Problem 2 - Active Calculus Multivariable

Question

If \(f(x, y)=3 x^{2}-1 y^{2},\) find the value of the directional derivative at the point (-1,-4) in the direction given by the angle \(\theta=\frac{2 \pi}{6}\).

Step-by-Step Solution

Answer

The value of the directional derivative at the point (-1,-4) in the direction given by the angle \(\frac{2\pi}{6}\) is \(D_{\bold{u}}f(-1, -4) = -3 + 4\sqrt{3}\).

Step 1: Compute the gradient vector of the function

We need to find the gradient vector of the function, which is given as a vector with its components being the partial derivatives of the function with respect to \(x\) and \(y\). So, we have: \(\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\) Let's calculate the partial derivatives: \(\frac{\partial f}{\partial x} = \frac{\partial(3x^2 - y^2)}{\partial x} = 6x\) \(\frac{\partial f}{\partial y} = \frac{\partial(3x^2 - y^2)}{\partial y} = -2y\) So, the gradient vector is: \(\nabla f(x, y) = (6x, -2y)\)

Step 2: Find the unit vector in the given direction

We are given the angle \(\theta = \frac{2\pi}{6}\). A unit vector in this direction can be represented as: \(\bold{u} = (\cos{\theta}, \sin{\theta})\) Using our given angle, we find that: \(\bold{u} = (\cos{\frac{2\pi}{6}}, \sin{\frac{2\pi}{6}}) = (\cos{\frac{\pi}{3}}, \sin{\frac{\pi}{3}}) = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\)

Step 3: Calculate the directional derivative

Now that we have the gradient vector and the unit vector, we can compute the directional derivative of the function at point (-1, -4) in the direction given by the angle \(\frac{2\pi}{6}\). The directional derivative can be found by taking the dot product of the two vectors: \(D_{\bold{u}}f(x, y) = \nabla f(x, y) \cdot \bold{u}\) Using the gradient vector \(\nabla f(x, y) = (6x, -2y)\) at point (-1, -4), the gradient vector becomes: \(\nabla f(-1, -4) = (6(-1), -2(-4)) = (-6, 8)\) Now, compute the dot product: \(D_{\bold{u}}f(-1, -4) = (-6, 8) \cdot \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = -6\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right) = -3 + 4\sqrt{3}\) So, the directional derivative at point (-1, -4) in the direction given by the angle \(\frac{2\pi}{6}\) is: \(D_{\bold{u}}f(-1, -4) = -3 + 4\sqrt{3}\)

Key Concepts

Multivariable Calculus

In multivariable calculus, we deal with functions that have more than one input variable. Unlike single-variable calculus, where we find the derivatives with respect to one variable, here we must consider how a function changes in multiple directions.

For instance, if we have a function such as f(x, y) = 3x^2 - y^2, it describes a surface in three-dimensional space. To understand how this function changes at a point, like (-1, -4), we explore not just along the x-axis or y-axis but in any direction across the xy-plane. This leads us to concepts like the gradient vector, directional derivative, and the importance of the dot product in computing the rate of change in a particular direction.

Gradient Vector

The gradient vector is fundamental in multivariable calculus. It is denoted by \(abla f\) and is composed of the partial derivatives of a function with respect to each of its variables.

\(abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\)

In simple terms, the gradient vector points in the direction of the steepest ascent of the function from any point, with its magnitude indicating how steep the ascent is. If you imagine hiking up a hill, following the gradient would be like going straight up the steepest path from your current position.

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one variable, holding the other variables constant. For a function like f(x, y), we would find two partial derivatives: \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\).

Let's take a closer look:

\(\frac{\partial f}{\partial x} = 6x\) shows how the function changes in the x-direction, while \(\frac{\partial f}{\partial y} = -2y\) shows changes in the y-direction. These partial derivatives help to construct the gradient vector, which in turn is used to find the directional derivative, indicating the rate of change of the function along any specified path.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. Symbolically, it's represented as \(\mathbf{a} \cdot \mathbf{b}\).

In the context of directional derivatives, we use the dot product to calculate how much of the gradient vector \(abla f\) is in the direction of the unit vector \(\mathbf{u}\). This operation provides us with the directional derivative, which expresses the rate of change of the function in the direction of \(\mathbf{u}\).

In our example, computing \(D_{\mathbf{u}}f(-1, -4)\) involved finding \(-6\left(\frac{1}{2}\right) + 8\left(\frac{\sqrt{3}}{2}\right)\), which yielded \(-3 + 4\sqrt{3}\). This gives the precise rate at which the function f(x, y) is increasing or decreasing at the point (-1, -4) in the direction specified by \(\theta\).