Problem 38 - Applied Calculus

Question

On the west coast of Canada, crows eat whelks (a shellfish). To open the whelks, the crows drop them from the air onto a rock. If the shell does not smash the first time, the whelk is dropped again. " The average number of drops, \(n,\) needed when the whelk is dropped from a height of \(x\) meters is approximated by $$n(x)=1+\frac{27}{x^{2}}.$$ (a) Give the total vertical distance the crow travels upward to open a whelk as a function of drop height, \(x\) (b) Crows are observed to drop whelks from the height that minimizes the total vertical upward distance traveled per whelk. What is this height?

Step-by-Step Solution

Answer

(a) Total distance is \(x + \frac{27}{x}\). (b) Minimum height is \(3\sqrt{3} \approx 5.2\) meters.

Step 1: Understand the Problem

We are given an equation for the average number of drops, \(n(x) = 1 + \frac{27}{x^2}\), where \(x\) is the drop height in meters. Our task is to determine the total vertical upward distance traveled by the crow (for part a) and then find the height \(x\) that minimizes this distance (for part b).

Step 2: Calculate Total Vertical Distance (Part a)

The total vertical distance the crow travels is the number of times it flies up times the height \(x\). Since the crow drops the whelk \(n(x)\) times, the total vertical distance is given by: \[ D(x) = x \cdot n(x) = x \left(1 + \frac{27}{x^2}\right) = x + \frac{27}{x} \]

Step 3: Find Derivative of Distance Function (Part b)

To minimize the total vertical distance, we need to find the critical points of \(D(x)\). First, compute the derivative of \(D(x)\):\[ D'(x) = \frac{d}{dx} \left( x + \frac{27}{x} \right) = 1 - \frac{27}{x^2} \]

Step 4: Solve for Critical Point

Set \(D'(x) = 0\) to find the critical points:\[ 1 - \frac{27}{x^2} = 0 \]Rearrange this equation:\[ \frac{27}{x^2} = 1 \]Multiply both sides by \(x^2\):\[ 27 = x^2 \]Take the square root of both sides to solve for \(x\):\[ x = \sqrt{27} = 3\sqrt{3} \approx 5.2 \text{ meters} \]

Step 5: Verify Minimum

To ensure this is a minimum, check the second derivative, \(D''(x)\):\[ D''(x) = \frac{d}{dx}\left(1 - \frac{27}{x^2}\right) = \frac{54}{x^3} \]Since \(D''(x) > 0\) for \(x > 0\), the function is concave up at \( x = 3\sqrt{3} \), confirming it is a minimum.

Key Concepts

Optimization

Optimization is a central concept in calculus that involves finding the maximum or minimum value of a function. In the exercise with crows and whelks, we are tasked with minimizing the total vertical distance a crow flies to open the whelk. Our main goal is to find the optimal drop height, which is the height from which the crow should ideally drop the whelk to minimize its effort.

To optimize, we first express the total vertical distance as a function of the drop height, resulting in the expression \( D(x) = x + \frac{27}{x} \). This function models the distance the crow flies upwards, factoring in both the height from which it flies and the multiple attempts needed to crack the whelk.

• By substituting different values for \( x \), you can see how the total distance changes.
• The goal is to find the value of \( x \) that leads to the least total distance.

Finding this minimum is crucial because it saves the crow energy and effort, which is a classic problem in real-world optimization scenarios.

Derivatives

Derivatives provide a powerful tool in calculus for studying the behavior of functions. In this problem, derivatives are used to find where the function for total vertical distance \( D(x) \) reaches its minimum. The derivative tells us the rate at which the total distance changes as the drop height \( x \) varies.

To minimize \( D(x) = x + \frac{27}{x} \), we compute its first derivative \( D'(x) \). The calculation shows: \[ D'(x) = 1 - \frac{27}{x^2} \]

We set this derivative to zero to find the critical point, which helps identify the minimum point of the function. In optimization, finding critical points is essential as they represent locations where the function's slope is zero, indicating potential maxima or minima.

• After solving \( D'(x) = 0 \), we find the potential minimum at \( x = 3\sqrt{3} \).
• To confirm it's a minimum, we use the second derivative test.

The second derivative \( D''(x) = \frac{54}{x^3} \) confirms that the function is concave up at \( x = 3\sqrt{3} \), ensuring a minimum point.

Mathematical Modeling

Mathematical modeling is the process of using mathematical equations and concepts to represent real-world situations. In this exercise, we create a model to describe the behavior of crows dropping whelks. The model aims to represent the relationship between the drop height \( x \) and the number of attempts needed by the crow, which affects the total vertical distance the crow travels.

Modeling starts with identifying key elements of the situation. Here, the crow's task is simplified into a function \( n(x) = 1 + \frac{27}{x^2} \), which provides the average number of drops as a function of height. This drives our total distance function \( D(x) \).

• The model captures the core dynamics between drop height and flying effort.
• It uses this relationship to predict the best drop height in terms of energy efficiency.

Mathematical models help us simplify complex systems into solvable problems, helping us make decisions based on quantitative analysis. In this case, the model supports the crow's strategy to minimize effort while accessing its food.