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 of most rapid decrease is \(-\frac{1}{\sqrt{5}}\hat{i} + \frac{2}{\sqrt{5}}\hat{j}\).

Step 1: Compute the Gradient of f

The gradient of a function \( f \) is a vector that points in the direction of the steepest increase. So, first, compute the gradient \( abla f(x, y) \). This involves finding the partial derivatives of \( f \). Given \( f(x, y) = 1 - x^2 - y^2 \), calculate:\[\frac{\partial f}{\partial x} = -2x \quad \text{and} \quad \frac{\partial f}{\partial y} = -2y\]Thus, the gradient is:\[abla f(x, y) = (-2x, -2y)\]

Step 2: Evaluate the Gradient at Point p

Evaluate the gradient \( abla f \) at the point \( \mathbf{p} = (-1, 2) \).Plug in \( x = -1 \) and \( y = 2 \) into the gradient:\[abla f(-1, 2) = (-2(-1), -2(2)) = (2, -4)\]

Step 3: Contrary Direction of Steepest Descent

The direction of steepest descent is the negative of the gradient. Since the gradient at \( \mathbf{p} \) is \((2, -4)\), the direction of the steepest descent is:\[-abla f(-1, 2) = -(2, -4) = (-2, 4)\]

Step 4: Normalize the Direction Vector

To find the unit direction vector in which \( f(x, y) \) decreases most rapidly, normalize the direction of descent vector \((-2, 4)\).The magnitude of the vector is:\[\|(-2, 4)\| = \sqrt{(-2)^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\]The unit direction vector is:\[\mathbf{u} = \left(\frac{-2}{2\sqrt{5}}, \frac{4}{2\sqrt{5}}\right) = \left(-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\]

Key Concepts

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to each variable independently. Let's consider a multivariable function like \(f(x, y) = 1 - x^2 - y^2\). To find out how this function changes as \(x\) or \(y\) change, we compute the partial derivatives.
- For \(x\), the change is given by \(\frac{\partial f}{\partial x} = -2x\), implying that changes in \(x\) affect the function linearly.
- Similarly, for \(y\), the change is \(\frac{\partial f}{\partial y} = -2y\). This tells us how much \(f\) changes when only \(y\) varies, keeping \(x\) constant.
Think of partial derivatives as taking slices of the function surface parallel to the coordinate planes, showing us the slope or instantaneous rate of change at any particular point.

Normalization

Normalization is a process of converting a vector into a unit vector while retaining its direction. This is useful in various calculations to simplify units and focus on direction rather than magnitude.
Given the vector \((-2, 4)\), its magnitude is calculated using the formula:
\[\|(-2, 4)\| = \sqrt{(-2)^2 + 4^2} = \sqrt{20} = 2\sqrt{5}\]
To normalize this vector, divide each component by the magnitude:

• The \(x\)-component: \(-\frac{2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}\)
• The \(y\)-component: \(\frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}}\)

By doing so, the direction remains unchanged, but now we have a unit vector \((\mathbf{u})\), which has a magnitude of 1.

Unit Vector

A unit vector is essential in representing direction, as it has a magnitude of 1 and points in a specific direction. For instance, suppose we determine the direction in which \(f(x, y)\) decreases fastest: this is represented by \(\mathbf{u} = \left(-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right)\).
The unit vector has the following properties:

• It simplifies understanding by removing the magnitude, enabling focus purely on direction.
• It is essential in unitless applications like rotations and derives from normalizing larger vectors.

A unit vector is particularly helpful in gradient descent as it makes calculations involving direction straightforward and precise.

Direction of Steepest Descent

The direction of steepest descent refers to the path where a function decreases most rapidly, opposite to the gradient.
The gradient \(abla f(x, y)\) points towards the direction of greatest increase, so to find the steepest descent, we travel in the opposite direction.
For \(f(x, y) = 1 - x^2 - y^2\) at point \((-1, 2)\), the gradient vector is \((2, -4)\). The steepest descent direction is \(-abla f = (-2, 4)\).

• This direction is crucial in optimization algorithms like gradient descent.
• It guides us to minimize a function by continuously updating in this direction.

Understanding the direction of steepest descent helps in problem-solving and improving optimization techniques.