Problem 25 - Calculus Early Transcendentals

Question

\(21-26\) Find the maximum rate of change of \(f\) at the given point and the direction in which it occurs. \(f(x, y, z)=\sqrt{x^{2}+y^{2}+z^{2}}, \quad(3,6,-2)\)

Step-by-Step Solution

Answer

The maximum rate of change is 1, in the direction \(\left(\frac{3}{7}, \frac{6}{7}, \frac{-2}{7}\right)\).

Step 1: Determine the Gradient of f

The gradient vector \( abla f \) gives the direction of the maximum rate of change of a function. To find \( abla f \), we compute the partial derivatives of \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \). These derivatives are:\[\frac{\partial f}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial f}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial f}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\]Thus, the gradient is:\[abla f = \left(\frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right)\]

Step 2: Evaluate the Gradient at the Given Point

Substitute the point \((3,6,-2)\) into the gradient:\[abla f(3, 6, -2) = \left(\frac{3}{\sqrt{3^2 + 6^2 + (-2)^2}}, \frac{6}{\sqrt{3^2 + 6^2 + (-2)^2}}, \frac{-2}{\sqrt{3^2 + 6^2 + (-2)^2}}\right)\]Simplify the denominator:\[\sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7\]So the gradient at the point is:\[abla f(3, 6, -2) = \left(\frac{3}{7}, \frac{6}{7}, \frac{-2}{7}\right)\]

Step 3: Determine the Maximum Rate of Change

The maximum rate of change of the function at a given point is the magnitude of the gradient vector evaluated at that point. Hence, compute the magnitude of \(abla f(3, 6, -2)\):\[\left|abla f(3, 6, -2)\right| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{-2}{7}\right)^2}\]Calculate this:\[\left|abla f(3, 6, -2)\right| = \sqrt{\frac{9}{49} + \frac{36}{49} + \frac{4}{49}} = \sqrt{\frac{49}{49}} = \sqrt{1} = 1\]Therefore, the maximum rate of change is 1.

Step 4: Find the Direction of Maximum Rate of Change

The direction in which the maximum rate of change occurs is given by the vector \(abla f(3, 6, -2)\):\[\mathbf{d} = \left(\frac{3}{7}, \frac{6}{7}, \frac{-2}{7}\right)\]This vector \(\mathbf{d}\) indicates the direction of the steepest ascent of the function.

Key Concepts

Gradient Vector

The gradient vector is an essential tool in multivariable calculus, representing the direction and rate of the steepest ascent of a function. It is denoted by \( abla f \), and provides the direction in which a function increases most rapidly. For a function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), the gradient vector is found by computing the partial derivatives with respect to each variable:

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

Thus, the gradient vector is \( abla f = \left(\frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}}\right) \).
This vector points in the direction of the highest rate of change of the function.

Partial Derivatives

Partial derivatives are a central concept in multivariable calculus, where they measure how a function changes as one of the variables is varied while keeping the others constant. In our exercise, the function is \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \).
To find how \( f \) changes as each variable changes independently, we take the derivative of \( f \) with respect to \( x, y, \) and \( z \):

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

These partial derivatives are combined to form the gradient vector, indicating how the function's value changes as the point moves in each coordinate direction.

Magnitude of Gradient

The magnitude of the gradient vector gives us the maximum rate of change of the function at a specific point. To compute this, we evaluate the gradient vector at the given point and determine its length.
For the function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \) evaluated at the point \((3, 6, -2)\), the gradient vector is \( abla f(3, 6, -2) = \left(\frac{3}{7}, \frac{6}{7}, \frac{-2}{7}\right) \). The magnitude is calculated as follows:\[|abla f(3, 6, -2)| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2 + \left(\frac{-2}{7}\right)^2} = \sqrt{1} = 1\]Thus, the magnitude is 1, indicating that at this point, the function changes at the fastest rate of 1 unit in the direction of the gradient.

Direction of Maximum Rate of Change

The direction of the maximum rate of change is given by the direction of the gradient vector. It represents the path along which our function increases most steeply. In the context of our exercise, for the function \( f(x, y, z) = \sqrt{x^2 + y^2 + z^2} \), when evaluated at the point \((3, 6, -2)\), this direction is represented by the vector:\[\mathbf{d} = \left(\frac{3}{7}, \frac{6}{7}, \frac{-2}{7}\right)\]This unit vector shows us the precise direction in which \( f \) ascends most rapidly from that point.
Understanding this direction aids in analyzing how changes in input variables affect the function's outcome, a crucial aspect in fields such as optimization and physics.