SydneyRoverP6B said:
Well we have a correct answer and Richard (quattro) is the first to do so. Well done Richard...
I used Calculus,...related rates and implicit differentiation to solve the problem. Time 12 mins and distance 32 root 13 km which approximates to 115.38km. Using Pythagorean triangles repeatedly to narrow down the distance will also yield the same outcome.
Hmmm....I'll have to find a tougher teaser next time.
Ron,
Here's my solution, which does match the others! I use vectors, as I find them very convenient!
For this problem, I shall assign t to have units of
minutes, as opposed to
seconds (again, for convenience).
At t = 0
Jet 1 is heading east at 12km/min. Let's assume that Jet 1 is at the origin at t = 0. Jet 1's displacement, which we we shall call
S1
S1 = 12t
î
Jet 2 is heading north at 8km/min and is 208 km away from Jet 1, at the same altitude/height. Jet 2's displacement, which we we shall call
S2
S2 = 208
î + 8t
ĵ
The displacement vector between the two is
S2 -
S1 = (208
î + 8t
ĵ) – (12t
î) = (208 – 12t)
î + 8t
ĵ
The magnitude of the vector is, as always, z^2 = x^2 + y^2
x = (208-12t), and y = (8t)
Implicit Differentiation time!
2z dz/dt = 2x dx/dt + 2y dy/dt, as 2 is a common factor, we can eliminate the 2’s to get :-
z dz/dt = x dx/dt + y dy/dt.
Now, we could divide through by z, but it’s not necessary here….
z is the distance between the planes, and therefore is non-zero. (If it was zero, it would mean an air crash had occurred! :shock
For max/min values, dz/dt = 0. We want the min value, so to make sure what we work out is the min value… we work out what (d/dt)^2 {z} is
z dz/dt = [(208 – 12t) x -12] + 64t = 208t – 2496 = 208(t – 12). dz/dt is 0 at t = 12 minutes
Differentiating again gives that (d/dt)^2 {z} = 208. As (d/dt)^2 {z} > 0, this means that at t = 12 mins, the minimum value of z is found. (There is no global maximum value, as after t = 12mins, the distance keeps on increasing.)
The Euclidean distance between the planes :-
||
S2 -
S1|| = [(208 – 12t)^2 + (8t)^2]^0.5
Ar t = 12 mins…
t =12 min , ||
S2 -
S1|| = [(208 - 144)^2 + (8 x 12)^2]^0.5 = [64^2 + 96^2]^0.5
As 64 and 96 are all multiples of 32, so… 64 = 32 x 2 and 96 = 32 x 3,
32 x [2^2 + 3^2]^0.5 = 32 x [4 + 9]^0.5 = 32 x 13^0.5 = 32 x 3.60555 (5 d.p) = 115.378km (3 d.p.)
The minimum distance between the planes is 115.378km, at time t = 12 minutes.