For some readers it may be useful to first examine previous blogs to do with this topic, e.g.

http://brane-space.blogspot.com/2010/06/looking-at-basic-differential-equations.html

and http://brane-space.blogspot.com/2010/06/back-to-diferential-equations.html

One of the most powerful applications is to rocket flight, for which a diagram of the type shown is often used, say to define the launch angle q

_{o}, as well as the final x, y- coordinates (e.g. at impact). Consider then the following problem:

Find the x- and y-coordinates of the points on the trajectory of a rocket launched at an angle of 80 degrees with an initial velocity of 100,000 f/s if the air resistance is 0.01mv. Find the value of x and y after 10 seconds.

Data: q = 80

^{o}, k = 0.01mv v = 10^{5}fs^{-1}and t = 10s
The differential equation can be written:

m(d

^{2}x/dt^{2}) = - k (dx/dt)/ v = - (0.01mv)x’/ v = - 0.01mx’
Similarly:

my” = -mg – (0.01mv)y’/v = -mg – 0.01my’

Yielding the simultaneous pair:

x” + 0.01x’ = 0

y” + 0.01 y’ = -g

We know:

x’

_{o}= v_{o}cos (q) = v_{o}cos (80) = (10^{5}) cos 80
y’

_{o}= v_{o}sin (q) = v_{o}sin (80) = (10^{5}) sin 80*General Solutions*:

x = c1 + c2 (e

^{-0.01t})
y = c3 + c4((e

^{-0.01t}) - 100 gt
x’ = -0.01c2 (e

^{-0.01t})
y’ = -0.01c4((e

^{-0.01t}) - 100 g
At t = 0, x = 0 so:

0 = c1 + c2 (e

^{-0.01t})
But:

x’

_{o}= v_{o}cos (80) = (10^{5}) cos 80 = -0. 01c2 (e^{-0.01t})
Then: c2 = - (10

^{7}) cos 80
So: 0 = c1 + ( - (10

^{7}) cos 80) or c1 = (10^{7}) cos 80
At t = 0, y = 0:

Then: 0 = c3 + c4 (e

^{-0.01t})
And: y’

_{o}= v_{o}sin (80) = (10^{5}) sin 80 = -0.01c4(e^{-0.01t}) - 100 g
So: c4 = -(10

^{7}) sin 80 – 10000g
\ c3 = - c4 (e

^{-0.01t}) = 10^{7}sin 80 + 10000g*To obtain the x and y-coordinates*:

First, the x-coordinate:

x = c1 + c2 (e

^{-0.01t}) = 10^{7}cos 80 – (10^{7}cos 80)( e^{-0.01t})
x = 10

^{7}cos 80 [1 - ( e^{-0.01t})] = 10^{7}(0.17365) [1 - ( e^{-0.01t})]
or: x = 1736500 [1 - ( e

^{-0.01t})]
The y-coordinate:

y = c3 + c4 ( e

^{-0.01t})
Or: y = 10

^{7}sin 80 + 10000g – (10^{7}sin 80 + 10000g(e^{-0.01t}))
y = 10

^{7}sin 80 + 10000g(1 - e^{-0.01t}) – 100 gt
y = [10

^{7}(0.9848) + 320,000](1 - e^{-0.01t}) – 100 gt
y = 1016800(1 - e

^{-0.01t}) – 100 gt*After 10 seconds*:

x = 1736500 [1 - ( e

^{-0.01t})] = x = 1736500 [1 - ( e^{-0.01(10)})]
x = 165, 200 ft. (» 31. 3 miles)

y = 1016800(1 - e

^{-0.01(10)}) – 100(32)(10)
y = 930,580 ft. ( » 177 miles)

Problem for the Math Maven:

Re-work the problem but obtain the coordinates in kilometers, instead of miles. (Hint: you will use an initial velocity of 30,300 m/s instead of 100,000 f/s. Also you will use g = 10 ms

^{-2 }instead of g = 32 fs^{-2}) What differences did you find in your final values - from those in the blog problem - and how might you account for them? (Convert either miles to km, or vice versa, to compare).
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