Problem 53 - Calculus: Early Transcendental Functions

Question

Compute the Laplacian \(\Delta f\) for \(f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}\).

Step-by-Step Solution

Answer

\(\Delta f = 0\)

Step 1: Calculate first order partial derivatives

Use the formula for first order partial derivatives to find \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\). Start by substituting \(f(x,y,z) = \sqrt{x^{2}+y^{2}+z^{2}}\) into the formula for first order partial derivatives: \[ \frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}},\]\[ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}},\]\[ \frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}\]

Step 2: Calculate second order partial derivatives

You now substitute the results from step 1 into the formula for second order partial derivatives:\[ \frac{\partial^{2} f}{\partial x^{2}} = \frac{y^2+z^2}{(x^2+y^2+z^2)^\frac{3}{2}},\]\[ \frac{\partial^{2} f}{\partial y^{2}} = \frac{x^2+z^2}{(x^2+y^2+z^2)^\frac{3}{2}},\]\[ \frac{\partial^{2} f}{\partial z^{2}} = \frac{x^2+y^2}{(x^2+y^2+z^2)^\frac{3}{2}}\].

Step 3: Compute the Laplacian of f

The Laplacian is given by the formula \(\Delta f = \frac{\partial^{2} f}{\partial x^{2}} + \frac{\partial^{2} f}{\partial y^{2}} + \frac{\partial^{2} f}{\partial z^{2}}\). Substituting the results from step 2 into this formula gives: \(\Delta f = \frac{y^2+z^2}{(x^2+y^2+z^2)^\frac{3}{2}} + \frac{x^2+z^2}{(x^2+y^2+z^2)^\frac{3}{2}} + \frac{x^2+y^2}{(x^2+y^2+z^2)^\frac{3}{2}} = 0\).

Key Concepts

Partial Derivatives

Partial derivatives are an essential tool used in calculus to describe how a function changes as one of its variables changes, while keeping the other variables constant. This means they help us understand the function's behavior in a multi-dimensional space.
The function given is \(f(x, y, z) = \sqrt{x^2 + y^2 + z^2}\).
To find the partial derivatives of this function, we differentiate \(f\) with respect to each variable (\(x\), \(y\), and \(z\)) separately, treating the other variables as constants:

• For \(x\), the partial derivative \(\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}\)
• For \(y\), the partial derivative \(\frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}\)
• For \(z\), the partial derivative \(\frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\)

This calculation helps identify the rate and direction of change of the function concerning each axis independently.

Second Order Derivatives

Second order derivatives provide insight into the curvature of the function surface; indicating concavity or convexity. In simple terms, they tell us how the rate of change itself is changing.
Given the partial derivatives from the first step, we take the derivative again with respect to the same variable to obtain the second order derivatives. This will provide us with:

• \(\frac{\partial^2 f}{\partial x^2} = \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\)
• \(\frac{\partial^2 f}{\partial y^2} = \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\)
• \(\frac{\partial^2 f}{\partial z^2} = \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\)

These calculations further investigate how each part of the function's rate of change behaves in the multidimensional space.

Laplacian Computation

The Laplacian, represented by \(\Delta\), is a differential operator that combines the second order partial derivatives. It gives a scalar value that can be interpreted as a measure of how the function diverges or converges at a point.
For the function \(f(x, y, z)\), the Laplacian is calculated by:

• Adding the second order partial derivatives: \(\Delta f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2}\)
• Substituting the values, we get \(\Delta f = \frac{y^2 + z^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} + \frac{x^2 + z^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}} + \frac{x^2 + y^2}{(x^2 + y^2 + z^2)^{\frac{3}{2}}}\)

When simplified, this results in \(\Delta f = 0\).
This implies the function is harmonic, meaning it satisfies Laplace’s equation and has no local maxima or minima; the function is in a state of equilibrium at any point in its domain.