Problem 33 - Calculus an Applied Approach

Question

Evaluate \(w_{x}, w_{y},\) and \(w_{z}\) at the point. $$ \text { Function } \quad \text { Point } $$ $$ w=\sqrt{x^{2}+y^{2}+z^{2}} \quad(2,-1,2) $$

Step-by-Step Solution

Answer

The values of \(w_x, w_y,\) and \(w_z\) at the point \((2,-1,2)\) are \(w_x = \frac{2}{3}, w_y = -\frac{1}{3}, w_z = \frac{2}{3}\) respectively.

Step 1: Find \(w_x\)

We first compute the partial derivative of \(w\) with respect to \(x\). By definition, \(w_x = \frac{\partial w}{\partial x}\). Using the rule for differentiating a square root function and the power rule, we obtain \(w_x = \frac{1}{2\sqrt{x^{2} + y^{2} + z^{2}}} \cdot 2x\) which simplifies to \(w_x = \frac{x}{\sqrt{x^2+y^{2}+z^2}}\). Evaluating this at the point \((2,-1,2)\), we get \(w_x = \frac{2}{3}\).

Step 2: Find \(w_y\)

Next, we compute the partial derivative of \(w\) with respect to \(y\). By definition, \(w_y = \frac{\partial w}{\partial y}\), and similarly, we get \(w_y = \frac{y}{\sqrt{x^2+y^{2}+z^2}}\). Evaluating this at the point \((2,-1,2)\), we get \(w_y = -\frac{1}{3}\).

Step 3: Find \(w_z\)

Finally, we compute the partial derivative of \(w\) with respect to \(z\). By definition, \(w_z = \frac{\partial w}{\partial z}\), and using the same method as before, we get \(w_z = \frac{z}{\sqrt{x^2+y^{2}+z^2}}\). Evaluating this at the point \((2,-1,2)\), we get \(w_z = \frac{2}{3}\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves computing the derivative of a function.
The derivative represents the rate at which a function is changing at any given point, providing insights into the function's behavior.

• For single-variable functions, differentiation involves finding the derivative with respect to one variable.
• For multivariable functions, partial derivatives are used to differentiate with respect to each variable separately.

In the context of the original exercise, we computed partial derivatives with respect to the variables \(x\), \(y\), and \(z\). This means we differentiated the function \(w = \sqrt{x^2 + y^2 + z^2}\) with each variable, treating the others as constants.
Each partial derivative gives the rate of change of \(w\) in the direction of the respective variable.
By finding \(w_x\), \(w_y\), and \(w_z\), we understand how \(w\) changes when moving slightly along the respective directions from the point \((2, -1, 2)\). Differentiation thus helps develop a deeper understanding of the behavior and geometry of multivariable functions.

Gradient

The gradient is an essential concept in multivariable calculus, representing a vector composed of all partial derivatives of a function.
This vector points in the direction of the greatest rate of increase of the function.

• The gradient of a function \(w(x, y, z)\) is denoted as \(abla w\).
• It consists of each of its partial derivatives: \( abla w = \left( \frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z} \right) \).

In our exercise, the gradient of \(w\) at the point \((2, -1, 2)\) combines the partial derivatives: \(\left(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\right)\).
This vector indicates the direction in which \(w\) increases the fastest and its magnitude indicates how steep that increase is.
By understanding the gradient, you can visualize how a multivariable function behaves in three-dimensional space, helping in optimization and finding maxima or minima.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions involving more than one variable.
It provides tools to analyze and visualize complex relationships in many fields, such as physics, engineering, and economics.

• In multivariable calculus, you learn to work with functions of several variables like \(w = \sqrt{x^2 + y^2 + z^2}\).
• You use partial derivatives and gradients to understand the rate of change and direction in which functions vary.
• This branch of mathematics includes topics like multiple integrals, vector calculus, and differential equations.

In practice, multivariable calculus helps solve real-world problems involving three-dimensional spaces or systems with multiple variables.
You often deal with surfaces and curves in space, optimizing over a domain using techniques like Lagrange multipliers.
By mastering multivariable calculus, you gain the ability to tackle complex systems and visualize interactions that are not apparent in single-variable contexts.
This makes it an invaluable tool for describing and influencing change in our multidimensional world.