Problem 28 - Calculus Early Transcendentals

Question

Direction of steepest ascent and descent Consider the following functions and points \(P\). a. Find the unit vectors that give the direction of steepest ascent and steepest descent at \(P\) b. Find a vector that points in a direction of no change in the function at \(P\). $$f(x, y)=x^{2}+4 x y-y^{2} ; P(2,1)$$

Step-by-Step Solution

Answer

a. The unit vectors in the direction of steepest ascent and steepest descent at point P are (0.8, 0.6) and (-0.8, -0.6), respectively. b. A vector that points in a direction of no change in the function at point P is (6, -8).

Step 1: Find the partial derivatives of the function with respect of x and y

To find the gradient vector, we first need to find the partial derivatives of \(f(x,y)\) with respect to \(x\) and \(y\). $$ \begin{aligned} \frac{\partial f}{\partial x} &= 2x + 4y \\ \frac{\partial f}{\partial y} &= 4x - 2y \end{aligned} $$

Step 2: Compute the gradient vector at point P

Now, we will plug the coordinates of point \(P(2,1)\) into these partial derivatives to find the gradient vector at \(P\). $$ \begin{aligned} \nabla f(P) &= \left(\frac{\partial f}{\partial x}(2,1), \frac{\partial f}{\partial y}(2,1) \right) \\ &=\left(2 \cdot 2 + 4 \cdot 1, 4 \cdot 2 - 2 \cdot 1\right) \\ &=(8, 6) \end{aligned} $$ So, the gradient vector at point \(P\) is \((8,6)\).

Step 3: Find the unit vector in the direction of steepest ascent and descent

The unit vector in the direction of the gradient vector (steepest ascent) is given by: $$ \text{Unit vector (ascent)} = \frac{(8, 6)}{\|(8, 6)\|} = \frac{(8, 6)}{10} = (0.8, 0.6) $$ The unit vector in the direction opposite to the gradient vector (steepest descent) is the negative of the unit vector in the gradient vector's direction: $$ \text{Unit vector (descent)} = -(0.8, 0.6) = (-0.8, -0.6) $$

Step 4: Find the direction of no change in the function

A vector \(\mathbf{v}\) that points in a direction of no change in the function must satisfy the following condition: $$ \nabla f(P) \cdot \mathbf{v} = 0 $$ We know that \(\nabla f(P) = (8, 6)\). Let \(\mathbf{v} = (a, b)\). Then, $$ (8, 6) \cdot (a, b) = 8a + 6b = 0 $$ We can pick any multiplier for \(a\) and \(b\). In this case, let \(a = 6\). Then, $$ 8 \cdot 6 + 6b = 0 $$ Solving for \(b\), we get: $$ b = -8 $$ So, a vector in the direction of no change in the function at point \(P\) is \((6, -8)\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as each individual variable changes, while keeping other variables constant. Imagine you're climbing up a hill with a steep slope.
As you move in the x-direction, for example, the partial derivative with respect to x tells you how steep the hill is at that point in the x-direction.

• The partial derivative with respect to x (\(\frac{\partial f}{\partial x}\)) looks at how the function changes as x changes, while y remains fixed.
• Similarly, the partial derivative with respect to y (\(\frac{\partial f}{\partial y}\)) indicates the change in the function as y changes, with x being held constant.

By computing the partial derivatives of the given function:\(\begin{aligned}\frac{\partial f}{\partial x} &= 2x + 4y \\frac{\partial f}{\partial y} &= 4x - 2y\end{aligned}\),we obtain a deeper understanding of how the function behaves around point P.
Finding these derivatives sets the stage for understanding the gradient vector, which encompasses these rates of change.

Steepest Ascent

The gradient vector plays a crucial role in determining the direction of steepest ascent. It is essentially a multi-dimensional rate of change, pointing in the direction where the function increases the fastest.
To calculate the steepest ascent direction, we firstly find the gradient vector, and then normalize it to a unit vector:

• At point P(2, 1), the gradient vector is calculated as (8, 6).
• The unit vector in the same direction becomes \(\frac{(8, 6)}{10} = (0.8, 0.6)\).

Think of this unit vector as a guide highlighting the sharpest uphill path at that point. For any potential optimization task, this path represents the effective fastest way to increase the function's value.

Steepest Descent

Just as there's a path that maximizes the increase in a function, there's also a path that maximizes the decrease. This is known as the steepest descent.
It is simply the opposite direction to the steepest ascent. By taking the negative of the gradient vector's unit vector, you find the direction where the function decreases most rapidly.

• The unit vector for descent mirrors the ascent but is negative: \(-(0.8, 0.6) = (-0.8, -0.6)\).

This vector guides you in the optimal direction to achieve the steepest drop in functional value, akin to efficiently navigating down a steep mountain slope.
For tasks involving minimization, this path is invaluable.

Direction of No Change

In some situations, it's helpful to move in a direction where there is no change in the function value. This occurs along paths that are perpendicular to the gradient vector.
To find such a direction, we look for vectors that satisfy the condition:\(abla f(P) \cdot \mathbf{v} = 0\).

• Here, let the vector be \((a, b)\), and the equation becomes: \(8a + 6b = 0\).

By choosing a value for one component (say, \(a = 6\)), you can solve for the other component, \(b = -8\).
Thus, the vector \((6, -8)\) represents a "level path" where the function does not change as you move along this vector.