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Differential Equation Word Problem?
Water flows out of a tank in the bottom. At time 't' minutes, the depth is 'x' metres. At any instant, the rate at which 't' is decreasing is proportional to the square root of the depth of water in the tank.
Part1
Write down a differential equation which models the situation.
Ok, no probs here: It is
dx/dt = -kx^0.5
Part2
When t=5, x=1
when t=0, x=2
Find t when x=0.5
ok, I know the answer but I can't quite figure it out.
I'm figuring you integrate both sides with respect to x.
On the RHS, I get (2/3)x^(3/2)+c, but I'm not quite sure about the LHS. I think I have to separate the variables but not quite sure how.
Help appreciated.
Thanks Robert, spot on!!
Thanks a Mill!
Thanks, I completely agree with your method. Only problem is I don't get the right answer. It turns out to be negative.
I get t = -12.07 when x=0.5
2 Answers
- ?Lv 41 decade agoFavorite Answer
Separate the variables. Treat dx/dt as dx divided by dt, where dx and dt are differentials.
dx/dt = -kx^0.5
dx = -kx^0.5 dt
dx / x^0.5 = -k dt
Integrate the LHS wrt x, and the RHS wrt t
2 x^0.5 = -kt + c
at t = 0, 2 * 2^0.5 = -k * 0 + c = c = 2 * 2^0.5 = 2.828
at t = 5, 2 * 1^0.5 = -k * 5 + 2.828
k= (2.828 - 2) / 5 = 0.166
So 2 x^0.5 = -0.166t + 2.828
2* (0.5)^0.5 = -0.166t + 2.828
gives t = 8.52
- nleLv 71 decade ago
dx/ sqrt(x) = -k dt
Integrate:
2*sqrt(x) = -kt +C
Now Use initial condition
when t =0 , x =2 then 2sqrt(2) = C
when t = 5 , x =1 then k = (2/5) [ sqrt(2) - 1 ]
Now plug in C and k into the original equation and solve for x
x (t) = 0 .00686291499 t^2- 0 .2343145748t +2.
