Question

Show that if \(f(x, y)=\frac{a x+b y}{c x+d y},\) where \(a, b, c,\) and \(d\) are real numbers with \(a d-b c=0,\) then \(f_{x}=f_{y}=0,\) for all \(x\) and \(y\) in the domain of \(f\). Give an explanation.

Short answer

Question: Verify that the partial derivatives of the function \(f(x, y) = \frac{a x + b y}{c x + d y}\) are equal to 0 for all \(x\) and \(y\) in the domain of \(f\), when the condition \(ad - bc = 0\) is met.

Step-by-step solution

(Step 1: Calculate \(f_x\) and \(f_y\))

Firstly, we need to find the partial derivatives of \(f\), \(f_x\) and \(f_y\). \(f_x = \frac{\partial f}{\partial x} = \frac{(a(c x + d y)) - ((a x + b y)c)}{(c x + d y)^2}\) and \(f_y = \frac{\partial f}{\partial y} = \frac{(b(c x + d y)) - ((a x + b y)d)}{(c x + d y)^2}\)

(Step 2: Simplify \(f_x\) and \(f_y\))

Now let's simplify these expressions: \(f_x = \frac{acx + ady - acx - bcy}{(cx + dy)^2}\) \(f_y = \frac{bcx + bdy - adx - bdy}{(cx + dy)^2}\)

(Step 3: Check the given condition)

We are given the condition \(ad - bc = 0\). Let's check what happens with our partial derivatives under this condition: Using the condition, \(f_x = \frac{acx + ady - acx - (ad)x}{(cx + dy)^2} = \frac{0}{(cx + dy)^2} = 0\) \(f_y = \frac{(bc)x + bdy - (bc)x - bdy}{(cx + dy)^2} = \frac{0}{(cx + dy)^2} = 0\)

(Step 4: Conclusion)

As we can see, under the given condition \(ad - bc = 0\), the partial derivatives \(f_x\) and \(f_y\) are both equal to 0 for all \(x\) and \(y\) in the domain of \(f\).

Key concepts

Multivariable Calculus

Multivariable Calculus extends the concepts of calculus to functions involving several variables. This means instead of dealing with simple, one-dimensional functions like \( f(x) \), we work with functions like \( f(x, y) \) or even more variables.

In this exercise, the function \( f(x, y) = \frac{ax + by}{cx + dy} \) incorporates two variables, \(x\) and \(y\). The real strength of multivariable calculus is that it allows us to understand how changing one variable affects the function while keeping other variables constant.

One of the key concepts here is the partial derivative, a type of derivative used in multivariable calculus. It measures the rate of change of a function concerning one variable while keeping the others constant. Understanding this concept is crucial for analyzing behavior in multidimensional spaces, a core element in many fields like physics, engineering, and economics.

Thus, multivariable calculus provides essential tools for dealing with real-world problems, which rarely depend on a single variable.

Functions of Multiple Variables

Functions of multiple variables like \( f(x, y) = \frac{ax + by}{cx + dy} \) have outputs determined by two or more inputs. These types of functions are common in real-life applications.

In these functions, each variable can independently influence the output. For instance, you might think about temperature affecting weather, where temperature is influenced by both latitude and altitude, each considered an independent variable.

• For functions with two variables, each point \((x, y)\) in the plane corresponds to a point in a three-dimensional space where the function is defined.
• This leads us to a function's surface, providing a 3D view that helps visualize how the function behaves over a domain.

Functions of multiple variables offer a more comprehensive look at systems and are extremely useful for modeling purposes. Grasping how they work gives insight into a wide range of complex systems.

Derivative Rules

Derivative rules are the backbone of calculus, dictating how we differentiate functions. When dealing with functions of multiple variables, we use partial derivatives to focus on one variable at a time.

In this exercise, you compute the partial derivatives \( f_x \) and \( f_y \) of \( f(x, y) = \frac{ax + by}{cx + dy} \).

For each variable, we apply the quotient rule, which states:

• If \( u(x, y) = \frac{p(x, y)}{q(x, y)} \), then the partial derivative \( u_x \) is \( \frac{q(x, y)p_x - p(x, y)q_x}{[q(x, y)]^2} \).
• The same rule applies for \( u_y \) with respect to \( y \).

By applying these rules, we ensure precise calculations of partial derivatives, which helps in understanding how changes in each input variable affect the output of the function.

In the given solution, using \( ad - bc = 0 \) dramatically simplifies the partial derivatives, confirming they equal zero across the domain, demonstrating the power of derivative rules in simplifying complex expressions.