Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

In what direction u does \(f(x, y)=1-x^{2}-y^{2}\) decrease most rapidly at \(\mathbf{p}=(-1,2) ?\)

Step-by-Step Solution

Answer

The direction is \( \mathbf{u} = \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \).

Step 1: Find the Gradient of f

First, we need to find the gradient \( abla f \) of the function \( f(x, y) = 1-x^2-y^2 \). The gradient is a vector of partial derivatives: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]Calculate each partial derivative:\[ \frac{\partial f}{\partial x} = -2x \]\[ \frac{\partial f}{\partial y} = -2y \]Thus, the gradient of \( f \) is:\[ abla f = (-2x, -2y) \]

Step 2: Evaluate the Gradient at Point p

We'll evaluate the gradient \( abla f \) at the point \( \mathbf{p} = (-1, 2) \).Substitute \( x = -1 \) and \( y = 2 \) into the gradient:\[ abla f(-1, 2) = (-2(-1), -2(2)) = (2, -4) \]So the gradient at \( \mathbf{p} \) is \( (2, -4) \).

Step 3: Determine the Direction of Maximum Decrease

The function decreases most rapidly in the direction opposite to the gradient. Thus, the direction of most rapid decrease is:\[ -abla f = -(2, -4) = (-2, 4) \]

Step 4: Normalize the Direction Vector

To find the unit vector in the direction of \( (-2, 4) \), we need to normalize it.Calculate the magnitude of \( (-2, 4) \):\[ \| (-2, 4) \| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \]Normalize the vector:\[ \mathbf{u} = \left( \frac{-2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}} \right) = \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]

Step 5: Final Step: Conclusion

The direction \( \mathbf{u} \) in which \( f(x, y) \) decreases most rapidly at \( \mathbf{p} = (-1, 2) \) is parallel to the normalized direction vector \( \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \).

Key Concepts

Partial Derivatives

A partial derivative is like a regular derivative, but it focuses on functions with more than one variable. In the function \( f(x, y) = 1-x^2-y^2 \), we see there's both an \( x \) and a \( y \). This means we can find how \( f \) changes if we change just \( x \) and keep \( y \) fixed, or vice versa. This is what partial derivatives can tell us.

• For \( x \), the partial derivative \( \frac{\partial f}{\partial x} = -2x \) shows how \( f \) changes with respect to \( x \).
• For \( y \), the partial derivative \( \frac{\partial f}{\partial y} = -2y \) explains how \( f \) shifts in regard to \( y \).

Using partial derivatives, we form a gradient vector \( abla f =( -2x, -2y) \). This gradient gives us a snapshot of how our function is changing in different directions at any given point.

Vector Normalization

Vector normalization is the process of scaling a vector to have a length of 1, essentially making it a unit vector. When considering a direction to move in a space, using a unit vector helps us focus solely on the direction itself rather than its magnitude. Let's consider the example vector \( (-2, 4) \).
To normalize a vector, we first compute its magnitude. The magnitude is the 'length' or 'size' of the vector, found using the formula: \[ \| (-2, 4) \| = \sqrt{(-2)^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \]
Afterward, we divide each component of the vector by this magnitude to get the unit vector: \[ \mathbf{u} = \left( \frac{-2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}} \right) = \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \]
This process ensures that the vector \( \mathbf{u} \) points in the same direction as \( (-2, 4) \) but with a length of just 1, which is useful for finding pure directional change.

Direction of Decrease

The direction of decrease for a function is crucial in optimization problems, where we want to minimize the function's output. For a given point \( \mathbf{p} = (-1, 2) \) and a function \( f(x, y) \), the direction of steepest descent, or most rapid decrease, is found directly opposite the gradient vector \( abla f \).
In our exercise, the gradient at point \( \mathbf{p} \) was calculated as \( (2, -4) \). To find the direction of decrease, we simply invert this gradient vector, resulting in \( (-2, 4) \).
This opposite vector indicates where \( f(x, y) \) decreases most rapidly. However, to find just the direction, not influenced by size, we normalize it to \( \left( \frac{-1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right) \). The unit vector now shows the direction in which to move from point \( \mathbf{p} \) to decrease the function \( f \) as efficiently as possible. This principle is foundational in techniques such as gradient descent used in various fields like machine learning.