Suppose the population of a colony of ants grows at the rate:
P’(x) = 1 + 5/sqroot(t+1)
If there are 250 ants after 120 hours, what was the population at t=0?
just crank it out:
dP/dt = 1 + 5/√(t+1)
dP = (1 + 5/√(t+1)) dt
P = t + 10√(t+1) + c
Now plug in (120,250) to find c, and then figure P(0)
To find the population at t=0, we can use the given information that the population after 120 hours is 250 ants.
The rate of change of the population is given by the derivative of P(x) with respect to x, which is P'(x). According to the given information, P'(x) = 1 + 5/√(t+1).
To find the population at t=0, we need to integrate P'(x) with respect to x. The indefinite integral of P'(x) will give us the population function P(x) of the colony, from which we can determine the population at t=0.
Let's solve this step by step:
Step 1: Integrate P'(x) with respect to x:
∫[P'(x) dx] = ∫[(1 + 5/√(t+1)) dx]
Step 2: Evaluate the indefinite integral to find P(x):
P(x) = ∫[(1 + 5/√(t+1)) dx]
= x + 10(√(t+1))+ C
Note: C represents the constant of integration.
Step 3: Use the given information to determine the value of C:
When x = 120, P(x) = 250
250 = 120 + 10(√(120+1)) + C
Solving for C, we have:
C = 250 - 120 - 10(√(120+1))
Step 4: Substitute C back into the population function P(x):
P(x) = x + 10(√(t+1)) + C
= x + 10(√(t+1)) + (250 - 120 - 10(√(120+1)))
Simplifying further:
P(x) = x + 10(√(t+1)) + 130 - 10(√(121))
Since we want to find the population at t=0, we substitute x = 0 into the population function:
P(0) = 0 + 10(√(0+1)) + 130 - 10(√(121))
= 0 + 10(1) + 130 - 10(11)
Simplifying the expression further:
P(0) = 10 + 130 - 110
= 30
Therefore, the population at t=0 is 30 ants.