Problem 27 - 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 \(P\). $$f(x, y)=x^{2}-4 y^{2}-9 ; P(1,-2)$$

Step-by-Step Solution

Answer

The unit vector for the steepest ascent is \(\left(\frac{1}{2\sqrt{17}}, \frac{4}{\sqrt{17}}\right)\), and the unit vector for the steepest descent is \(\left(-\frac{1}{2\sqrt{17}}, -\frac{4}{\sqrt{17}}\right)\). b. Which vector represents the direction of no change in the function at point P(1, -2)? A direction of no change can be either \((16, 2)\) or \((-16, -2)\).

Step 1: Compute the gradient of the function

Find the partial derivatives of the function with respect to \(x\) and \(y\). The gradient is the vector containing these two partial derivatives, i.e., \(\nabla f(x,y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). $$\frac{\partial f}{\partial x} = 2x,$$ $$\frac{\partial f}{\partial y} = -8y.$$ So, $$\nabla f(x,y) = \left(2x, -8y\right).$$

Step 2: Find the gradient vector at the given point

Evaluate the gradient vector of the function at the given point \(P(1,-2)\). $$\nabla f(1,-2) = \left(2(1), -8(-2)\right) = (2, 16).$$

Step 3: Find the unit vectors for the steepest ascent and descent

Normalize the gradient vector to find the unit vector in the direction of steepest ascent. $$u_{\text{ascent}} = \frac{\nabla f(1, -2)}{||\nabla f(1, -2)||} = \frac{(2, 16)}{\sqrt{2^2 + 16^2}} = \frac{(2, 16)}{4\sqrt{17}}=\left(\frac{1}{2\sqrt{17}}, \frac{4\sqrt{17}}{2\sqrt{17}}\right)=\left(\frac{1}{2\sqrt{17}}, \frac{4}{\sqrt{17}}\right).$$ The unit vector in the direction of the steepest descent is the negation of the unit vector in the direction of steepest ascent. $$u_{\text{descent}} = -u_{\text{ascent}} = \left(-\frac{1}{2\sqrt{17}}, -\frac{4}{\sqrt{17}}\right).$$

Step 4: Find a vector in the direction of no change

A vector in the direction of no change will be orthogonal to the gradient vector. To find such a vector, we can simply swap the two components and negate one of them. So, the direction of no change would be either \((16,2)\) or \((-16,-2)\). Hence, the answer is: a. \((\frac{1}{2\sqrt{17}}, \frac{4}{\sqrt{17}})\) for steepest ascent, and \((-\frac{1}{2\sqrt{17}}, -\frac{4}{\sqrt{17}})\) for steepest descent. b. A direction of no change can be either \((16, 2)\) or \((-16, -2)\).

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept when working with multivariable functions. They help us understand how a function changes as one of its input variables changes, while keeping the others constant. Imagine a function that depends on two variables, like position on a hill based on latitude and longitude. Partial derivatives give you a slope, but only in one direction at a time.

For a function like \(f(x, y) = x^2 - 4y^2 - 9\), we calculate its partial derivatives with respect to \(x\) and \(y\). This means we're looking at how \(f\) changes if only \(x\) changes (keeping \(y\) fixed), and vice versa.

• The partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x} = 2x\), which tells us how the function changes as \(x\) increases or decreases.
• The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y} = -8y\), showing how changes in \(y\) affect the function.

These derivatives help form the gradient, a vector that points in the direction of the steepest ascent from any given point.

Gradient Vector

The gradient vector is like a compass that points in the direction where a multivariable function increases the fastest. It's made up of all the partial derivatives of the function, thereby summarizing the steepest ascent at any point.

For our function \(f(x, y) = x^2 - 4y^2 - 9\), the gradient \(abla f(x, y)\) is a vector \((2x, -8y)\). Let's see how this works practically. At the point \(P(1,-2)\), calculate the gradient:

• Replace \(x\) with 1 to get \(2(1) = 2\)
• Replace \(y\) with -2 to get \(-8(-2) = 16\)

So, the gradient vector at \(P(1,-2)\) is \((2, 16)\). This vector directs us to the steepest ascent. If you imagine standing on a slope, the gradient tells you which way to head uphill the fastest.

To trek back down (i.e., find the steepest descent), negate the ascent vector. Unit vectors are then calculated to standardize these directions, ensuring they have a length of one. These unit vectors serve as precise guides for ascent and descent, crucial for methods like Gradient Descent in optimization problems.

Direction of No Change

A direction of no change indicates a path along which the function's value remains constant. For a function \(f(x, y)\), this direction is orthogonal to the gradient vector. This means it forms a right angle with the gradient, analogous to walking along a contour line on a map where the altitude stays the same.

To find such a vector, take the components of the gradient and swap them, changing the sign of one. For our gradient \((2, 16)\) at \(P(1,-2)\), valid directions of no change could be \((16, 2)\) or \((-16, -2)\). Both vectors are orthogonal to \((2, 16)\), ensuring the function's value doesn't vary when you move in these directions.

• Orthogonal vectors are particularly important in understanding level curves and maintaining functions in optimization scenarios.
• They can guide us to explore alternative paths or maintain equilibrium in dynamic systems.

Recognizing and using the direction of no change is essential for unraveling complex systems where multiple factors interplay.