Hey guys! Today, we're diving into a classic multivariable calculus problem: finding the directional derivative. This might sound intimidating, but don't worry, we'll break it down step by step. We're given the function f(x, y, z) = z³ - x²y, a point (-1, 3, 2), and a direction vector v = <2, -5, 5>. Our mission is to figure out how much f changes at that point if we move in the direction of v. Let's get started!
Understanding Directional Derivatives
Before we jump into the calculations, let's take a moment to understand what the directional derivative actually means. Imagine you're standing on a hill represented by the function f(x, y, z). The directional derivative tells you the rate of change of the altitude (f) as you move in a specific direction (v). It's like asking, "If I take a step in this direction, how much will I go uphill or downhill?" This concept extends the idea of partial derivatives, which only consider changes along the x, y, or z axes. The directional derivative allows us to explore changes in any arbitrary direction. Now, to make this calculation, we need a couple of key ingredients: the gradient of the function and a unit vector in the direction we're interested in. We'll calculate these in the following sections. Remember, the directional derivative is a scalar value, not a vector. It represents the magnitude of the change in the function in the given direction, not a direction itself. Thinking about this geometrically can really help solidify the concept before we dive into the algebra.
Calculating the Gradient
Our first step in finding the directional derivative is to compute the gradient of the function f(x, y, z) = z³ - x²y. The gradient, denoted by ∇f, is a vector that points in the direction of the greatest rate of increase of the function. It's essentially a collection of all the partial derivatives of f. So, let's find those partial derivatives. The partial derivative with respect to x, denoted by ∂f/∂x, is found by treating y and z as constants and differentiating with respect to x. Applying the power rule, we get ∂f/∂x = -2xy. Similarly, the partial derivative with respect to y, ∂f/∂y, is found by treating x and z as constants, giving us ∂f/∂y = -x². Finally, the partial derivative with respect to z, ∂f/∂z, is found by treating x and y as constants, resulting in ∂f/∂z = 3z². Now, we can assemble these partial derivatives into the gradient vector: ∇f = <-2xy, -x², 3z²>. This vector field tells us the direction of the steepest ascent at any point (x, y, z) in space. To find the gradient at our specific point (-1, 3, 2), we simply substitute these values into the gradient vector. This gives us ∇f(-1, 3, 2) = <-2(-1)(3), -(-1)², 3(2)²> = <6, -1, 12>. This vector, <6, -1, 12>, represents the direction of the steepest increase of f at the point (-1, 3, 2). However, we're not interested in the steepest increase; we want to know the rate of change in the direction of the vector v. For that, we'll need a unit vector.
Finding the Unit Vector
Now that we have the gradient, the next step is to find a unit vector in the direction of v = <2, -5, 5>. A unit vector is simply a vector with a magnitude (or length) of 1. We need a unit vector because the directional derivative measures the rate of change per unit distance in the given direction. To find a unit vector, we first need to calculate the magnitude of v. The magnitude of a vector <a, b, c> is given by the formula ||v|| = √(a² + b² + c²). In our case, ||v|| = √(2² + (-5)² + 5²) = √(4 + 25 + 25) = √54 = 3√6. So, the magnitude of v is 3√6. To get the unit vector, we divide each component of v by its magnitude. This gives us the unit vector u = v / ||v|| = <2/(3√6), -5/(3√6), 5/(3√6)>. We can simplify this a bit by rationalizing the denominators, but for now, this form is perfectly fine for our calculations. This unit vector u points in the same direction as v, but it has a length of 1. This is crucial because when we take the dot product of the gradient and this unit vector, we'll get the rate of change of f per unit distance in the direction of v. Remember, the unit vector ensures that the directional derivative represents the change in f along the specified direction, independent of the magnitude of the direction vector. Now we have all the pieces of the puzzle. We have the gradient at the point, which tells us the direction of steepest ascent, and we have the unit vector, which points in the direction we're interested in. All that's left is to put them together.
Calculating the Directional Derivative
Finally, we're ready to calculate the directional derivative. The directional derivative of f at the point (-1, 3, 2) in the direction of v is given by the dot product of the gradient of f at that point and the unit vector in the direction of v. Mathematically, this is expressed as Duf(-1, 3, 2) = ∇f(-1, 3, 2) · u, where u is the unit vector we calculated earlier. We already found that ∇f(-1, 3, 2) = <6, -1, 12> and u = <2/(3√6), -5/(3√6), 5/(3√6)>. Now, let's compute the dot product. The dot product of two vectors <a, b, c> and <d, e, f> is given by a d + b e + c f. So, Duf(-1, 3, 2) = <6, -1, 12> · <2/(3√6), -5/(3√6), 5/(3√6)> = (6)(2/(3√6)) + (-1)(-5/(3√6)) + (12)(5/(3√6)) = 12/(3√6) + 5/(3√6) + 60/(3√6) = (12 + 5 + 60) / (3√6) = 77 / (3√6). We can rationalize the denominator by multiplying the numerator and denominator by √6, which gives us Duf(-1, 3, 2) = (77√6) / (3 * 6) = (77√6) / 18. This is our final answer! The directional derivative of f(x, y, z) = z³ - x²y at the point (-1, 3, 2) in the direction of v = <2, -5, 5> is (77√6) / 18. This value tells us the rate of change of f as we move in the direction of v at that specific point. A positive value indicates that f is increasing, while a negative value would indicate that f is decreasing. And there you have it, guys! We successfully navigated the concept of directional derivatives and found the solution to our problem. Remember, the key is to break down the problem into smaller steps: find the gradient, find the unit vector, and then take the dot product. With a little practice, you'll be directional derivative pros in no time!
Conclusion
In this article, we tackled the problem of finding the directional derivative of the function f(x, y, z) = z³ - x²y at the point (-1, 3, 2) in the direction of the vector v = <2, -5, 5>. We first understood the concept of directional derivatives and how they represent the rate of change of a function in a specific direction. Then, we broke down the problem into manageable steps. We calculated the gradient of the function, which points in the direction of the steepest ascent. We found the unit vector in the direction of v, ensuring that we were measuring the rate of change per unit distance. Finally, we computed the dot product of the gradient and the unit vector, which gave us the directional derivative. The result, (77√6) / 18, tells us the rate at which the function f is changing at the point (-1, 3, 2) as we move in the direction of the vector v. Understanding directional derivatives is crucial in multivariable calculus and has applications in various fields, including physics, engineering, and computer graphics. By mastering these concepts, you'll be well-equipped to tackle more complex problems involving rates of change in multiple dimensions. Keep practicing, and you'll become a directional derivative master! Remember, the journey through calculus is about understanding the fundamental concepts and applying them step by step. So, keep exploring, keep learning, and keep having fun with math!