Two airplanes (at the same height) are flying away from an airport at a right angle to each other. Plane X is 190 miles from the airport, plane Y is 250 miles from the airport, plane X is traveling at a speed of 490 mi/hr, and plane Y is traveling at a speed of 540 mi/hr. How fast is the distance between them increasing?
(DO not assume that the planes departed at the same time)
2007-04-01 07:36:44 · 4 answers · asked by Nicky 2 in Science & Mathematics ➔ Mathematics
call distance between x, y is z
we have x=190, y=250,
use pythagorean theorem to find z
z^2=x^2+y^2
z^2=190^2+250^2
z= 314 mile
z^2=x^2+y^2
take derivative both sides
2zdz/dt= 2xdx/dt+2ydy/dt
2*314dz/dt= 2(190)(490)+2(250)(540)
628dz/dt= 456200
dz/dt=456200/628= 726 mi/hr
good luck
2007-04-01 07:50:25 · answer #1 · answered by Helper 6 · 0⤊ 0⤋
How would you calculate the distance between these planes? It would be d = Sqrt[ (190 + 490*t)^2 + (250 + 540*t)^2 ]. Now take the derivative of d with respect to t. You get dd/dt = (22810 + 53170t)) / Sqrt[ 5317t^2 + 4562t + 986]. Then just plug in t = 0, and we get dd/dt(0) = 11405 * Sqrt[2/493] = (approx.) 726.4 mi/hr.
Hope this helps.
Steve
2007-04-01 14:49:17 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The distance between them is d = sqrt(x^2 + y^2) where x and y are the distance each plane is from the airport.
dd/dt = 1/2 * 1/sqrt(x^2 + y^2) * (2x dx/dt + 2y dy/dt)
Sub in...
dd/dt = 1/2 * 1/sqrt(190^2 + 250^2) * (2*190*490 + 2*250*540)
Time for a calculator!
2007-04-01 14:40:38 · answer #3 · answered by tedfischer17 3 · 1⤊ 0⤋
it depends where on earth this is. close the poles ? or close between the two poles ?
2007-04-01 14:41:10 · answer #4 · answered by gjmb1960 7 · 0⤊ 2⤋