Solving real life problem with differential equation


by Woolyabyss
Tags: differential, equation, life, real, solving
Woolyabyss
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#1
Oct30-13, 12:08 AM
P: 80
1. The problem statement, all variables and given/known data
A car starts from rest.When it is at a distance s from its starting point,its speed is v and it acceleration is a = (25v + v^3).

Show that dv = (25 + v^2)ds and find its speed when s = 0.01

2. The attempt at a solution

a = v(dv/ds) = (25v + v^3) divide both sides by v and cross multiply s

dv = (25 + v^2) ds


1/(25+v^2) dv = 1 ds

integrating both sides

(1/5)tan-inverse(v/5) = s

using limits s = 0 when v = 0 and s = 0.01 when v=v

1/5(tan-inverse(v/5) = .01

tan-inverse(v/5) = .05

v/5 = tan.05

v = 5tan.05 = .0044


My book says the answer is 1.28m/s. I think I might have gone wrong with the limits?
Any help would be appreciated.

If I solve the last part using radians the answer is still only 0.25
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haruspex
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#2
Oct30-13, 06:00 AM
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You should certainly use radians. Standard formulae such as the one you used to integrate to get arctan assume radians. I also get 0.25. The book answer matches 5 tan(.25). Maybe it was supposed to be s =.05.
Woolyabyss
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#3
Oct30-13, 07:18 AM
P: 80
Quote Quote by haruspex View Post
You should certainly use radians. Standard formulae such as the one you used to integrate to get arctan assume radians. I also get 0.25. The book answer matches 5 tan(.25). Maybe it was supposed to be s =.05.
Thanks it might be wrong so I think I'll move on to the next question then and when I first started differentiation I remember my book making a very big deal about only using radians in calculus but it just never seems to stick, thanks anyway.

Ray Vickson
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Oct30-13, 11:17 AM
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Solving real life problem with differential equation


Quote Quote by Woolyabyss View Post
1. The problem statement, all variables and given/known data
A car starts from rest.When it is at a distance s from its starting point,its speed is v and it acceleration is a = (25v + v^3).

Show that dv = (25 + v^2)ds and find its speed when s = 0.01

2. The attempt at a solution

a = v(dv/ds) = (25v + v^3) divide both sides by v and cross multiply s

dv = (25 + v^2) ds


1/(25+v^2) dv = 1 ds

integrating both sides

(1/5)tan-inverse(v/5) = s

using limits s = 0 when v = 0 and s = 0.01 when v=v

1/5(tan-inverse(v/5) = .01

tan-inverse(v/5) = .05

v/5 = tan.05

v = 5tan.05 = .0044


My book says the answer is 1.28m/s. I think I might have gone wrong with the limits?
Any help would be appreciated.

If I solve the last part using radians the answer is still only 0.25
There is something seriously wrong with the original question: if the acceleration is a = v^3 + 25 v, then v(t) obeys the differential equation
[tex] \frac{dv}{dt} = v^3 + 25 v, [/tex]
whose possible solutions are ##v(t) = v_1(t), v_2(t) \text{ or } v_3(t)##, where
[tex] v_1(t) = 0\:\: \forall t \\
v_2(t) = \frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }\\
v_3(t) = -\frac{ 5e^{25(t+c)} }{ \sqrt{1 - e^{50(t+c)} } }
[/tex] and where c is a constant. For solutions v_2 and v_3 there are no values of c that make v(0) = 0; in fact, there is no t at all that makes v(t) = 0, so the car could never, ever be at rest! It can, of course, be at rest for solution v_1(t), but in that case it remains at rest forever.
haruspex
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#5
Oct30-13, 03:48 PM
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Quote Quote by Ray Vickson View Post
There is something seriously wrong with the original question:
Good point. If a = v*f(v) and v(t0) = 0 then a and all higher derivatives are zero at t = t0. No movement can occur.
To make the question work, the initial condition needs to be lim s→0 v = 0, or somesuch. A 'real world' example is an object nudged away from unstable equilibrium, like a pencil balanced on its point.


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