Higher Level only. It is 3.11 with the parameter renamed to t and given units, which is worth saying out loud: nothing new is being learned about lines, only about what the dial means.
Press Run and let it loop twice before saying anything. The joining segment shortens, turns green for an instant at t = 3, and lengthens again. The dot on the curve traces the dip at the same moment.
Then drag the slider slowly from 2.8 to 3.4 and ask them to call out where it is shortest. They will land on 3, and the naive 3.33 is marked in red right beside it, half a centimetre higher up the curve.
The thing worth pointing at explicitly: the two tracks cross and the boats are never at the crossing. A is there at t = 3.33 and B at t = 4.44. If a class leaves with one sentence from this sub-topic, it should be that one.
| Question | Answer |
|---|---|
| 1. A's speed | |(3, 4)| = 5. |
| 2. Distance apart at t = 2 | AB = (4, −2), so √20 = 4.47. |
| 3. Time of closest approach | Vertex of 10t² − 60t + 100, so t = 3. |
| 4. The least distance | AB = (1, −3), so √10 = 3.16. |
| 5. Why the crossing point is not it | A. A is there at t = 3.33, B at t = 4.44. |
Question 2 exists to make the table real before the algebra arrives. A student who has computed 4.47 by hand at t = 2 has a reason to believe the symmetric table, and the symmetry is the cheapest check in the whole sub-topic.
1 markAB as a function of t, both components.
1 markThe distance or its square as a function of t.
1 markThe minimising time.
1 markThe least distance, with units.
Everything hangs on the first mark. A student who writes AB with one component, or with the subtraction the wrong way round and then takes a magnitude, will still get a plausible-looking quadratic, and the error only surfaces in the final number. Insist on a column vector written out.
| They wrote | What happened |
|---|---|
| 7 for the speed | Added the components. Chase it back to 3.10, not here. |
| 4 on question 2 | Used the x-component of AB alone. Very common, because 10 − 3t is the component people write first. |
| 20 on question 2 | Gave the squared distance. Cheap to fix and worth pointing out that the units would be km². |
| 3.33 on question 3 | One component reaching zero, read as the closest approach. They solved 10 − 3t = 0, which is one component reaching zero, not the distance reaching a minimum. |
| 4.44 on question 3 | Found when B reaches the crossing point. A more sophisticated version of the same confusion. |
| 6 on question 3 | Used −b/a rather than −b/(2a). An algebra slip, not a vectors one; separate it clearly or they will blame the topic. |
| 10 on question 4 | Gave the squared distance, which here is exactly 10 and so looks like a clean answer. It is also the gap at t = 0, which is a coincidence worth naming so nobody builds a theory on it. |
| 1 on question 4 | The first component of AB at t = 3. They have found the right time and stopped one step early. |
An answer of 10 on question 4 is the one to watch, because it is right as a squared distance, right as the starting gap, and wrong as the answer. Ask what the units are and it resolves itself.
"Why square it?" Because a square root does not change where a minimum is, and a quadratic has a vertex you can write down. If the square of the distance is least at t = 3, so is the distance. Say it once in those words and the technique stops looking like a trick.
"Can I differentiate instead?" Yes, and it is the same answer: d/dt(10t² − 60t + 100) = 20t − 60 = 0 gives t = 3. For a class that has done Topic 5 this is the more natural route and it generalises to the variable-velocity case, where the distance is not quadratic and the vertex formula has nothing to offer.
"Do they collide?" Only if both components of AB vanish at the same t. Here 10 − 3t = 0 needs t = 3.33 and −t = 0 needs t = 0, so no. The structure of that answer is worth drilling: two equations, two values of t, and a sentence saying they differ.
"Is velocity just speed with a direction?" Yes, and the reason to insist on the distinction is that questions ask for one and students give the other. "Find the speed" wants 5. "Find the velocity" wants (3, 4). Both appear in the same question often enough that guessing costs marks.
| Stage | What to do |
|---|---|
| Demonstrate | Graph √((10-3x)²+x²) and find the minimum: (3, 3.16). Then, deliberately, graph 10-3x and find its root: 3.33. Do the wrong one second and on purpose, so the class sees that it produces a perfectly reasonable graph and a perfectly wrong answer. That is the inoculation. |
| Where they stick | Typing the square root over the whole expression. Both machines need the bracket closed before the second term is added, and a missing bracket gives √(10−3x)² + x², which graphs happily and minimises somewhere else. Have them check the drawn curve starts at 10 when x is 0; if it does not, the brackets are wrong. |
| The check | The symmetry. Compute the gap one unit either side of the claimed minimum. If 2 and 4 give the same distance, 3 is the vertex. It is two substitutions and it is proof, not a hint. |
No trigonometric function appears, so the angle mode does not matter here. It matters at 3.13 and in any question that gives a velocity as a speed and a bearing rather than as components.
| Step | What |
|---|---|
| 1 | r = r₀ + tv, named against the 3.11 equation they already have. Nothing new, just vocabulary. |
| 2 | Velocity against speed, with one question of each on the board. |
| 3 | Ask when the boats are closest, before any method. Collect guesses; 3.33 will come up. |
| 4 | Fill the distance table together at t = 0, 1, 2, 3, 4. The symmetry does the arguing. |
| 5 | AB = (10 − 3t, −t), then the squared distance, then the vertex. |
| 6 | Figure, Run, then the slider through 2.8 to 3.4. |
| 7 | The crossing point: A at 3.33, B at 4.44. The collision test as a separate shape of answer. |
Do not say "find where the paths cross". It is the answer to a different question and students remember phrases. If a question genuinely asks for an intersection of paths, say "where the tracks cross, which may be at different times" every time, until the qualification is attached to the phrase.
Do not say "the closest approach is when the velocities are equal". It is not, and it sounds plausible enough to stick. Here the velocities are never equal and there is still a perfectly good minimum. The condition that does hold is that AB is perpendicular to the relative velocity at the closest moment, which is worth mentioning only to a class that has met 3.13 already.