Question

Suppose \(M\) and \(N\) are continuous and have continuous partial derivatives \(M_{y}\) and \(N_{x}\) that satisfy the exactness condition \(M_{y}=N_{x}\) on an open rectangle \(R\). Show that if \((x, y)\) is in \(R\) and $$ F(x, y)=\int_{x_{0}}^{x} M\left(s, y_{0}\right) d s+\int_{y_{0}}^{y} N(x, t) d t, $$ then \(F_{x}=M\) and \(F_{y}=N\).

Short answer

Question: Prove that if the functions M and N are continuous, have continuous partial derivatives, and satisfy the exactness condition \(M_y = N_x\), then the function F(x, y) defined as $$F(x, y) = \int_{x_0}^x M(s, y_0) ds + \int_{y_0}^y N(x, t) dt$$ satisfies \(F_x = M\) and \(F_y = N\).

Step-by-step solution

Compute the partial derivative of F with respect to x

Differentiate F with respect to x, using the fundamental theorem of calculus for the first term and the chain rule for the second term: $$ F_x(x, y) = \frac{\partial}{\partial x}\left(\int_{x_0}^x M(s, y_0) ds\right) + \frac{\partial}{\partial x}\left(\int_{y_0}^y N(x, t) dt \right) $$ $$ F_x(x, y) = M(x, y_0) + \int_{y_0}^y \frac{\partial N}{\partial x}(x, t) dt $$

Compute the partial derivative of F with respect to y

Differentiate F with respect to y, using the chain rule for both terms: $$ F_y(x, y) = \frac{\partial}{\partial y}\left(\int_{x_0}^x M(s, y_0) ds\right) + \frac{\partial}{\partial y}\left(\int_{y_0}^y N(x, t) dt \right) $$ $$ F_y(x, y) = \int_{x_0}^x \frac{\partial M}{\partial y}(s, y_0) ds + N(x, y) $$

Use the exactness condition

Now, we incorporate the exactness condition \(M_y = N_x\). This means that for every point (x, y) in R, $$ \frac{\partial M}{\partial y}(s, y_0) = \frac{\partial N}{\partial x}(x, t) $$ Thus, we can substitute this into our expressions for the partial derivatives of F: $$ F_x(x, y) = M(x, y_0) + \int_{y_0}^y \frac{\partial M}{\partial y}(s, y_0) ds $$ $$ F_y(x, y) = \int_{x_0}^x \frac{\partial N}{\partial x}(x, t) ds + N(x, y) $$

Show that \(F_x = M\) and \(F_y = N\)

Now, we just need to show that the partial derivatives we calculated above reduce to M and N: For \(F_x(x, y)\): $$ F_x(x, y) = M(x, y_0) + \int_{y_0}^y \frac{\partial M}{\partial y}(s, y_0) ds $$ Since the integration is with respect to y, the second term is a constant with respect to x and its partial derivative with respect to x is zero. Therefore: $$ F_x(x, y) = M(x, y_0) $$ Since M is continuous and has continuous partial derivatives, we can substitute \(y_0\) with y, obtaining: $$ F_x(x, y) = M(x, y) $$ For \(F_y(x, y)\): $$ F_y(x, y) = \int_{x_0}^x \frac{\partial N}{\partial x}(x, t) ds + N(x, y) $$ Since the integration is with respect to x, the first term is a constant with respect to y and its partial derivative with respect to y is zero. Therefore: $$ F_y(x, y) = N(x, y) $$ In conclusion, we showed that \(F_x = M\) and \(F_y = N\).

Key concepts

Partial Derivatives

When we talk about partial derivatives, we're focusing on how a function changes as we adjust one of its variables, keeping all others constant. This is crucial when dealing with functions of multiple variables. Imagine you have a function F(x, y), which depends on both x and y.
The partial derivative with respect to x, denoted by \( \frac{\partial F}{\partial x} \), measures how F changes as x changes, while y remains unchanged. Similarly, \( \frac{\partial F}{\partial y} \), tells us how F changes with respect to y when x is fixed.
Partial derivatives are foundational in understanding behaviors in multi-dimensional spaces. They help us explore how a small change in one variable contributes to a change in the function's value.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) serves as a bridge between differentiation and integration. It has two parts; the first part tells us how a definite integral of a function relates to its antiderivative.

• Part 1 establishes that if you have a function \( f \) that is continuous over an interval, and \( F \) is an antiderivative of \( f \), then \( \int_a^b f(x) \) is just \( F(b) - F(a) \).
• Part 2 states that if \( F(x) \) is the integral \( \int_a^x f(t) dt \), then the derivative of this integral with respect to x is simply f(x).

In this problem, the FTC helps us differentiate the integral parts of \( F(x, y) \), beautifully showing how differentiation and integration counterbalance each other.

Chain Rule

The Chain Rule is an essential tool in calculus that helps us differentiate composite functions. It suggests that we manage the differentiation of functions that are nested within each other by identifying and differentiating the outer and inner functions separately.
Consider a function \( h(x) = f(g(x)) \). To find \( h'(x) \), the chain rule tells us to take the derivative of the outer function \( f \) evaluated at \( g(x) \) and multiply it by the derivative of the inner function \( g \). Mathematically, it looks like:

• \( h'(x) = f'(g(x)) \cdot g'(x) \).

In our exercise, the chain rule aids in differentiating \( F(x, y) \) especially when dealing with integrals that involve one of the variables nested inside.

Integration Techniques

Integration techniques are crucial when handling integrals, especially when they're not straightforward to solve directly. Different techniques can simplify the integration process.

• Substitution: Sometimes, changing variables can make an integral easier to solve. This involves replacing a complex part with a simple variable.
• Integration by parts: Useful when dealing with the product of two functions. It uses the formula: \( \int u dv = uv - \int v du \).
• Partial Fraction Decomposition: Helps when integrating rational functions by expressing them as a sum of simpler fractions.

In our solution, we rely on basic integration methods to evaluate the integrals \( \int_{x_{0}}^{x} M(s, y_{0}) ds \) and \( \int_{y_{0}}^{y} N(x, t) dt \). Understanding these techniques can make exact differential equations much more approachable.