Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(z=f(x, y)\) and \(\partial z / \partial x=\partial z / \partial y,\) then \(z=c(x+y)\)

Short answer

The statement is false. A function such as \(z = xy\) stands as a counterexample, where \(\partial z / \partial x\) is not always equal to \(\partial z / \partial y\), and even when they are equal (when \(x = y\)), \(z\) can't be rewritten as a function of \((x + y)\).

Step-by-step solution

Analysis of the given statement

Here, it's suggested that if the partial derivatives of two variables \(x\) and \(y\) for a function \(z\) are equal, then \(z\) can be written as a function of \((x + y)\). Engineering a counterexample is the most effective way to determine if this statement is universally valid.

Develop a counterexample

Consider a function, \(z = xy\). This implies that \(\partial z / \partial x = y\) and \(\partial z / \partial y = x\). Here, unless \(x = y\), the condition \(\partial z / \partial x = \partial z / \partial y\) is not satisfied. However, even when \(x = y\), \(z = xy\) cannot be simplified to a function of \((x + y)\). Thus, this function constitutes a counterexample.

Conclude and provide explanation

The supposed equivalence is then confirmed to be false. The condition \(\partial z / \partial x = \partial z / \partial y\) doesn't inherently mean that \(z\) is a function of \((x + y)\). Our tested function \(z = xy\) proves successfully contrary to the given statement.

Key concepts

Counterexamples in Calculus

In calculus, understanding the concept of a counterexample is crucial when working with statements and theorems. A counterexample is a specific case that disproves a proposed general rule or law. In the context of the given exercise, the statement suggests that if the partial derivatives of a function with respect to two different variables are equal, then the function must be expressible solely as a function of the sum of those variables.

In practice, finding a counterexample involves identifying a function that meets the initial conditions but fails to comply with the conclusion. The solution provided utilizes the function z = xy to demonstrate that the assumption is false. Even if the partial derivatives with respect to x and y are equal under certain conditions (namely, when x = y), this does not guarantee that z can be reduced to c(x + y). Hence, the significance of counterexamples is in showing that a statement cannot be universally applied and that one must be cautious when making generalizations in calculus.

Partial Differentiation

Partial differentiation is a fundamental concept in multivariable calculus, which allows us to find the rate of change of a function with respect to one of its variables while keeping the others constant. The notation \(\partial z / \partial x\) represents the partial derivative of z with respect to x.

In our exercise, z represents a function of x and y, and finding the partial derivatives \(\partial z / \partial x\) and \(\partial z / \partial y\) means analyzing how the function z changes as either x or y varies independently. The equality of these partial derivatives in specific instances does not imply that they represent the gradient of a function dependent solely on x + y. This is a common misunderstanding among students, which the counterexample method helps to clarify.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions of multiple variables. It includes concepts such as partial derivatives, gradient vectors, and multiple integrals. It's a leap from analyzing one-dimensional situations to exploring the curvature and behavior of surfaces in higher dimensions.

The exercise presented touches upon a key aspect of multivariable calculus: functions of two variables, x and y. It challenges the notion that the equality of partial derivatives simplifies the function to a single-variable dependency. This is a nuanced concept because, in multiple dimensions, the relationship between variables can't always be captured in a single expression like c(x + y). Understanding this helps students appreciate the complexity of functions in multivariable calculus, where each variable can influence the function in its unique way.