Problem 88 - A Graphical Approach to Precalculus with Limits

Question

"Mido" Simon is in his bass boat, the Katie, 3 miles from the nearest point on the shore. He wishes to reach his camp at Maggie Point, 6 miles farther down the shoreline. If Mido's motor is disabled, and he can row his boat at a rate of 4 mph and walk at a rate of 5 mph, find the least amount of time, to the nearest hundredth of an hour, that he will need to reach the camp. (Picture cant copy)

Step-by-Step Solution

Answer

The least amount of time needed is 1.65 hours.

Step 1: Define the Problem

Mido needs to get from his current location to his camp at Maggie Point. He has to row 3 miles perpendicular to the shore and then walk 6 miles along the shore. We need to find the point where Mido should land his boat to minimize his total travel time.

Step 2: Express the Total Time

The total time to reach his camp will be the sum of rowing time and walking time. Let \( x \) be the distance Mido rows parallel to the shore. Thus, the distance to row is \( \sqrt{3^2 + x^2} \) miles. The remaining distance on the shore will be \( 6 - x \) miles. The rowing time is \( \frac{\sqrt{3^2 + x^2}}{4} \) and the walking time is \( \frac{6-x}{5} \). The total time function is: \[ T(x) = \frac{\sqrt{9 + x^2}}{4} + \frac{6 - x}{5} \]

Step 3: Differentiate the Time Function

To find the minimum time, we need to find the derivative of the time function \( T(x) \) and set it to zero. Differentiating, we use the chain rule for the first term: \[ T'(x) = \frac{x}{4\sqrt{9+x^2}} - \frac{1}{5} \]

Step 4: Solve for Critical Points

Set the derivative \( T'(x) \) to zero to find the critical points. \[ \frac{x}{4\sqrt{9+x^2}} = \frac{1}{5} \] Multiply through by \( 20\sqrt{9+x^2} \) to clear the fractions:\[ 5x = 4\sqrt{9+x^2} \] Square both sides to eliminate the square root:\[ 25x^2 = 16(9 + x^2) \] Simplify and solve for \( x \):\[ 25x^2 = 144 + 16x^2 \] \[ 9x^2 = 144 \] \[ x^2 = 16 \] \[ x = 4 \] (only positive value since \( x \) represents a distance).

Step 5: Verify the Minimum Time

Check the second derivative to confirm \( x = 4 \) gives a minimum. The second derivative is positive, confirming a local minimum. Calculate the total minimum time using \( x = 4 \):\[ T(4) = \frac{\sqrt{9 + 4^2}}{4} + \frac{6 - 4}{5} \] \[ T(4) = \frac{\sqrt{25}}{4} + \frac{2}{5} \] \[ T(4) = \frac{5}{4} + \frac{2}{5} \] \[ T(4) = 1.25 + 0.4 = 1.65 \] Thus, the least time required is 1.65 hours.

Key Concepts

Rate of Change

Understanding the rate of change is crucial when solving optimization problems like Mido's travel scenario. The rate of change essentially measures how a quantity changes relative to another. In this case, we're looking at how Mido's travel time changes as he chooses different points along the shoreline to dock.
When we calculate time based on distance and speed, we're inherently dealing with rate of change. The rowing and walking rates translate directly into time through the equation \( \text{time} = \frac{\text{distance}}{\text{speed}} \). This is what allows us to formulate the total travel time equation \( T(x) \).
By analyzing how quickly the total time \( T(x) \) decreases or increases as Mido changes his path, we identify the most efficient route. This concept of rate of change, when linked with derivative concepts, guides us to find that specific point to minimize travel time.

Derivative

The derivative is a fundamental concept in calculus, helping us understand the rate at which something changes. In Mido's case, we need to find the minimum travel time, and derivatives are incredibly useful for this purpose.
By taking the derivative of the total travel time function \( T(x) \), we acquire \( T'(x) \). This derivative tells us how the total time changes with respect to the distance \( x \) that Mido rows parallel to the shore. The expression \( T'(x) = \frac{x}{4\sqrt{9+x^2}} - \frac{1}{5} \) illuminates the exact trade-off between rowing and walking time.
Finding when \( T'(x) = 0 \) helps us spot the critical points where this change in time either levels off to a minimum or other stationary points. At such a critical point, it is likely that we have found the optimal solution to Mido's travel problem.

Critical Points

In optimization problems, critical points are where you potentially find your extremities - maximums, minimums, or saddle points. To minimize Mido's travel time, we must focus on these critical points.
A critical point is found when the derivative of a function equals zero or is undefined. For Mido's time function \( T(x) \), solving \( \frac{x}{4\sqrt{9+x^2}} = \frac{1}{5} \) leads us to a critical point. Through solving this equation, we derive \( x = 4 \), indicating where his rowing distance best balances the rowing and walking times.
Verifying a critical point with the second derivative confirms the nature of our initial finding. In this exercise, the positive second derivative indicates a local minimum, validating this critical point \( x = 4 \) as yielding the least travel time.

Calculus

Calculus provides us with the tools to solve complex optimization problems like finding Mido's fastest route. It extends beyond simple arithmetic, delving into changes and rates of change, allowing us to model real-world scenarios intricately.
The application of calculus in this exercise involves determining the time function \( T(x) \) through distances defined by \( \sqrt{9 + x^2} \) and other calculations. With calculus, we're not just finding answers but understanding how these calculations dynamically adjust with changes in inputs, such as distance.
Derivatives, as a part of calculus, help us isolate and evaluate critical points, ensuring that the solutions we find are optimized. In Mido's case, calculus equips us to calculate the total travel time realistically and efficiently, aiding him in minimizing his journey to Maggie Point and demonstrating the practical utility of mathematical insights.