Directional Derivative Of F(x, Y) = X³y² At (1, -3)

Hey guys! Let's dive into a super interesting problem today: finding the directional derivative of a function. Specifically, we're going to tackle the function f(x, y) = x³y². We'll figure out its directional derivative in the direction of the vector ⟨3, 2⟩ at the point (1, -3). Buckle up, because we're about to get mathematical!

Understanding Directional Derivatives

Before we jump into the calculations, let's make sure we're all on the same page about what a directional derivative actually is. Imagine you're standing on a hill, and the hill's height is represented by our function f(x, y). The directional derivative tells you how steeply the hill slopes in a particular direction you choose to walk. It's like asking, "If I walk this way, how much will my altitude change?"

The directional derivative is a crucial concept in multivariable calculus, showing the rate of change of a function along a specific direction. Unlike partial derivatives, which only consider changes along the x and y axes, directional derivatives give us a more comprehensive view of the function's behavior. This is super useful in fields like physics (understanding how temperature changes in a room) and engineering (optimizing the shape of an airplane wing).

To really nail this, we need to remember a couple of key ingredients:

1. The Gradient: This is a vector made up of the partial derivatives of our function. Think of it as a map pointing in the direction of the steepest ascent. We'll calculate this first.
2. A Unit Vector: We need to make sure our direction vector (⟨3, 2⟩ in this case) has a length of 1. This is because the directional derivative measures the rate of change per unit distance. We'll normalize our vector before using it.

Calculating the Gradient

Okay, let's get our hands dirty with some calculations. The first step is to find the gradient of f(x, y) = x³y². Remember, the gradient, denoted by ∇f, is a vector containing the partial derivatives of f with respect to x and y.

1. Partial Derivative with Respect to x (∂f/∂x)

To find ∂f/∂x, we treat y as a constant and differentiate f(x, y) with respect to x. So,

∂f/∂x = ∂/∂x (x³y²) = 3x²y²

2. Partial Derivative with Respect to y (∂f/∂y)

Similarly, to find ∂f/∂y, we treat x as a constant and differentiate f(x, y) with respect to y. So,

∂f/∂y = ∂/∂y (x³y²) = 2x³y

3. The Gradient Vector

Now we can assemble the gradient vector:

∇f(x, y) = <∂f/∂x, ∂f/∂y> = <3x²y², 2x³y>

The gradient vector is super important. It tells us the direction of the greatest rate of increase of the function at any given point. Think of it as the compass that always points uphill on our imaginary hill. By calculating the partial derivatives correctly, we ensure that our compass is accurate, leading us to the correct directional derivative.

Evaluating the Gradient at (1, -3)

We're interested in the directional derivative at the point (1, -3), so we need to evaluate the gradient at this point. Let's plug in x = 1 and y = -3 into our gradient vector:

∇f(1, -3) = <3(1)²(-3)², 2(1)³(-3)> = <3(1)(9), 2(1)(-3)> = <27, -6>

So, the gradient of f at the point (1, -3) is the vector ⟨27, -6⟩. This vector points in the direction of the steepest ascent of the function at that specific location.

Finding the Unit Vector

Next up, we need to find the unit vector in the direction of ⟨3, 2⟩. A unit vector has a magnitude (or length) of 1, which is essential for the directional derivative calculation. To get a unit vector, we divide the original vector by its magnitude.

1. Calculate the Magnitude

The magnitude of a vector ⟨a, b⟩ is given by √(a² + b²). So, the magnitude of ⟨3, 2⟩ is:

||⟨3, 2⟩|| = √(3² + 2²) = √(9 + 4) = √13

2. Normalize the Vector

Now, we divide the vector ⟨3, 2⟩ by its magnitude √13 to get the unit vector, which we'll call u:

u = ⟨3, 2⟩ / √13 = <3/√13, 2/√13>

The unit vector is our normalized direction. It ensures that when we calculate the directional derivative, we are only considering the change in the function's value per unit distance traveled in the specified direction. Normalizing the vector is a critical step, preventing us from overestimating or underestimating the rate of change.

Calculating the Directional Derivative

Alright, we're in the home stretch! We have the gradient at the point (1, -3), which is ⟨27, -6⟩, and we have the unit vector in the direction of ⟨3, 2⟩, which is <3/√13, 2/√13>. Now we can finally calculate the directional derivative.

The directional derivative, denoted by D_{uf(x, y), is calculated as the dot product of the gradient and the unit vector:}

D_{uf(x, y) = ∇f(x, y) · u}

In our case, this means:

D_{uf(1, -3) = ⟨27, -6⟩ · <3/√13, 2/√13>}

1. Compute the Dot Product

To calculate the dot product, we multiply the corresponding components of the vectors and add them together:

⟨27, -6⟩ · <3/√13, 2/√13> = (27)(3/√13) + (-6)(2/√13) = 81/√13 - 12/√13 = 69/√13

2. Rationalize the Denominator (Optional)

Sometimes, it's nice to have a rational denominator. We can multiply the numerator and denominator by √13 to achieve this:

(69/√13) * (√13/√13) = (69√13) / 13

So, the directional derivative of f(x, y) = x³y² in the direction of ⟨3, 2⟩ at the point (1, -3) is 69/√13, or (69√13) / 13 if you prefer a rationalized denominator.

Conclusion

The directional derivative tells us the rate of change of the function f(x, y) = x³y² at the point (1, -3) in the direction of the vector ⟨3, 2⟩. We found this rate of change to be 69/√13. That's pretty neat, right?

We've covered a lot in this explanation, including understanding the concept of directional derivatives, calculating partial derivatives to find the gradient, normalizing vectors to get unit vectors, and finally computing the directional derivative using the dot product. Each step is crucial, and mastering them will make you a directional derivative pro!

So, the final answer is 69/√13 (or (69√13) / 13). Hope this breakdown helped you guys understand directional derivatives a little better. Keep up the awesome work, and happy calculating!