Problem 81 - Calculus Early Transcendentals

Question

The domain of $$f(x, y)=e^{-1 /\left(x^{2}+y^{2}\right)}$$ excludes \((0,0) .\) How should \(f\) be defined at (0,0) to make it continuous there?

Step-by-Step Solution

Answer

Answer: The value of the function at (0,0) that makes it continuous there is 0.

Step 1: Analyze the given function

We can observe that the given function \(f(x, y) = e^{-\frac{1}{x^2+y^2}}\) is undefined at \((0,0)\). We have to find a value for \(f(0,0)\) such that the limit of \(f(x, y)\) as \((x, y) \to (0,0)\) exists.

Step 2: Express \(f(x, y)\) in polar coordinates

Let's express our function in polar coordinates using \(x = r\cos\theta\) and \(y = r\sin\theta\). Substituting these expressions into \(f(x, y)\), we get: $$f(r, \theta) = e^{-\frac{1}{r^2(\cos^2\theta + \sin^2\theta)}}$$ Since \(\cos^2\theta + \sin^2\theta = 1\), our expression simplifies to; $$f(r, \theta) = e^{-\frac{1}{r^2}}$$

Step 3: Evaluate the limit as \((x, y)\) approaches \((0,0)\)

We want to find the limit as \(r \to 0\) for our function \(f(r, \theta)\): $$\lim_{r \to 0}e^{-\frac{1}{r^2}}$$ As \(r \to 0\), the term \(\frac{1}{r^2} \to \infty\). Hence the term \(e^{-\frac{1}{r^2}} \to e^{-\infty} = 0\).

Step 4: Define \(f(0,0)\) for continuity

Now that we found the limit, to make the function continuous, we define the function value at \((0,0)\) as follows: $$f(0,0) = 0$$ With this definition, the function \(f(x, y) = e^{-\frac{1}{x^2+y^2}}\) is continuous at the origin \((0,0)\) as its limit exists and equals the value of the function at that point.

Key Concepts

Limits and Continuity

In multivariable calculus, limits and continuity extend the ideas of single-variable functions into multiple dimensions. For a function of two variables, like our given function \( f(x, y) = e^{-1/(x^2+y^2)} \), evaluating the limit as \((x, y)\) approaches any point, such as \((0,0)\), requires us to approach from every possible direction in the plane.

The concept of a limit here means that as the points \((x, y)\) get infinitely close to \((0,0)\), the value of \(f(x, y)\) should approach some real number. For the function to be continuous at \((0,0)\), this limit must be both existent and equal to \(f(0,0)\).

• A function is continuous if there is no interruption in its graph; i.e., you don't need to lift your pencil while drawing it.
• With multivariable calculus, we consider the approach from all paths: straight lines, curves, or even spirals.
• We use transformation methods, such as polar coordinates, to identify whether the approach path influences the function.

Understanding limits and continuity is crucial because it tells us about the behavior of a function in response to changes in variable inputs.

Polar Coordinates

Expressing functions in polar coordinates can simplify the process of evaluating limits and understanding the behavior near a specified point, particularly when dealing with functions like \( f(x, y) = e^{-1/(x^2+y^2)} \).

Polar coordinates transform \((x, y)\) into \((r, \theta)\), where \(r\) is the radial distance from the origin, and \(\theta\) is the angle with the positive x-axis. This can be useful to analyze functions dependent on the distance to the origin, as is the case here.

• Polar coordinates are defined as \(x = r\cos\theta\) and \(y = r\sin\theta\).
• For a rotationally symmetric function like \(e^{-1/r^2}\), using polar coordinates helps in identifying that its behavior is solely dependent on \(r\), not \(\theta\).
• This makes it easier to evaluate the limit, as the function doesn’t change based on directional angles.

By converting to polar coordinates, we transformed a possibly complicated 2D problem into a simpler 1D problem, aiding in evaluating the limit as \(r\to 0\).

Exponential Functions

Exponential functions have an intriguing property: they exhibit extremely rapid growth or decay based on their exponents. In our function \(f(x, y) = e^{-1/(x^2+y^2)}\), the component \(-1/(x^2+y^2)\) plays a critical role in determining the behavior of the exponential function.

As \(x^2+y^2\) approaches zero, \(1/(x^2+y^2)\) approaches infinity, pushing \(-1/(x^2+y^2)\) towards negative infinity. This results in \(e\) raised to a very large negative number, which approaches zero. Hence the limit, as we found, becomes zero. Let's highlight some properties:

• Exponential decay occurs rapidly; in this case, the decay accelerates with smaller values of \(r\).
• Exponential functions are continuous and differentiable at all points except where we explicitly define discontinuities, as seen with points where \(x\) or \(y\) are zero.
• The decay or growth rate in functions like \(e^{-x}\) or \(e^{-1/x^2}\) is important in applications across calculus and real-world scenarios such as physics or finance.

Understanding how exponential functions react to operations on their exponents is crucial, especially when establishing continuity or evaluating limits in calculus.