I was having a chat with an American friend who asked the following question, remembered from their high school calculus class. I’m fairly good with maths, but the question is an unfamiliar one and I find myself frustrated. So, please enlighten me:

Boat A leaves the port at noon, travelling North at 8 knots.
Boat B leaves the port an hour later, travelling at 30 degrees North by East (in other words, at a bearing of 030 degrees), also at 8 knots.

At what rate is the distance between the boats increasing at 5pm?

I have two methods - one using vector algebra and one using the cosine rule - which yield two different answers. I’m assured that the solution requires the use of implicit differentiation, which seems to be true also.

This seems like a bit of an oblique way to have us do your homework, which we generally don’t do around here. Why don’t you show us the work from your two methods, and we can try to help you see where you’re going wrong?

You need four equations. Position of boat 1 in X and Y, and position of boat 2 in X and Y as a function of time. When you have those four equations you need to use the distance formula. Distance = sqrt[(X1-X2)^2+(Y1-Y2)^2]. Since you want the rate of change of that distance you need to take it’s derivative with respect to time.

There are ways to do this which require no real calculus. For example, to be vague about it, take a frame of reference where one boat is fixed at the origin, and then keep in mind that movement of a vector “perpendicular to its radius” instantaneously adds nothing to its magnitude; only movement of a vector of a vector “along its radius” matters for magnitude.

I’m 24 years old and a long time math enthusiast. I last did homework ooo…about 3 years ago

But in the interests of full disclosure, here’s my first method:

position vector of boat A = (0,8(t+1))

position vector of boat B = (8tsin30, 8tcos30)

Assuming we start the clock when B leaves port.

Then, the magnitude of the position vector seperating the boats is X

X^2 = (8tsin30)^2 + (8tcos30 - 8(t+1))^2

From here I simplify and differentiate d/dt(X^2) = 2X.dX/dt

But the algebra appears clumsy and this is after all a high school calculus problem; I’m pretty sure I’m overcomplicating it.

The other method I’ve worked with is to work out the distance at 5pm by using the cosine rule, then differentiating the cosine rule equation to give me the rate of change. This seems a little more practical, but I’m still not happy with the answer.

I really am just a guy trying to figure out a math problem that should be well within my capabilities! If the posting of full methods is not appropriate, I would appreciate a bottom line numerical answer, so I can at least know what I’m aiming for

SaintCad has it. At all times the boats lie at the vertices of a 30-75-75 triangle whose apex is the port. The ratio of the separation speed to either boat’s speed is the ratio of the base of the triangle to one of its two equal sides.

This is my feeling as well…Boat B leaves port one hour later than A. This means that the triangle is not isosceles.

Another thing that’s bugging me here; the question asks at what rate the distance between the boats is changing at 5pm. Surely this value is not dependant on time once both boats are moving?

What I’m getting at here is that if you are going at a constant speed one way and I am going at a constant speed the other way, then surely our relative speed is constant?

Your relative speed is constant, but that doesn’t mean the distance between you increases at a constant rate. Consider, for example, what would happen if my position at time T was <1, T>. I’d be moving up vertically at a constant speed of 1, yet my distance from the origin would not be increasing at a constant rate (since sqrt(1 + T^2) does not have a constant derivative).

That is the method I used and got what I think is the right answer. The algebra isn’t too ugly. I had a lot of terms involving the sin(30) but since that’s 1/2, it makes it all pretty clean. I assume that was the intent of whoever designed the problem in the first place.

IOMDave, You had the right solution. Just carry it out to the end. It comes out to be 1.98945 knots for me. It’s a pretty straight-forward instantaneous rate of change calculus problem requiring some implicit differentiation.

Well, I got a different number. Here’s my calculus-free solution:

First, for (what I would consider) convenience’s sake, let’s use units which divide all the 8s out of the problem; we’ll bring them back at the end. (So, our basic speed unit will be equal to 8 knots, our basic distance unit will be equal to 8 nautical miles, and our basic time unit will remain 1 hour). Consider a reference frame which travels with boat B (thus, boat B will always be at the origin). At 1 PM, boat A is at <0, 1> (in our units). It is moving at a velocity (in this reference frame) of <0, 1> - <1/2, sqrt(3)/2> = <-1/2, 1 - sqrt(3)/2>. Thus, its location at 5 PM, four hours later, is <0, 1> + 4 * <-1/2, 1 - sqrt(3)/2> = <-2, 5 - 2 * sqrt(3)>. As I said above, only that part of its velocity which is “lined up” with its direction from the origin contributes to the instantaneous rate of change of its distance from the origin; thus, we get the answer by normalizing its position and taking the dot product with the velocity. [Another way to put this is that the answer is the magnitude of its velocity times the cosine of the angle between its velocity and its position]. Its position has magnitude sqrt(41 - 20 * sqrt(3)) and the dot product of its position with its velocity is -1/2 * -2 + (1 - sqrt(3)/2) * (5 - 2 * sqrt(3)) = 9 - 9 * sqrt(3)/2. Thus, we find that the final answer is (9 - 9 * sqrt(3)/2)/(sqrt(41 - 20 * sqrt(3))) * 8 knots, which is approximately 3.82526 knots.

ETA: And using calculus to do it for confirmation, I get the same number (3.82526…). So I’m pretty confident about it, though I may just have made an equivalent mistake both times.