Problem 14 - Calculus for Biology and Medicine

Question

In Problems 9-14, compute the directional derivative of \(f(x, y)\) at the given point in the indicated direction. $$ f(x, y)=\frac{1}{\sqrt{x^{2}+y^{2}}} \text { at }(0,1) \text { in the direction }\left[\begin{array}{r} 4 \\ -1 \end{array}\right] $$

Step-by-Step Solution

Answer

The directional derivative is \(\frac{1}{\sqrt{17}}\).

Step 1: Find the Gradient of the Function

The gradient of a scalar function \(f(x, y)\) is given by \(abla f = \left[\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right]\). For the function \(f(x, y) = \frac{1}{\sqrt{x^2 + y^2}}\), calculate the partial derivatives: \(\frac{\partial f}{\partial x} = -\frac{x}{(x^2 + y^2)^{3/2}}\) and \(\frac{\partial f}{\partial y} = -\frac{y}{(x^2 + y^2)^{3/2}}\). Thus, the gradient is \(abla f = \left[-\frac{x}{(x^2 + y^2)^{3/2}}, -\frac{y}{(x^2 + y^2)^{3/2}}\right]\).

Step 2: Evaluate the Gradient at the Point

Evaluate the gradient \(abla f\) at the given point \((0, 1)\). Substituting \(x = 0\) and \(y = 1\) into the gradient, we get \(abla f(0,1) = \left[-\frac{0}{(0^2 + 1^2)^{3/2}}, -\frac{1}{(0^2 + 1^2)^{3/2}}\right] = [0, -1]\).

Step 3: Normalize the Direction Vector

Normalize the direction vector \(\mathbf{d} = \begin{bmatrix} 4 \ -1 \end{bmatrix}\) by dividing it by its magnitude. The magnitude \(|\mathbf{d}|\) is \(\sqrt{4^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}\). Thus, the normalized direction vector is \(\mathbf{\hat{d}} = \begin{bmatrix} \frac{4}{\sqrt{17}} \ -\frac{1}{\sqrt{17}} \end{bmatrix}\).

Step 4: Compute the Directional Derivative

The directional derivative of \(f\) in the direction of \(\mathbf{\hat{d}}\) is given by \(D_{\mathbf{\hat{d}}} f = abla f \cdot \mathbf{\hat{d}}\). Substituting the gradient and normalized direction vector, we have \([0, -1] \cdot \begin{bmatrix} \frac{4}{\sqrt{17}} \ -\frac{1}{\sqrt{17}} \end{bmatrix} = 0\cdot\frac{4}{\sqrt{17}} + (-1)\cdot\left(-\frac{1}{\sqrt{17}}\right) = \frac{1}{\sqrt{17}}\).

Key Concepts

Gradient

In the world of calculus, the gradient of a function is like a compass that guides us towards the direction of steepest ascent. For a function of two variables, such as our function \( f(x, y) = \frac{1}{\sqrt{x^2 + y^2}} \), the gradient is composed of its partial derivatives. It is written as \( abla f = \left[ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right] \). Each element of this gradient vector represents the rate of change of the function with respect to one of its variables.

In our exercise, we calculated the partial derivatives:

• \( \frac{\partial f}{\partial x} = -\frac{x}{(x^2 + y^2)^{3/2}} \)
• \( \frac{\partial f}{\partial y} = -\frac{y}{(x^2 + y^2)^{3/2}} \)

This gradient vector, \( \left[-\frac{x}{(x^2 + y^2)^{3/2}}, -\frac{y}{(x^2 + y^2)^{3/2}} \right] \), tells us not just how the function changes, but also the rate and direction of this change at any given point \((x, y)\).

By evaluating the gradient at a specific point, like \((0, 1)\), we can determine the precise directional rate of change at that point, which becomes critical for finding the directional derivative.

Partial Derivatives

Partial derivatives help us examine how a function changes as we vary just one of its variables, with the others held constant. In our problem, we encountered a function \( f(x, y) = \frac{1}{\sqrt{x^2 + y^2}} \), where the goal was to compute how the function changes with either \( x \) or \( y \) independently of one another.

The partial derivative \( \frac{\partial f}{\partial x} \) allows us to assess how the function changes as we vary \( x \), while \( y \) remains fixed. Similarly, \( \frac{\partial f}{\partial y} \) focuses on changes as \( y \) varies and \( x \) stays the same.

These partial derivatives are significant because they construct the components of the gradient vector \( \left[ \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right] \). Understanding the changes in each independent direction on a two-dimensional plane empowers us to map out how the function behaves in any direction. This understanding is vital in numerous fields, including engineering and physics, where such precision helps predict outcomes in complex scenarios.

Vector Normalization

Vectors play a fundamental role in mathematics, especially when working with directional derivatives. It's not enough just to know the direction in which you're moving; the magnitude has a profound effect on your calculations as well. That's where vector normalization comes into play.

Consider the direction vector \( \mathbf{d} = \begin{bmatrix} 4 \ -1 \end{bmatrix} \) from our exercise. This vector points towards a direction, but to use it in our directional derivative calculation, it must be a unit vector. This means it should have a magnitude of 1, ensuring it merely indicates direction without affecting the scale of our computations due to its length.

To normalize a vector \( \mathbf{d} \), divide each of its components by its magnitude \( |\mathbf{d}| \). The magnitude of a vector \( \mathbf{d} = \begin{bmatrix} a \ b \end{bmatrix} \) is calculated as \( \sqrt{a^2 + b^2} \). For our direction vector:

• Magnitude: \( \sqrt{4^2 + (-1)^2} = \sqrt{17} \)
• Normalized vector: \( \mathbf{\hat{d}} = \begin{bmatrix} \frac{4}{\sqrt{17}} \ -\frac{1}{\sqrt{17}} \end{bmatrix} \)

This normalized vector ensures the directional derivative accurately reflects the rate of change in the specific, intended direction, without skewing the results due to the vector's original length.