The Canoe Trip, Variation 1

Task

Mike likes to canoe. He can paddle 150 feet per minute. He is planning a river trip that will take him to a destination about 30,000 feet upstream (that is, against the current). The speed of the current will work against the speed that he can paddle.

1. Let $s$ be the speed of the current in feet per minute. Write an expression for $r(s)$, the speed at which Mike is moving relative to the river bank, in terms of $s$.

2. Mike wants to know how long it will take him to travel the 30,000 feet upstream. Write an expression for $T(s)$, the time in minutes it will take, in terms of $s$.

3. What is the vertical intercept of $T$? What does this point represent in terms of Mike’s canoe trip?

4. At what value of $s$ does the graph have a vertical asymptote? Explain why this makes sense in the situation.

5. For what values of $s$ does $T(s)$ make sense in the context of the problem?

IM Commentary

The purpose of this task is to give students practice constructing functions that represent a quantity of interest in a context, and then interpreting features of the function in the light of that context. It can be used as either an assessment or a teaching task. Variation 2 of this task does not use function notation, and focuses more on the numerical and graphical behavior of the function near its vertical asymptote.

As a teaching task it provides an opportunity to discuss mathematical models, their interpretation, and their limits. For example, teachers could ask if it makes sense for $s$ to be negative. This might correspond to a flow of water moving in the same direction as Mike, and indeed the equation in the solution gives the correct answer in that case.

More fanciful, and requiring a longer discussion, is the question of whether it makes sense to consider values of $s$ larger than 150. If $s=300$, for example, a naive application of the formula predicts that Mike will arrive at his destination in $-200$ minutes! It is reasonable to say that negative times do not make sense and to exclude values of $s$ greater than 150. However, value $-200$ could also be interpreted as referring to an event that takes place 200 minutes before the trip starts. If Mike had been at his destination 200 minutes ago, then a river which was flowing at 300 feet per minute against his direction of travel would push him precisely the 30,000 feet from his destination that the problem began with.

Solution

1. The current is working against Mike’s paddling, pushing him back, so we subtract the speed of the current, $s$, from Mike’s paddling velocity, $150$, to get how fast he is actually moving, $r(s)$. So, with units of feet per minute, we have $r(s)=150-s$.

2. Since Mike is traveling with a constant velocity $r(s)$, measured in feet per minute, and $T(s)$ is the number of minutes it will take Mike to travel 30,000 feet, we know that $r(s)\cdot T(s) = 30,000$: the units on the left are (feet per minute) $\times$ (minutes), giving units of feet on the right hand side. Using $r(s)=150-s$, we have $$30,000=(150-s)\cdot T(s).$$ We solve for $T(s)$ to obtain $$T(s)=\frac{30,000}{150-s}.$$

3. A graph of $T(s)=\frac{30,000}{150-s}$ is shown below
   
   

   
   
   
   The vertical intercept is $T(0) = 200$. That means that Mike’s trip would take 200 minutes if the river’s current was 0 feet/min (perfectly still).

4. The vertical asymptote is at $s = 150$. This makes sense because if the river's current was 150 feet per minute then it would be as strong as Mike and so he wouldn't be able to move. If the current is just a little bit less than 150 then Mike can barely move and so $T(s)$, the time it take's Mike to complete the trip will be very large, as shown on the graph.

5. From d), we know that for Mike to be able to complete his trip, $s$ must be less than $150$. The problem implies the speed of the current is greater than zero, so we have a natural domain of $0\lt s\lt 150$.