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?
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 plug in t=0
To find the population at t=0, we need to integrate the given rate equation P'(x) with respect to time, t, and then solve for the constant of integration by using the given information.
Given:
P’(x) = 1 + 5/√(t+1)
We need to integrate P’(x) to find P(x).
Integrating the equation, we have:
∫[P’(x)] dx = ∫[1 + 5/√(t+1)] dx
The integral of 1 with respect to t is t, and the integral of 5/√(t+1) can be found using the substitution method. Let's substitute u = √(t+1) and find du.
u = √(t+1)
Squaring both sides: u^2 = t + 1
Differentiating both sides with respect to t: 2u du = dt
Dividing both sides by 2u: du = dt/(2u)
Now we can rewrite the integral using the substitution u = √(t+1):
∫[1 + 5/√(t+1)] dx = ∫[1 + 5/u] * dt/(2u)
Simplifying the integral:
= ∫(1/2u) dt + ∫(5/u^2) dt
= (1/2) ∫(1/u) dt + 5 ∫(1/u^2) dt
Integrating with respect to t:
= (1/2) ln |u| - 5/u + C
Now, we need to substitute back u = √(t+1) and solve for C, the constant of integration.
Substituting u = √(t+1):
= (1/2) ln |√(t+1)| - 5/√(t+1) + C
= (1/2) ln √(t+1) - 5/√(t+1) + C
= (1/2) ln (√(t+1)) - (5/√(t+1)) + C
Now, we can use the given information that at t = 120 hours, the population is 250 ants.
P(120) = (1/2) ln (√(120+1)) - (5/√(120+1)) + C = 250
We can solve this equation for C.
C = 250 - (1/2) ln (√(120+1)) + (5/√(120+1))
Now, we have the value of the constant of integration, C. We can substitute it back into the equation to find the population at t = 0.
P(0) = (1/2) ln (√(0+1)) - (5/√(0+1)) + C
= (1/2) ln (√1) - (5/√1) + C
= (1/2) ln (1) - (5/1) + C
= 0 - 5 + C
= C - 5
Therefore, the population at t = 0 is C - 5.