This is a homework question from a precalculus class that I'm a TA for.

Boat $A$ is initially at position $(1,4)$ and moves at a constant velocity $\langle 3,5 \rangle$. Boat $B$ is at position $(7,2)$ and moves at a constant velocity of $\langle 1,10 \rangle$. Do the paths of the boats ever cross? If so where? Will the boats collide?

I wanted to write up a thorough solution to this exercise for my class, and figured I'd post it online to help anyone else who may wander across it.

  • $\begingroup$ What is it that you want your students to practice? The simplest solution (I think) is to move to a frame in which one of the boats is at rest at the origin and see if the resulting forward ray from the other boat reaches the origin, i.e., that the difference in positions is a negative multiple of the difference in velocities. $\endgroup$ – amd Oct 25 '18 at 1:20
  • $\begingroup$ @amd That's probably the simplest solution computationally, but I don't think it's very intuitive. In the situation both boats are moving, so within the calculations both boats should be moving too. The students are struggling to become comfortable thinking in terms of vectors anyways, and I don't think they are ready to make the mental leap of shifting the frame of reference from the origin to one of the boats. I'll probably share this thought with any of the students who have a good grasp on working with vectors though, because it is a very useful thought. $\endgroup$ – Mike Pierce Oct 25 '18 at 14:51

Let $P_A(a)$ denote the position of the boat $A$ at time $a$, and let $P_B(b)$ denote the position of boat $B$ at time $b$. From the initial positions and velocities given, we have: $$ \begin{align} P_A(a) &= (1,4) + a\langle 3,5 \rangle &\qquad P_B(b) &= (7,2) + b\langle 1,10 \rangle \\ &= (3a+1, 5a+4) &\qquad &= (b+7,10b+2) \end{align} $$ Now these equations give the paths of the boats starting at time $a=b=0$. The paths of the boats cross only if at some time $a$ and some time $b$ after each starts moving they have the same position. In terms of those equations, the paths of the boats will cross if there are positive times $a$ and $b$ such that $P_A(a) = P_B(b)$. Now the boats collide if not only is there a location where their paths cross, but if they are at that location at the same time. So the boats collide if $P_A(a) = P_B(b)$ for some positive $a$ equal to $b$. So we can proceed by setting $P_A = P_B$: $$ (3a+1, 5a+4) = (b+7,10b+2) \implies \begin{cases} 3a+1=5a+4 \\ b+7=10b+2 \end{cases}\ \implies \begin{cases} 3a-b=6 \\ 5a-10b=-2 \end{cases}\,, $$ This, being a system of linear equations, has at most a single solution, which we can calculate to be $a = \frac{62}{25}$ and $b = \frac{36}{25}$. These are both positive times, so the paths of the boats do cross, but since this is the only solution and $a \neq b$, the boats do not collide. To find the actual coordinates where they do cross will be the location of boat $A$ at time $a=\frac{62}{25}$ (which should equal the location of $B$ at time $b=\frac{36}{25}$ if we've done our calculations correctly), which we can calculate: $$ P_A\left(\frac{62}{25}\right) = \left(3\cdot\frac{62}{25}+1, 5\cdot\frac{62}{25}+4\right) = \left(\frac{211}{25} , \frac{410}{25} \right)\,. $$

  • $\begingroup$ Why not use the same time parameter and see if the positions are ever equal? That’s a single equation with only one variable to check, basically equivalent to trying to solve the above system by back-substitution. Using two different parameters seems more appropriate when simultaneity isn’t important, such as when you’re simply computing the intersection of the two lines. As in my original comment to your question, it comes down to what you’re trying to illustrate, which isn’t clear to me from the question itself. $\endgroup$ – amd Oct 25 '18 at 18:19
  • $\begingroup$ @amd Part of the problems does ask where (if) the paths of the boats cross though. $\endgroup$ – Mike Pierce Oct 25 '18 at 21:07
  • $\begingroup$ Ah, yes, missed that part of the question. $\endgroup$ – amd Oct 25 '18 at 21:10
  • 1
    $\begingroup$ Yeah, that’s a very common beginner mistake with parametric equations. It might also be worth pointing out to your students what happens when you do use a common parameter: the correct conclusion is that the boats don’t collide, not that their paths don’t intersect. $\endgroup$ – amd Oct 25 '18 at 21:20
  • 1
    $\begingroup$ I would probably write the functions $P_A$ and $P_B$ originally with $t$ as the argument to both, then switch to $t_1$ and $t_2$ when looking for whether they cross. $\endgroup$ – Toby Bartels Nov 6 '18 at 22:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.