Morphing Triangle Problem?

Question

[b]1. Two boats start out 800 miles apart with boat A directly to the west of boat B. At the same time both boats start moving with boat A travelling to the east at 40mph while boat B travels north at 20mph. Determine if the distance between the boats is increasing, decreasing, or not changing after the following travel times: (a) 7 hours (b) 16 hours (c) 25 hours [/b]



[b]2. I attempted to break apart their distances travelled by using component vector analysis. The distance between the two boats is the hypotenuse of whatever triangle they produce on the graph at a certain time. We want to find d'.[/b]



[b]3. d'=? I park boat A at origin, therefore its cartesian coordinates are (0, 0), so boat B must be at (0, 800)

For part (a) I use vector components: A in x-direction: 400(7)=280 x-direction
A is (280, 0) B is (800, 140)
Base distance is 800-280=520 Height is just 140, how do I proceed to find h' ? Similar triangle don't work bcuz boats have diff. rates.

Answer 1

This problem was deliberately made easy by having the two boats on orthogonal (perpendicular) courses.

It happens to be easier still if you put B at the origin O. Then A is traveling along the X axis, B is traveling along the Y axis, and AOB is a right angle so the distance between A and B is given by the length of the hypotenuse. This means we can treat AO and OB as scalar lengths. Then:

D = sqrt(AO^2 + OB^2)

where:

AO = -800 + 40t
OB = 20t

But what we really want is the derivative of D w.r.t. time

We need to use the chain rule: if H(t) = F(G(t)) then
dH/dt = (dF/du)(du/dt) where

u = G(t) so du/dt = dG/dt

In this case:
F(u) = sqrt(u)
G(t) = AO(t)^2 + OB(t)^2

So we compute the formula for H'(t) - the rate at which the distance changes, and then evaluate that formula at t = 7, t = 16, etc. If H'(t) is positive, the distance is increasing, etc.

In the more general case, AO and OB are the vectors (not just scalars) of interest, the angle between them isn't 90 degrees, and it doesn't matter where we put the origin.

So now, the vector from ship A to ship B is the vector D

D = AO + OB

and we want to find the magnitude of the vector D.

AO and OB get represented as two-dimensional vectors. If we want to put the origin at the starting point of A, with the positive X axis going east we get:

AO = <-40t, 0>
OB = <800, 20t>

(Note that AO goes from A to O; we can also use OA by reversing all the signs)

We add the two vectors component by component to get the vector D(t):

D(t) = <800 - 40t, 20t>

The magnitude of a vector is given by the square root of the sum of the squares of the components, so, we are back to computing H'(t). In this case, we get the exact same answer, as we should.