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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 \(P\) $$f(x, y)=2 \sin (2 x-3 y) ; P(0, \pi)$$

Short Answer

Expert verified
Answer: The direction of steepest ascent at point P is (-8, 6), the direction of steepest descent is (8, -6), and the direction of no change is (6, 8).

Step by step solution

01

Compute the gradient vector

Compute the partial derivatives of f(x, y) with respect to x and y and form the gradient vector, represented by ∇f(x, y): $$ \frac{\partial f}{\partial x} = 4 \cos (2x - 3y) \\ \frac{\partial f}{\partial y} = -6 \cos (2x - 3y) $$ Now, form the gradient vector: $$ \nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) = (4 \cos (2x-3y), -6 \cos (2x-3y)) $$
02

Evaluate the gradient vector at point P

Substitute the coordinates of point P(0, π) into the gradient vector: $$ \nabla f(0, \pi) = (4 \cos (2(0)-3(\pi)), -6 \cos (2(0)-3(\pi))) = (4 \cos (-3\pi), -6 \cos (-3\pi)) $$ Since cos(x) is even and cos(-x) = cos(x), we get: $$ \nabla f(0, \pi) = (4 \cos(3\pi), -6 \cos(3\pi)) $$ Calculating the values, we get: $$ \nabla f(0, \pi) = (-8, 6) $$
03

Find the directions of steepest ascent and descent

The direction of steepest ascent at point P is given by the gradient vector at P, while the direction of steepest descent is the opposite of that: Direction of steepest ascent at P: $$(-8, 6)$$ Direction of steepest descent at P: $$(8, -6)$$
04

Find the direction of no change

To find a vector that points in a direction of no change in the function at point P, we need a vector that is orthogonal to the gradient vector: $$ (-8, 6) \cdot (x, y) = 0 \\ -8x + 6y = 0 $$ A possible direction of no change at P is: $$ (6, 8) $$ Finally, let's summarize our results: a) Direction of steepest ascent at P: $$(-8, 6)$$ Direction of steepest descent at P: $$(8, -6)$$ b) Direction of no change in the function at P: $$(6, 8)$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direction of Steepest Ascent
Imagine you're hiking up a hill, and you want to reach the peak as quickly as possible. The fastest way up the hill is to follow the path of steepest ascent. In the world of calculus, this idea applies to functions of two variables, like the one in our exercise.

The direction of steepest ascent is represented by the gradient vector, denoted as \( abla f(x, y) \). The gradient vector points in the direction where the function increases most rapidly. To find it, you calculate the partial derivatives of the function with respect to both variables, and then evaluate these derivatives at a specific point.

In our exercise, the direction of steepest ascent at point \( P(0, \pi) \) comes out to be \( (-8, 6) \). This means that if you move in this direction from point \( P \), the function \( f(x, y) \) will increase most quickly. It's very much like climbing the steepest part of our metaphorical hill.
Direction of Steepest Descent
Now, suppose you've enjoyed the view from the peak, and you're ready to walk down. You'd probably look for the fastest route to the bottom. In terms of functions and gradients, we call this the direction of steepest descent. The mathematics is beautifully symmetrical here, because this direction is simply the negative of the gradient vector.

Hence, if the direction of steepest ascent is \( (-8, 6) \), the direction of steepest descent from point \( P(0, \pi) \) will be the opposite: \( (8, -6) \). It's the quickest way down the hill, the path where the function decreases most abruptly.
Partial Derivatives
To fully understand gradient vectors and directions of ascent and descent, you must be familiar with partial derivatives. Imagine you're dealing with a function that has several variables, like temperature over a region, which depends on both time and location. How do you know how the temperature changes at a specific time as you move from one place to another?

That's where partial derivatives come in. You hold all variables constant except one, and see how the function changes with respect to that one variable. Mathematically, we use the symbol \( \frac{\partial}{\partial x} \) for the partial derivative with respect to \( x \), and \( \frac{\partial}{\partial y} \) for \( y \). In our exercise, we use these to find the components of the gradient vector, which guide us to the steepest ascent and descent.
Directional Derivatives
What if you're not traveling exactly up or down the hill? What if you're moving diagonally or in some arbitrary direction? This leads us to directional derivatives. Think of the directional derivative as the rate at which the function changes as you move in a specific direction.

Mathematically, you calculate it by taking the dot product of the gradient vector and a unit vector in the direction you're interested in. The directional derivative tells you how quickly the function is increasing or decreasing as you move away from a point in that particular direction. In our exercise, we discussed a special case of directional derivatives — a direction of no change which results in a directional derivative of zero.

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