Problem 22 - The Calculus with Analytic Geometry

Question

Given \(f(x, y, z)= \begin{cases}\frac{x y^{2} z}{x^{4}+y^{4}+z^{4}} & \text { if }(x, y, z) \neq(0,0,0) \\ 0 & \text { if }(x, y, z)=(0,0,0)\end{cases}\) (a) Show that \(D_{1} f(0,0,0), D_{2} f(0,0,0)\), and \(D_{3} f(0,0,0)\) exist; (b) make use of the fact that differentiability implies continuity to prove that \(f\) is not differentiable at \((0,0,0)\).

Step-by-Step Solution

Answer

All partial derivatives exist at (0,0,0). However, function is not continuous at (0,0,0), hence not differentiable.

Step 1: Definition of Partial Derivatives

Recall that the partial derivative of a function with respect to a variable at a point is defined as \[ D_{1} f(0,0,0) = \frac{\frac{\text{d} f}{\text{d} x}\bigg|_{(x,y,z)=(0,0,0)} - f(0,0,0)}{x} \bigg|_{(x,y,z) \to (0,0,0)} \]

Step 2: Compute Partial Derivative with Respect to x

First, compute \( D_{1} f(0,0,0) \): For \( (x, y, z) eq (0,0,0) \), substitute \( f(x,y,z) \) and simplify: \[ D_{1} f(0,0,0) = \frac{\text{d}}{\text{d} x} \bigg|_{(x,y,z)=(0,0,0)} \frac{x y^2 z}{x^4 + y^4 + z^4} \] Set \( (x,y,z) = (h, 0, 0) \) \[ \frac{h \times 0^2 \times 0}{h^4 + 0^4 + 0^4} = 0 Therefore, \text{ as h approaches 0, } D_{1} f(0,0,0) = \frac{0 - 0}{h} = 0 \] Hence, \( D_{1} f(0,0,0) = 0 \).

Step 3: Compute Partial Derivative with Respect to y

Next, compute \( D_{2} f(0,0,0) \): For \( (x, y, z) eq (0,0,0) \), substitute \( f(x,y,z) \) and simplify: \[ D_{2} f(0,0,0) = \frac{\text{d}}{\text{d} y} \bigg|_{(x,y,z)=(0,0,0)} \frac{x y^2 z}{x^4 + y^4 + z^4} \] Set \( (x,y,z) = (0, h, 0) \) \[ \frac{0 \times h^2 \times 0}{0^4 + h^4 + 0^4} = 0 Therefore, \text{ as h approaches 0, } D_{2} f(0,0,0) = \frac{0 - 0}{h} = 0 \] Hence, \( D_{2} f(0,0,0) = 0 \).

Step 4: Compute Partial Derivative with Respect to z

Finally, compute \( D_{3} f(0,0,0) \): For \( (x, y, z) eq (0,0,0) \), substitute \( f(x,y,z) \) and simplify: \[ D_{3} f(0,0,0) = \frac{\text{d}}{\text{d} z} \bigg|_{(x,y,z)=(0,0,0)} \frac{x y^2 z}{x^4 + y^4 + z^4} \] Set \( (x,y,z) = (0, 0, h) \) \[ \frac{0 \times 0^2 \times h}{0^4 + 0^4 + h^4} = 0 Therefore, \text{ as h approaches 0, } D_{3} f(0,0,0) = \frac{0 - 0}{h} = 0 \] Hence, \( D_{3} f(0,0,0) = 0 \).

Step 5: Verify Continuity at (0,0,0)

To verify continuity, check that: \[ \text{lim}_{(x,y,z) \to (0,0,0)} f(x,y,z) = f(0,0,0) \] where \( f(0,0,0) = 0 \). Let's approach along each axis, the limit is 0. However, along path \( x=t , y=t, z=t \): \[ f(t,t,t) = \frac{t \times t^2 \times t}{3t^4} = \frac{t^3}{3t^4} = \frac{1}{3t} \] which tends to infinity as \( t \to 0 \). Hence, limit does not uniformly approach 0.

Step 6: Conclusion

From Step 5, f(x,y,z) is not continuous at (0,0,0). Since differentiability implies continuity, \( f \) is not differentiable at \( (0,0,0) \).

Key Concepts

Continuity and Differentiability of Multivariable Functions

Understanding the concepts of continuity and differentiability in the context of multivariable functions is crucial. A function is said to be continuous at a point if the limit of the function as it approaches the point from any direction is equal to the function's value at that point. In mathematical terms, for a function \( f(x, y, z) \) to be continuous at \((a, b, c)\), the following must hold: \[\lim_{(x, y, z) \to (a, b, c)} f(x, y, z) = f(a, b, c)\] Differentiability, on the other hand, is a stronger condition than continuity. A function is differentiable at a point if it is locally linearizable at that point—this means it can be approximated by a linear function within a small neighborhood around the point. Importantly, if a function is differentiable at a point, it is always continuous at that point. But continuity alone does not guarantee differentiability, as illustrated in the given exercise problem at the point \((0,0,0)\). To conclude the differentiability at \((0,0,0)\) we need to establish the continuity first, which fails for our specific function.

Exploring Multivariable Functions

In multivariable calculus, a function can depend on multiple variables. For instance, the given function \( f(x, y, z) \) depends on three variables—\(x\), \(y\), and \(z\). This means its behavior and values are determined by three independent parameters. The function is defined piecewise: \[f(x,y,z) = \begin{cases}\frac{x y^{2} z}{x^{4}+y^{4}+z^{4}} & \text {if } (x, y, z) e (0,0,0) \ 0 & \text {if } (x, y, z) = (0,0,0) \end{cases}\] This definition means the function behaves differently at the origin \((0,0,0)\) compared to other points in the space. When working with multivariable functions, it is important to understand how they vary along different directions and planes. Whereas univariate functions vary along a single direction (the x-axis, for instance), multivariable functions can change along multiple axes simultaneously, adding complexity to their analysis.

Partial Derivative Computation

Partial derivatives are a fundamental tool in multivariable calculus used to measure how a function changes as one of its variables changes while the others are held constant. For a function \( f(x, y, z) \), the partial derivative with respect to \(x\) at a point is computed as follows: \[D_{1} f(x, y, z) = \frac{\partial f}{\partial x}\bigg|_{(x,y,z)}\] For the function given in the exercise: \( f(x, y, z) = \frac{x y^{2} z}{x^{4} + y^{4} + z^{4}} \), we calculate the partial derivatives at \((0,0,0)\) for \(x\), \(y\) and \(z\) independently. Partial derivative with respect to \(x\) was shown as: \[ \frac{d}{dx} \bigg|_{(x,y,z)=(0,0,0)} \frac{x y^2 z}{x^4 + y^4 + z^4} \] It was derived that \( D_{1} f(0,0,0) = 0 \) similarly for \( D_{2} f(0,0,0) \) and \( D_{3} f(0,0,0) \) respecting variables \( y \) and \( z \) respectively. Despite all partial derivatives at \( (0,0,0) \) being zero, the function was shown to not be continuous at this point, hence not differentiable. This is a classic example where partial derivatives exist at a point but the function fails to be differentiable there.