Calc problem

[kim jean]

In a certain barbarous land, two neighboring tribes have hated one another from time immemorial. Being barbarous people, their powers of belief are strong, and a solemn curse pronounced by the medicine man of the first tribe deranges the members of the second tribe and drives them to murder and suicide. If the rate of change of the population, P, of the second tribe is -P^(1/2) people per week, and if the population is 676 when the curse is uttered, when will they all be dead?

 

[tkhunny]

I found an old algebra text not long ago. It had problems about dropping bombs and flying the route to accomplish the task. I thought, "Wow! You can't write problems like THAT anymore!" I guess we CAN write worse ones.

 

[opticaltempest]

We're given the rate of change of the population in the problem. We want to know how many weeks must pass until the population reaches zero. To find that we need to find the function that describes our population at any given time.

Letting t = time (in weeks), P = the population,

We want to find the time at which the population is zero.
For what values of t does P(t) = 0.

We are given dP/dt = -t^(1/2)

To find the population function we need to integrate

integral[-t^(1/2)] dt

Evalutating the above integral you find P(t)=(-2/3)t^(3/2) + C.

Next we have to solve for C.

In the problem we are told the initial population is 676 tribesmen.
So our population function at time zero must be 676.
P(0) = 676

Replace t with 0 and set equal to 676 in order to find the value of C

(-2/3)0^(3/2) + C = 676

C = 676

So our population function is P(t) = (-2/3)t^(3/2) + 676

Now solving (-2/3)t^(3/2) + 676 = 0 will give us the value of t when
the population reaches zero.

EDIT: It looks like Rahian2k and Gene are correct. I assumed -P^(1/2) was
a typo and really -t^(1/2), which would have made it a simpler problem. I hope I didn't cause too much confusion. [/b]

 

[rahian2k]

[deleted]

 

[Gene]

How 'bout
dP/P^(1/2) =
2(1/2)dP/P^(1/2 =
2d(P^(1/2) = -dt
Integrating
2P^(1/2)=-t+c
(as Rahi said)
P = (c-t)²/4 =
(c²-2ct+t²)/4
P(0)=c²/4=676
c=52 giving
P = (52²-2*52*t+t²)/4
P=0 at t=52 weeks
(but P=.25 at 51 weeks and
1 at 50 weeks)