((1 / y) d x - (x / y^{2}) d y\)

Here, M(x,y) = 1/y and N(x,y) = -x/y^2. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = -1/y^2\) and \(\frac{\partial N}{\partial x} = -1/y^2\). These are equal, so it is an exact differential.

Check (d) \(y d x + 2 x d y\)

Here, M(x,y) = y and N(x,y) = 2x. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 1\) and \(\frac{\partial N}{\partial x} = 2\). These are not equal, so it is not an exact differential.

Check (e) \(\cos x d x - \sin y d y\)

Here, M(x,y) = \cos x and N(x,y) = -\sin y. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 0\) and \(\frac{\partial N}{\partial x} = 0\). These are equal, so it is an exact differential.

Check (f) \((x^2 + y) d x + (x + y^2) d y\)

Here, M(x,y) = x^2 + y and N(x,y) = x + y^2. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 1\) and \(\frac{\partial N}{\partial x} = 1\). These are equal, so it is an exact differential.

Check (g) \(x d x + \sin y d y\)

Here, M(x,y) = x and N(x,y) = \sin y. Compute partial derivatives: \ \ \(\frac{\partial M}{\partial y} = 0\) and \(\frac{\partial N}{\partial x} = 0\). These are equal, so it is an exact differential.

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one particular variable changes, while keeping all other variables constant. If you have a function of two variables, say, \(f(x,y)\), there are two partial derivatives: \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).
The notation \(\frac{\partial f}{\partial x}\) represents the rate of change of \(f\) with respect to \(x\), while holding \(y\) constant. Conversely, \(\frac{\partial f}{\partial y}\) represents the rate of change with respect to \(y\), with \(x\) held constant.
Calculating partial derivatives is a lot like taking normal derivatives, but you treat other variables as if they are constants. For example in the function \(f(x, y) = x^2y + y^3\), the partial derivative with respect to \(x\), \(\frac{\partial f}{\partial x}\), would be \(2xy\), as \(y\) is treated as a constant. Whereas, for \(\frac{\partial f}{\partial y}\), it would be \(x^2 + 3y^2\) because \(x\) is treated as a constant.

Exact Differentials Condition

An exact differential is a specific kind of differential form in multivariable calculus. It indicates that a function is well-behaved and can be integrated without ambiguity between paths.
For a differential \( M(x, y)dx + N(x, y)dy \) to be exact, it must satisfy the exactness condition:
\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
This condition ensures that \( M \) and \( N \) come from a single scalar potential function \(F(x, y)\) for which \( dF = Mdx + Ndy \).
To use the condition, just compute the partial derivatives \( \frac{\partial M}{\partial y} \) and \( \frac{\partial N}{\partial x} \) and check if they match. If they do, you have an exact differential. This process is illustrated in the solution steps 2-8 of our original exercise. For instance, checking \(6 x^{5} d x + d y\), we find both \( M=6x^5 \) and \( N=1 \) have zero derivatives with respect to \( y \) and \( x \) respectively, confirming it. Checking each part is as simple as ensuring the derivatives align. If they don't, the differential is not exact.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions of multiple variables. It includes topics such as partial derivatives, multiple integrals, and vector calculus. This extension allows solving more complex problems found in fields like engineering and physics.

One of the key topics in multivariable calculus is the gradient, denoted by \( abla f \). The gradient is a vector of all the partial derivatives of a function. For a function \( f(x, y) \), its gradient would be \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
Understanding the concept of the gradient is crucial for fields like optimization and in determining the direction of steepest ascent or descent.

In the context of exact differentials, multivariable calculus helps us verify if a given differential form can be integrated directly. By utilizing partial derivatives, we can confirm the potential functions, simplifying the integration process. For example, verifying the condition \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) lets us know if our differential form is exact, making it easier to handle integrations.

• Partial derivatives allow functions to be analyzed in terms of individual variables.
• Exact differentials act as a bridge to find potential functions for integration.
• Multivariable calculus provides the tools to solve complex problems in higher dimensions.