Solving a system in polar coordinates

In summary, the solution to the given system of equations is r(t) = \frac{xe^{t}}{1-x+xe^{t}} and \theta(t) = t, and the process of solving it involves finding the individual solutions for each equation and then using partial fractions to integrate them. The final values of the constants can be determined by plugging in the initial conditions.
  • #1
Somefantastik
230
0
Hey Everybody.

for the system:

[tex] r' = r(1-r) [/tex]
[tex] \theta' = 1 [/tex]

with

[tex]r(0) = x; \theta(0) = 0 [/tex];

the answer is

[tex]r(t) = \frac{xe^{t}}{1-x+xe^{t}} [/tex]

[tex]\theta(t) = t [/tex]

This answer was given in class as part of a process, and I can't remember how that answer is calculated. Can someone help me?
 
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  • #2
Somefantastik said:
Hey Everybody.

for the system:

[tex] r' = r(1-r) [/tex]
[tex] \theta' = 1 [/tex]

with

[tex]r(0) = x; \theta(0) = 0 [/tex];

the answer is

[tex]r(t) = \frac{xe^{t}}{1-x+xe^{t}} [/tex]

[tex]\theta(t) = t [/tex]

This answer was given in class as part of a process, and I can't remember how that answer is calculated. Can someone help me?
Pretty basic. You have two completely separate equations- just solve each one separately. [itex]\theta'= 1[/itex] so, integrating, [itex]\theta= t+ C[/itex]. Since [itex]\theta(0)= 0+ C= 0[/itex], C= 0 and [itex]\theta(t)= t[/itex]. [itex]r'= r(1- x)[/itex] is only slightly harder:[itex]dr/dt= r(1-r)[/itex] can be rewritten as
[tex]\frac{dr}{r(1- r)}= dt[/tex]
and that can be integrated by "partial fractions".
 
  • #3
HallsofIvy said:
Pretty basic. You have two completely separate equations- just solve each one separately. [itex]\theta'= 1[/itex] so, integrating, [itex]\theta= t+ C[/itex]. Since [itex]\theta(0)= 0+ C= 0[/itex], C= 0 and [itex]\theta(t)= t[/itex]. [itex]r'= r(1- x)[/itex] is only slightly harder:[itex]dr/dt= r(1-r)[/itex] can be rewritten as
[tex]\frac{dr}{r(1- r)}= dt[/tex]
and that can be integrated by "partial fractions".

[tex] \int\frac{dr}{r(1-r)} \ = \ \int dt [/tex]

where [tex] \int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1} [/tex]

so [tex] ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2} [/tex]

[tex] 2r -1 + C_{3} = e^{t} + C_{4} [/tex]

[tex] r = \frac{e^{t} + C + 1}{2} [/tex]

[tex] r(0) = \frac{e^{0} + C + 1}{2} = x [/tex]

[tex] C = x-1 [/tex];

[tex] r = \frac{e^{t} + x}{2} [/tex];

I'm still not getting the right answer.
 
Last edited:
  • #4
Somefantastik said:
[tex] \int\frac{dr}{r(1-r)} \ = \ \int dt [/tex]

where [tex] \int\frac{dr}{r(1-r)} \ = \ \int\left(\frac{1}{r} \ + \ \frac{1}{1-r} \right)dr \ = \ ln(r) \ - \ ln(1-r) + C_{1} [/tex]

so [tex] ln(r) \ - \ ln(1-r) + C_{1} = t + C_{2} [/tex]

[tex] 2r -1 + C_{3} = e^{t} + C_{4} [/tex]
No. ln(r/(1-r))= t+ C' (I've combined your C1 and C2).

Now r/(1-r)= Cet. (C is eC')
So r= Cet- rCet, (1- Cet)r= Cet, and
[tex]r(t)= \frac{Ce^{t}}{1- Ce^{t}}[/tex]
Since r(0)= x,
[tex]r(0)= \frac{C}{1- C}= x[/tex]
so C= (1- C)x, C= x- Cx, C+ Cx= C(1+ x)= x so
[tex]C= \frac{x}{1+x}[/tex]

[tex] r = \frac{e^{t} + C + 1}{2} [/tex]

[tex] r(0) = \frac{e^{0} + C + 1}{2} = x [/tex]

[tex] C = x-1 [/tex];

[tex] r = \frac{e^{t} + x}{2} [/tex];

I'm still not getting the right answer.
 
  • #5
Thanks so much, I get it now.
 

What are polar coordinates?

Polar coordinates are a system used to locate a point in a plane by its distance from a fixed point (known as the pole or origin) and its angle from a fixed reference line (known as the polar axis).

Why is it useful to solve systems in polar coordinates?

Solving systems in polar coordinates can be useful in situations where the problem involves circular or symmetrical shapes. It can also simplify calculations involving trigonometric functions.

How do you convert from Cartesian to polar coordinates?

To convert from Cartesian coordinates (x,y) to polar coordinates (r,θ), you can use the following equations: r = √(x² + y²) and θ = arctan(y/x). Keep in mind that the angle θ is measured counterclockwise from the positive x-axis.

What is the process for solving a system in polar coordinates?

The first step in solving a system in polar coordinates is to convert all equations from Cartesian form to polar form. Then, use substitution and/or elimination to solve for the variables. Finally, convert the polar solutions back to Cartesian form if necessary.

Can you solve any system using polar coordinates?

No, not all systems can be solved using polar coordinates. It is most effective when dealing with circular or symmetrical shapes. In some cases, it may be more efficient to solve the system using Cartesian coordinates.

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