Expert: Socrates - 9/16/2006

Question
I a having difficulty with the following problems can you give me some guidance.  Thanks.

1. A newly opened housing community stocks its lake with 1000 fish at time t=0.(time is in months)  From that time on, the population of the fish increases a the rate of cubrt(F) fish per month when there are F fish in the lake.  How many fish will be in the lake in 6 months? Round your answer to the nearest fish.

2. This one is really giving me heck.

A particle has an acceleration of a(t) = 9-t^2.  Find s(t) if v(0)=6 and s(0)=6.

Answer
Problem 1) is much more difficult than 2). Unless you are told the form of the function F in advance, you must solve a little differential equation.


1)

Let F(t) be the number of fish at time t

F(0) = 1,000

the population of the fish increases a the rate of
cubrt(F) fish per month when there are F fish in the lake

This means dF/dt = F^1/3

so we can write

dF = F^1/3 dt

F^-1/3 dF = 1 dt

finding the anti derivative of both sides gives

(3/2)F^2/3 = t + C

F^2/3 = (2/3)(t + C)

F(t) = (2/3)^3/2 (t+C)^3/2

Since F(0) = 1000,

1000 = (2/3)^3/2 (C)^3/2

1000 = (.5443)(C)^3/2

1,837.2221 = (C)^3/2

1,837.2221^2/3 = C

150.0057 = C

So we now know that

F(t) = (.5443) (t+150.0057)^3/2

Letting t=6

F(6) = (.5443) (156.0057)^3/2

F(6) = (.5443)(1,948.5461)

F(6) = 1,060.59364

rounding to the nearest whole number of fish

F(6) = 1,061


2)

a(t)= 9 - t^2

v(t) = 9t - (1/3)t^3 + C

Since v(0) = 6

6 =  C


So now we know that

v(t) = 9t - (1/3)t^3 + 6

Then

s(t) = (9/2)t^2 - (1/12)t^4 + 6t + D

Since s(0) = 6

6 =  D


The answer is

s(t) = (9/2)t^2 - (1/12)t^4 + 6t + 6