Integration of a rational function

dan

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Mar 25, 2015
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Hello. I've been doing my differential equations coursework recently and I am so close to finishing it but I have one last bit I'm stuck on.

The coursework aim is to model 2 equations, representing the velocity of an aeroplane that has just landed (at t=0). All we are given is a table of particular values for the velocity of the plane over 26 seconds (at 1s intervals), and the mass of the plane (120 000 kg).

Initially, the plane is slowed by just air resistance, then by air resistance and a CONSTANT breaking force.. so before breaks are on:
v' = -kv2 (air resistance is proportional to velocity)
Solving this gives me 1/v = kt - c
at t=0, v=96 => c = -1/96
I used parameters v=55,t=9 (given to me in the table) to find the value of k (these gave the smallest error across all the particular values I am supposed to be modelling)
I have to get it to the nearest integer (that's how the data is given to me), and with these parameters I managed to model all of the data correctly before t=9 (This is when I worked out that the breaks were turned on). My value for k was 41/47520 (=8.6279461x10-4)
My problem is after t=9.
I decided that since it is a constant breaking force, the deceleration associated with that constant braking force will also be constant. So...
v' = -kv - B
rearranging and seperating variable:
INTEGRAL OF(1/kv2+B)dv = INTEGRAL OF(1)dt
Obviously the right hand side is easy to integrate, however the left hand side is the problem.
I tried using the fact that INTEGRAL OF(1/a2+x2)dx = (1/a)arctan(x/a) + c
This gave me (although I may have made a mistake somewhere)...
(kB)-1/2arctan(v(k/B)1/2) + c = -t
when v=0, t=26 => c = -26
so... finally I have
(kB)-1/2arctan(v(k/B)1/2) - 26 = -t
(remember, I know the value of k already, B is the only unknown constant).
How on earth do I work out what B is?! (without use of trial and improvement/trial and error)
Thank you for your time!

P.S... Here is the table of values I was given:
t - v
0 - 96
1 - 89
2 - 82
3 - 77
4 - 72
5 - 68
6 - 64
7 - 61
8 - 58
9 - 55
10 - 50
11 - 46
12 - 41
13 - 38
14 - 34
15 - 31
16 - 27
17 - 24
18 - 21
19 - 18
20 - 16
21 - 13
22 - 10
23 - 8
24 - 5
25 - 3
26 - 0
 
Some hints:

For differential equations of the type
v' = -kv - B
or, rearranging and changing the -B to b,
v' + k v = b
The solution is of the form of a homogenous solution plus a particular solution.

We first find the homogenous solution, i.e.
v' + k v = 0
which can be solved in several ways, i.e. an integrating factor, separation of variables, ... Call that solution f(t). A particular solution can be found in several ways also but for first order linear equations I like the variation of parameters method. That is assume a solution of the form
v(t) = f(t)g(t)
If we do the differentiation using the chain rule we are left with an easy (in this case) solution for g(t). How easy the solution will be depends on b. If b were a function of t, the solution might not have a closed form solution.
 
Sorry I posted the second part wrong... it was supposed to be v' = -kv2-B , not v' = -kv-B
 
Sorry I posted the second part wrong... it was supposed to be v' = -kv2-B , not v' = -kv-B
I'm not sure there is another way. You could try a least squares approach but I think you would end up with the same problem. BTW: If you let a = (kB)1/2, then since B is an "arbitrary constant" so is a and you can write
v = a-1 k tan[a (26-t)]
which is a little easier to handle.

Not sure how well it would do but, since
B = - (v' + k v2),
you could do a numerical derivative, i.e. vj'=vj+1-vj since the time step is one (or some other numerical derivative) and do a least squares on that. That is, letting rj = - (vj' + k vj2):
E = \(\displaystyle \Sigma (B-r_j)^2\)
and minimize E wrt B.
 
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