Problem 35 - 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. \({ }^{5}\) The average number of drops, \(n\), needed when the whelk is dropped from a height of \(x\) meters is approximated by $$n(x)=1+\frac{27}{x^{2}} \text { . }$$ (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

Crows drop whelks from approximately 5.2 meters to minimize travel distance.

Step 1: Determine the total number of drops

First, the function for the average number of drops needed is given by \( n(x) = 1+\frac{27}{x^2} \). This function will help us calculate how many times the crow drops the whelk.

Step 2: Calculate the total distance traveled

Since each drop requires the crow to fly up again, the total distance the crow travels, \( D(x) \), is \( n(x) \times x \). This gives us \( D(x) = x \times \left(1 + \frac{27}{x^2}\right) = x + \frac{27}{x} \).

Step 3: Find the minimum height

To minimize the total vertical distance \( D(x) \), we need to find the derivative \( D'(x) \) and set it to zero. The derivative is \[ D'(x) = 1 - \frac{27}{x^2} \]. Setting \( D'(x) = 0 \) gives \[ 1 = \frac{27}{x^2} \].

Step 4: Solve for \( x \)

Solve the equation from Step 3: \( x^2 = 27 \). Therefore, \( x = \sqrt{27} = 3\sqrt{3} \approx 5.2 \) meters.

Key Concepts

Applied Calculus

Applied calculus is a branch of mathematics that uses calculus principles to solve real-life problems. In this particular problem, we observe how crows efficiently open whelks. It's a practical example of how calculus can be applied in nature. By understanding the functional relationship between variables—in this case, the height from which the crow drops the whelk and the effort it has to expend—we can derive meaningful insights. For instance:

• We created a formula to predict the number of drops required to break the whelk, showing abstract math's relevance in concrete situations.
• From there, we derived further calculations to understand total effort, reflecting typical tasks in applied calculus.

This process shows how applied calculus can analyze and optimize efforts, leading to reduced energy expenditure and enhanced efficiency.

Derivatives

In calculus, derivatives are fundamental for determining the rate of change of a function. They form the backbone for finding extrema of functions. In this problem, we need derivatives to find the optimal height for the crow's drops.

• The vertical distance travelled is given by the function: \[ D(x) = x + \frac{27}{x} \]
• To minimize this distance, we calculate its derivative: \[ D'(x) = 1 - \frac{27}{x^2} \]
• The zeroes of this derivative will indicate potential minima or maxima of the original function.

By setting the derivative equal to zero, we isolate the necessary condition for extremum, solving for the drop height that minimizes the crow's flying distance. This exemplifies the role of derivatives in optimization.

Minimization Problems

Minimization problems are crucial in calculus when trying to reduce a quantity to its lowest value. Here, the objective was to find the height that minimizes the crow's vertical travel distance when opening a whelk.

• Start with the function describing the total distance traveled: \[ D(x) = x + \frac{27}{x} \]
• Using the derivative, as calculated: \[ D'(x) = 1 - \frac{27}{x^2} \]
• Set the derivative equal to zero and solve for \( x \), leading to: \( x = 3\sqrt{3} \approx 5.2 \) meters.

Through this process, we find the optimal strategy for the crows: dropping from this height minimizes energy expenditure. This represents how minimization problems can guide decisions and strategies in natural processes.