Solving real life problem with differential equation

In summary: If it starts exactly balanced, it will remain exactly balanced and not tumble.From the given problem, the acceleration is given asa = 25v + v^3We can write this asv^3 + 25v - a = 0This is a cubic equation in v with three possible roots. We can solve this using Cardano's method which gives usv = (A + B + C)/3whereA = -25/(3*3^(1/2))*(3b^2 - ac)^(1/2)B = (2/3)*(-b + (1+i*3^(1/2))a^(1/2))C = (2/3
  • #1
Woolyabyss
143
1

Homework Statement


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) ds1/(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 = .0044My 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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  • #2
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.
 
  • #3
haruspex said:
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.
 
  • #4
Woolyabyss said:

Homework Statement


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.
 
  • #5
Ray Vickson said:
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.
 

What is a differential equation?

A differential equation is a mathematical equation that relates a function to its derivatives. It represents the rate of change or the relationship between a function and its changes over time.

How can differential equations be used to solve real-life problems?

Differential equations can be used to model and solve a variety of real-life problems, such as predicting population growth, understanding the spread of diseases, and analyzing the movement of objects in physics.

What are the steps involved in solving a real-life problem with differential equations?

The first step is to identify the variables and their relationships in the problem. Then, formulate a differential equation that represents this relationship. Next, solve the differential equation using appropriate mathematical techniques. Finally, interpret the solution in the context of the real-life problem.

What are some common techniques used to solve differential equations?

Some common techniques used to solve differential equations include separation of variables, substitution, and the use of integrating factors. Other techniques such as Laplace transforms and numerical methods can also be used for more complex problems.

What are some limitations of using differential equations to solve real-life problems?

Some limitations of using differential equations include the assumption of continuous and smooth functions, which may not accurately represent real-life systems. In addition, the initial conditions and parameters used in the differential equation may not be entirely accurate, leading to errors in the solution.

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