Problem 9 - A First Course in Real Analysis

Question

Show that the relation \(F\left(x_{1}, x_{2}, y\right)=0\) yields \(y\) as a function of \(\left(x_{1}, x_{2}\right)\) in a neighborhood of the given point \(P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)\). Denoting this function by \(f\), compute \(f_{, 1}\) and \(f_{, 2}\) at \(P\). \(F\left(x_{1}, x_{2}, y\right) \equiv x_{1}+x_{2}+y-e^{x_{1} x_{2} y}=0 ; P\left(x_{1}^{0}, x_{2}^{0}, y^{0}\right)=\left(0, \frac{1}{2}, \frac{1}{2}\right)\)

Step-by-Step Solution

Answer

The Implicit Function Theorem assures the existence of y as a function of (x1, x2) since Fy is non-zero at P. At P, the partial derivatives f_{,1} and f_{,2} are computed as -1 and -0.5, respectively.

Step 1: Verify the Implicit Function Theorem Conditions

To show that y is a function of (x1, x2), we should first check that F satisfies the conditions of the Implicit Function Theorem at point P. These conditions require that F is continuously differentiable at P and that the partial derivative of F with respect to y (denoted as Fy) is non-zero at P.

Step 2: Calculate the Partial Derivative Fy at P

Find the partial derivative of F with respect to y and evaluate it at the given point P. Fy is given by the derivative of F(x1, x2, y) with respect to y. The partial derivative Fy at a point P is calculated by differentiating F with respect to y and then substituting the coordinates of P into the result.

Step 3: Check that Fy is Non-zero at P

Verify that the value of Fy at point P is non-zero. This confirmation will ensure that the Implicit Function Theorem’s conditions are met, allowing us to conclude that there exists a neighborhood around P where y can be expressed as a function of (x1, x2).

Step 4: Compute the Partial Derivatives f_{,1} and f_{,2}

Once it is established that y can be expressed as a function of (x1, x2), compute the partial derivatives of this function, which are denoted as f_{,1} and f_{,2}. Use the chain rule to derive these partial derivatives from the original equation F(x1, x2, y) = 0.

Step 5: Evaluate the Partial Derivatives at P

Finally, evaluate the partial derivatives f_{,1} and f_{,2} at the point P to find their values at this specific point. This step requires substituting the coordinates of point P into the expressions for f_{,1} and f_{,2} that were obtained in the previous step.

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are crucial for understanding the rate of change of a function with respect to one variable, while keeping the others constant. To visualize this, imagine a gently sloping hill; the partial derivative tells you how steep the hill is if you walk directly north, without veering east or west.

When calculating the partial derivative of a function like \( F(x_1, x_2, y) \) with respect to \( y \), denoted as \( F_y \), we differentiate \( F \) solely with respect to \( y \) and treat \( x_1 \) and \( x_2 \) as constants. Applying this to our exercise, the partial derivative \( F_y \) at point \( P \) reflects how \( F \) changes in the \( y \)-direction at that specific location in the function's domain. Evaluating \( F_y \) at \( P \) is key to using the Implicit Function Theorem to solve for \( y \) as a function of \( x_1 \) and \( x_2 \).

Continuously Differentiable Functions

A function that is continuously differentiable means that not only does the function have a derivative, but that derivative is a continuous function of the variables involved. This concept is lovely in its elegance because it implies that no abrupt changes or 'jumps' occur in the slope of the function—much like a smoothly paved road that curves without any sharp edges or potholes.

When we talk about the Implicit Function Theorem, we need our function \( F \) to be this kind of smooth traveler. This smoothness ensures that small changes in \( x_1 \), \( x_2 \), and \( y \) will not result in unexpected jumps in the value of \( F \). In practical terms, for every point close to \( P \), the behavior of \( F \) will be predictable, allowing us to confidently state that \( y \) can indeed be expressed as a function of \( x_1 \) and \( x_2 \), within a neighborhood around \( P \).

Chain Rule

The chain rule is a superstar in calculus, helping us navigate the relationships between variables that are linked in a multi-layered fashion. When we differentiate composite functions, the chain rule handily tells us how to unpack each link in the chain.

It shines in our context when we have \( y \) expressed implicitly as a function of \( x_1 \) and \( x_2 \) through \( F(x_1, x_2, y) = 0 \). Even though \( y \) does not appear neatly on one side of the equation, the chain rule allows us to find its relationship with \( x_1 \) and \( x_2 \). In essence, it's like decoding a secret message; it tells us how a change in \( x_1 \) or \( x_2 \) trickles down to affect \( y \). When computing \( f_{,1} \) and \( f_{,2} \), which represent the partial derivatives of \( y \) as a function of \( x_1 \) and \( x_2 \) at point \( P \), we apply the chain rule to disentangle the intertwined dependencies of these variables.

Multivariable Calculus

Multivariable calculus expands the tools of single-variable calculus into the lush garden of functions with several variables. We're no longer strolling along a single path but exploring a landscape with hills, valleys, and curved surfaces.

Within this landscape, our function \( F(x_1, x_2, y) \) represents not just a curve, but a surface or even higher-dimensional shape. To understand such a world, we need our powerful set of tools, like partial derivatives and the chain rule, to analyze how functions change in different directions. Remember, each variable adds another dimension to our calculus adventure.

In the context of our exercise, multivariable calculus allows us to find a function \( f \) that can map every point \( (x_1, x_2) \) in some neighborhood to a point \( y \), creating a three-dimensional relationship derived from the two-dimensional input of \( (x_1, x_2) \). It's like finding the perfect fitting key for a complex lock; each turn of \( x_1 \) and \( x_2 \) will uniquely determine the position of \( y \) according to the Implicit Function Theorem.