Problem 29 - 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 \(P\). $$f(x, y)=x^{4}-x^{2} y+y^{2}+6 ; P(-1,1)$$

Step-by-Step Solution

Answer

Answer: The unit vector for the steepest ascent direction is \(\left[-\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right]\), for the steepest descent direction is \(\left[\frac{4}{\sqrt{17}}, -\frac{1}{\sqrt{17}}\right]\), and a vector pointing in a direction of no change at point \(P\) is \([1, 4]\).

Step 1: Compute Partial Derivatives

First, find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\): $$\frac{\partial f}{\partial x} = 4x^3 - 2xy = 4x(x^2 - \frac{y}{2})$$ $$\frac{\partial f}{\partial y} = -x^2 + 2y$$

Step 2: Evaluate Partial Derivatives at Point P

Next, plug the coordinates of point \(P(-1,1)\) into the partial derivatives: $$\left.\frac{\partial f}{\partial x}\right|_{(-1,1)} = 4(-1)((-1)^2 - \frac{1}{2}) = -4$$ $$\left.\frac{\partial f}{\partial y}\right|_{(-1,1)} = -(-1)^2 + 2(1) = 1$$ So, the gradient of the function at point \(P\) is: $$\nabla f = \left[ -4, 1 \right]$$

Step 3: Find the Unit Vectors for Steepest Ascent and Descent

Now, find the unit vector in the direction of the gradient: $$\textbf{u}_{ascent} = \frac{\nabla f}{||\nabla f||} = \frac{[-4, 1]}{\sqrt{(-4)^2 + 1^2}} = \frac{[-4, 1]}{\sqrt{17}} = \left[ -\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right]$$ For the steepest descent, simply take the opposite direction: $$\textbf{u}_{descent} = -\textbf{u}_{ascent} = \left[\frac{4}{\sqrt{17}}, -\frac{1}{\sqrt{17}} \right]$$

Step 4: Find a Direction of No Change

To find a direction of no change in the function, we need to find a vector that is orthogonal to the gradient. We can use the property of the dot product that two vectors are orthogonal if their dot product is equal to zero: $$[-4, 1] \cdot [x, y] = 0$$ We can solve this for \(x\) and \(y\): $$-4x + y = 0$$ $$y = 4x$$ So any vector with its components related by a factor of 4 will have no change in the function at point \(P\). One example would be: $$\textbf{u}_{noChange} = [1, 4]$$ This gives us the final results: Steepest ascent direction (unit vector): \(\left[-\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}\right]\) Steepest descent direction (unit vector): \(\left[\frac{4}{\sqrt{17}}, -\frac{1}{\sqrt{17}}\right]\) Direction of no change in the function at \(P\): \([1, 4]\)

Key Concepts

Partial Derivatives

Partial derivatives help us understand how the function changes as we modify one variable at a time.They reflect the rate of change of the function with respect to one of its variables, keeping the others constant.
In the context of a function like \( f(x, y) = x^4 - x^2y + y^2 + 6 \), partial derivatives allow us to assess how changes in \(x\) or \(y\) alone affect the overall function.For example:

• \( \frac{\partial f}{\partial x} = 4x^3 - 2xy = 4x(x^2 - \frac{y}{2}) \) which tells us about changes when only \(x\) varies.
• \( \frac{\partial f}{\partial y} = -x^2 + 2y \) which informs us about changes when only \(y\) varies.

At a specific point, such as \( P(-1, 1) \), these expressions allow us to compute exact changes in function values.This calculation helps us when determining the gradient, which in turn helps identify directions of steepest ascent and descent.

Steepest Ascent

The steepest ascent direction tells us where the function increases most rapidly from a given point.To find this direction, we use the gradient, \( abla f \), which consists of the partial derivatives.For example, at point \( P(-1, 1) \), the gradient is \([-4, 1]\) indicating the direction where the function increases the fastest.

To form a unit vector, which maintains the direction but standardizes its length to 1, we calculate:

• \( \textbf{u}_{ascent} = \frac{abla f}{||abla f||} = \frac{[-4, 1]}{\sqrt{17}} = \left[ -\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right] \)

The unit vector \( \left[ -\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}} \right] \) points us in the direction of the steepest ascent according to the gradient.

Steepest Descent

The steepest descent is simply the mirror image of steepest ascent; it's the fastest path downward for the function.To determine this direction, we take the negative of the gradient's unit vector.This indicates a complete inversion of the ascent direction.

For our function at \( P(-1, 1) \):

• The gradient's direction is \([-4, 1]\), giving us the unit vector \([-\frac{4}{\sqrt{17}}, \frac{1}{\sqrt{17}}]\).
• Then, the descent direction is \( \textbf{u}_{descent} = -\textbf{u}_{ascent} = \left[ \frac{4}{\sqrt{17}}, -\frac{1}{\sqrt{17}} \right] \).

This opposite vector leads us precisely towards the steepest descent.

Direction of No Change

Sometimes, a direction exists where the function does not change; this is known as a direction of no change.Perpendicular to the gradient, this direction ensures that the rate of change is zero - highlighting equilibrium along this vector.
To find such a direction at point \( P \), we look for vectors orthogonal to the gradient using the dot product:

• The condition \( [-4, 1] \cdot [x, y] = 0 \) leads us to \(-4x + y = 0\).
• Solving, we find \( y = 4x \), so any vector \([x, 4x]\) will have no change.

A simple example vector, \([1, 4]\), perfectly captures the direction of no change at our point of interest.