Problem 33 - Applied Calculus

Question

Is there a function \(f\) which has the following partial derivatives? If so what is it? Are there any others? $$ \begin{array}{l} f_{x}(x, y)=4 x^{3} y^{2}-3 y^{4} \\ f_{y}(x, y)=2 x^{4} y-12 x y^{3} \end{array} $$

Step-by-Step Solution

Answer

Yes, the function is \( f(x, y) = x^4y^2 - 3xy^4 + C \), where \( C \) is any constant.

Step 1: Check Consistency with Clairaut's Theorem

For a function to exist with these partial derivatives, it must satisfy the condition that the mixed partial derivatives are equal. In other words, the derivative of \( f_x \) with respect to \( y \) must be equal to the derivative of \( f_y \) with respect to \( x \). Calculate these derivatives to verify:\[ \frac{\partial}{\partial y}(f_{x}) = \frac{\partial}{\partial y}(4x^3y^2 - 3y^4) = 8x^3y - 12y^3 \]\[ \frac{\partial}{\partial x}(f_{y}) = \frac{\partial}{\partial x}(2x^4y - 12xy^3) = 8x^3y - 12y^3 \]Both derivatives are equal, so a function \( f \) can exist.

Step 2: Integrate with Respect to x

To find \( f(x, y) \), integrate \( f_x(x, y) = 4x^3y^2 - 3y^4 \) with respect to \( x \):\[\int (4x^3y^2 - 3y^4) \, dx = y^2 \int 4x^3 \, dx - 3y^4 \int 1 \, dx\]\[= y^2 \left( x^4 \right) - 3y^4x + g(y)\]So, \( f(x, y) = x^4y^2 - 3xy^4 + g(y) \), where \( g(y) \) is an arbitrary function of \( y \).

Step 3: Determine Function g(y)

Use the expression found for \( f(x, y) \) to differentiate with respect to \( y \) and compare with \( f_y(x, y) = 2x^4y - 12xy^3 \):\[\frac{\partial}{\partial y} (x^4y^2 - 3xy^4 + g(y)) = x^4(2y) - 3x(4y^3) + g'(y)\]\[= 2x^4y - 12xy^3 + g'(y)\]Equating this to \( f_y(x, y) \):\[2x^4y - 12xy^3 + g'(y) = 2x^4y - 12xy^3\]It follows that \( g'(y) = 0 \). Therefore, \( g(y) \) must be a constant. So, \( g(y) = C \), where \( C \) is a constant.

Step 4: Write the Final Function

With \( g(y) = C \), our function \( f(x, y) \) becomes:\[ f(x, y) = x^4y^2 - 3xy^4 + C \]This is the function satisfying the given partial derivatives.

Step 5: Conclusion on Uniqueness

Since the integration with respect to \( x \) resulted in an arbitrary constant \( C \), the family of functions that satisfy the given conditions is \( f(x, y) = x^4y^2 - 3xy^4 + C \), where \( C \) is any constant. Therefore, there are infinitely many functions differing only by a constant.

Key Concepts

Clairaut's Theorem

Clairaut's Theorem is a fundamental rule in calculus, helpful in identifying whether a function with given partial derivatives actually exists. The essence of the theorem lies in its symmetry principle for mixed partial derivatives. This principle states that if the mixed partial derivatives of a function are continuous, they must be equal.

In simpler terms, if you differentiate a function first with respect to one variable and then with another, the order shouldn't matter—the results should be the same.

• This was essential in verifying the consistency of the problem’s function. By showing that \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \), the original solution confirms that the requirements set by Clairaut's Theorem are satisfied.
• This means a function \( f(x, y) \) can indeed be constructed from the given partial derivatives.

Clairaut's Theorem simplifies verifying potential functions' existence by providing a quick check for consistency.

Mixed Partial Derivatives

Mixed Partial Derivatives are partial derivatives taken with respect to different variables. For instance, in our original exercise, we took derivatives with respect to both \( x \) and \( y \) for functions \( f_x \) and \( f_y \).

These allow us to understand how a function's rate of change behaves when influenced by multiple variables at once.

• The process involves calculating these derivatives to check the function's consistency against Clairaut's Theorem.
• In our example, \( \frac{\partial}{\partial y}(f_{x}) \) resulted in \( 8x^3y - 12y^3 \) and \( \frac{\partial}{\partial x}(f_{y}) \) gave the same result, confirming that mixed partial derivatives are equal.

Recognizing this equality is crucial. When two different orders of partial derivatives yield the same result, it strengthens the validation for the function's derivation process.

Integral Calculus

Integral Calculus plays a vital role in constructing a function from its derivatives. This branch of calculus is concerned with the reverse process of differentiation, known as integration.

In constructing \( f(x, y) \), we relied on integrating the partial derivative \( f_x(x, y) = 4x^3y^2 - 3y^4 \) with respect to \( x \). This involves:

• First, integrating \( y^2 \int 4x^3 \, dx - 3y^4 \int 1 \, dx \) to get the intermediate function.
• We obtained \( x^4y^2 - 3xy^4 + g(y) \), with \( g(y) \) as an arbitrary function of \( y \).
• Finally, solving for \( g(y) \) knowing that its derivative must equate to 0 led to the conclusion that \( g(y) = C \).

Therefore, integral calculus not only helps find a function from its derivatives but also identifies multiple such functions differentiated by a constant. Integration is key in closing the loop of this mathematical construction process.