Problem 12 - A First Course in Real Analysis

Question

Suppose that \(F(x, y, z)=0\) is such that the functions \(z=f(x, y), x=g(y, z)\), and \(y=h(z, x)\) all exist by the Implicit function theorem. Show that $$ f_{1} \cdot g_{, 1} \cdot h_{, 1}=-1 $$ This formula is frequently written $$ \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}=-1 $$

Step-by-Step Solution

Answer

\(f_{1} \times g_{, 1} \times h_{, 1} = -1\)

Step 1: Differentiate F Implicitly with respect to x

Since F(x, y, z) is a function of three variables and we want to find \frac{ormal{d}z}{ormal{d}x}, we will differentiate F with respect to x while treating y as a constant. We use the chain rule to get: \( F_x + F_z \frac{ormal{d}z}{ormal{d}x} = 0 \). Solve for \( \frac{ormal{d}z}{ormal{d}x} \) to get \( \frac{ormal{d}z}{ormal{d}x} = -\frac{F_x}{F_z} \). This is \( f_1 \).

Step 2: Differentiate F Implicitly with respect to y

Similar to step 1, we differentiate F with respect to y while treating z as a constant, leading to: \( F_y + F_x \frac{ormal{d}x}{ormal{d}y} = 0 \). We solve for \( \frac{ormal{d}x}{ormal{d}y} \) and find \( \frac{ormal{d}x}{ormal{d}y} = -\frac{F_y}{F_x} \). This is \( g_{, 1} \).

Step 3: Differentiate F Implicitly with respect to z

Finally, we differentiate F with respect to z while treating x as constant, yielding: \( F_z + F_y \frac{ormal{d}y}{ormal{d}z} = 0 \). Solving for \( \frac{ormal{d}y}{ormal{d}z} \), we get \( \frac{ormal{d}y}{ormal{d}z} = -\frac{F_z}{F_y} \). This is \( h_{, 1} \).

Step 4: Combine the Derived Partial Derivatives

Now we will multiply the partial derivatives found in steps 1, 2, and 3. By substituting the expressions for \( f_{1} \), \( g_{, 1} \), and \( h_{, 1} \), we obtain: \(f_{1} \times g_{, 1} \times h_{, 1} = \big(-\frac{F_x}{F_z}\big) \times \big(-\frac{F_y}{F_x}\big) \times \big(-\frac{F_z}{F_y}\big) = -1\).

Key Concepts

Partial Derivatives

Understanding partial derivatives is essential in multivariable calculus, where functions depend on more than one variable. Imagine you are at a hill station, and you want to know how steep the hill is if you walk directly north. This is what a partial derivative measures — the rate at which a function changes as one variable changes, while keeping all other variables constant. In the exercise, we talk about F(x, y, z) being a function of three variables. The notation f_1 or \( \frac{\partial z}{\partial x} \) represents the partial derivative of z with respect to x, with y held constant.

When calculating partial derivatives, we differentiate with respect to one variable and treat the others as constants. This is akin to examining a slice of a three-dimensional space along one axis and ignoring the rest. It's like studying the contours of that hill station from a map, focusing only on the northward path, regardless of the landscape in other directions.

Multivariable Calculus

In the realm of multivariable calculus, we sail through seas where functions are not limited to just one input and output but can fluctuate with many variables at play. It's just like watching the weather pattern in a city—it doesn't change based on just the time of day or just the season. It changes due to a complicated dance between various factors, including temperature, pressure, and humidity, to name a few.

In this context, the Implicit Function Theorem is like your weather forecast, indicating how changes in several variables relate to each other under certain conditions. In our exercise, F(x, y, z) = 0 may describe a surface in three-dimensional space, and the theorem assures us that we can solve for z in terms of x and y, just as one might predict temperature based on time and humidity. These solutions form functions like z = f(x, y), showing the interconnectedness of variables in our multivariable weather system.

Chain Rule

Now let's turn to the chain rule, a compass that guides us through the web of interconnected variables. Much like following a thread through a complex tapestry, the chain rule helps us unravel how a change in one variable affects another, via a third variable, in a function of several variables.

For instance, when you eat ice cream too fast, and you get a brain freeze, the pain doesn't just go from ice cream to head—it goes from ice cream to throat to head, with the intensity of the brain freeze depending on the path it takes. Similarly, in our exercise, when differentiating F(x, y, z) with respect to x, we're looking at how x changes as we change z, and then how z changes as x changes—essentially following the path of change through the variables. Applying the chain rule allowed us to find the relationship \( f_1 \times g_{, 1} \times h_{, 1} = -1 \), akin to tracing the effect of the ice cream path from your mouth to your head.