Problem 300 - Calculus 3

Question

For the following exercises, find the maximum rate of change of \(f\) at the given point and the direction in which it occurs. $$ f(x, y)=\sqrt{x^{2}+2 y}, \quad(4,10) $$

Step-by-Step Solution

Answer

Maximum rate of change is \(\frac{\sqrt{17}}{6}\) in direction \(\left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right)\).

Step 1: Find the Gradient

The first step to find the maximum rate of change is to calculate the gradient of the function \( f(x, y) \). The gradient is a vector consisting of the first partial derivatives with respect to \( x \) and \( y \). The function given is \( f(x, y)=\sqrt{x^{2}+2y} \).Compute the partial derivative with respect to \( x \): \[ f_x(x, y) = \frac{d}{dx}(x^2 + 2y)^{1/2} = \frac{x}{\sqrt{x^2 + 2y}} \]Compute the partial derivative with respect to \( y \): \[ f_y(x, y) = \frac{d}{dy}(x^2 + 2y)^{1/2} = \frac{1}{\sqrt{x^2 + 2y}} \]The gradient is:\[ abla f(x, y) = \left( \frac{x}{\sqrt{x^2 + 2y}}, \frac{1}{\sqrt{x^2 + 2y}} \right) \].

Step 2: Evaluate the Gradient at the Given Point

Next, we need to evaluate the gradient at the point \((4, 10)\).Substitute \(x = 4\) and \(y = 10\) into the expressions for \( f_x \) and \( f_y \):\[ f_x(4, 10) = \frac{4}{\sqrt{4^2 + 2 \times 10}} = \frac{4}{\sqrt{16 + 20}} = \frac{4}{\sqrt{36}} = \frac{4}{6} = \frac{2}{3} \]\[ f_y(4, 10) = \frac{1}{\sqrt{4^2 + 2 \times 10}} = \frac{1}{\sqrt{36}} = \frac{1}{6} \]Thus, the gradient at \((4, 10)\) is:\[ abla f(4, 10) = \left( \frac{2}{3}, \frac{1}{6} \right) \].

Step 3: Find the Magnitude of the Gradient

The maximum rate of change is given by the magnitude of the gradient vector at the given point.Calculate the magnitude:\[ ||abla f(4, 10)|| = \sqrt{ \left( \frac{2}{3} \right)^2 + \left( \frac{1}{6} \right)^2 } = \sqrt{ \frac{4}{9} + \frac{1}{36} } \]To simplify:\[ \frac{4}{9} = \frac{16}{36}, \quad \text{so} \quad \frac{16}{36} + \frac{1}{36} = \frac{17}{36} \]\[ ||abla f(4, 10)|| = \sqrt{ \frac{17}{36} } = \frac{\sqrt{17}}{6} \]

Step 4: Determine the Direction of the Maximum Rate of Change

The direction of the maximum rate of change is given by the normalized (unit) gradient vector at the given point.From Step 2, the gradient vector is \( \left( \frac{2}{3}, \frac{1}{6} \right) \).The unit vector is:\[ \text{unit vector} = \frac{1}{||abla f(4, 10)||} \cdot \left( \frac{2}{3}, \frac{1}{6} \right) = \frac{6}{\sqrt{17}} \cdot \left( \frac{2}{3}, \frac{1}{6} \right) \]Multiply the vector:\[ = \left( \frac{6}{\sqrt{17}} \cdot \frac{2}{3}, \frac{6}{\sqrt{17}} \cdot \frac{1}{6} \right) = \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right) \]Thus, the direction is \( \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right) \).

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a crucial concept for studying functions with multiple variables. They help us understand how a function changes as one of its input variables is varied, holding the others constant.
For a function like \( f(x, y) = \sqrt{x^2 + 2y} \), we can find its partial derivatives by differentiating first with respect to one variable while treating others as constants.

• The partial derivative with respect to \( x \), noted as \( f_x(x, y) \), indicates how the function \( f \) changes as \( x \) changes, with \( y \) constant.
• The partial derivative with respect to \( y \), noted as \( f_y(x, y) \), shows the change in \( f \) when \( y \) changes, keeping \( x \) fixed.

These derivatives form the gradient vector, which points in the direction of the fastest increase of the function at a given point. The gradient encompasses both partial derivatives into a vector \( abla f(x, y) = \left( f_x(x, y), f_y(x, y) \right) \), providing valuable information about changes in all directions.

Maximum Rate of Change

The maximum rate of change of a function at a particular point is a key concept in multivariable calculus. It represents the steepest slope or the fastest ascent of the function around that point. This concept is determined by the gradient vector—specifically, the magnitude of this vector.
To compute it, we take the gradient vector \( abla f(x, y) = \left( \frac{2}{3}, \frac{1}{6} \right) \) evaluated at our point of interest, such as \((4, 10)\).
The magnitude of the gradient, \( ||abla f(4, 10)|| = \frac{\sqrt{17}}{6} \), already tells us our function's maximum rate of change at \((4, 10)\).

• The magnitude provides a scalar value indicating how intensely the function increases at that point.
• This rate is critical for understanding how steeply the landscape of the function changes.

Understanding this rate can prove beneficial in fields like engineering and physics, where knowing the rate of change helps solve complex real-world problems.

Unit Vector

A unit vector is an essential tool in vector calculus, used to describe the direction of a vector while disregarding its magnitude. In the context of finding the direction of maximum change, the unit vector derived from the gradient vector tells us the path of steepest ascent.
To calculate the unit vector, we take the gradient vector and normalize it. This means dividing each component by the magnitude of the gradient vector itself.
Thus, for our function at point \((4, 10)\), the gradient is \( abla f(4, 10) = \left( \frac{2}{3}, \frac{1}{6} \right) \).

• Normalization involves using its magnitude, \( ||abla f(4, 10)|| = \frac{\sqrt{17}}{6} \), to convert our gradient into a unit vector.
• The resulting unit vector is \( \left( \frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right) \).

This unit vector clearly outlines the precise direction in which the rate of change is at its maximum at that point. Meanwhile, the unit vector's purpose is purely directional, showcasing where the steepest increase occurs without scaling.