The rate of growth dP/dt of a population of bacteria is proportional to the square root of t, where p is the population size and t is the time in days (0<equal t <equal 10). That is, dp/dt = ksqrt(t). The initial size of the population is 500. After 1 day the population has grown to 600. Estimate the population after 7 days.
I know that f(0)=500 and f(1)=600.
The first thing I would do would be to find the integral of ksqrt(t)dt, right? I don't understand what that k means. Do I just leave it there?
This is what I have so far, but I think it is wrong.
dp/dt = ksqrt(t)
dp=integral(ksqrt(t)dt)
P= 2/3k(t^3/2)
Now I'm stuck. Can someone help point me in the right direction?
To solve this problem, let's start by finding the value of the constant "k" using the given information. We know that after 1 day, the population has grown from 500 to 600. We can use this information to find the value of "k".
Given that dp/dt = ksqrt(t), we can integrate both sides of the equation with respect to t:
∫(dp/dt) dt = ∫(ksqrt(t)) dt
Integrating both sides gives us:
∫ dp = k∫ sqrt(t) dt
Integrating the left side with respect to p and the right side with respect to t, we have:
p = k * (2/3)t^(3/2) + C
Where C is a constant of integration. Since we know that f(0) = 500, we can use this information to find the value of C:
500 = k * (2/3) * (0)^(3/2) + C
500 = C
Therefore, the equation becomes:
p = k * (2/3)t^(3/2) + 500
Now, we can use the known value of f(1) = 600 to solve for the value of k:
600 = k * (2/3)*(1)^(3/2) + 500
600 = k * (2/3) + 500
100 = k * (2/3)
k = 150/2
k = 75
The equation becomes:
p = 75 * (2/3)t^(3/2) + 500
Now, we can use this equation to estimate the population after 7 days:
p(7) = 75 * (2/3)*(7)^(3/2) + 500
Simplifying the equation:
p(7) = 75 * (2/3)*(7 * sqrt(7)) + 500
p(7) ≈ 75 * (2/3)*(14 * 2.646) + 500
p(7) ≈ 75 * (2/3)*(37.044) + 500
p(7) ≈ 75 * (24.696) + 500
p(7) ≈ 1852.2 + 500
p(7) ≈ 2352.2
Therefore, the estimated population size after 7 days would be approximately 2352.2.
To solve this problem, you need to use both the given information and the differential equation provided.
First, we are given that the initial population size is 500, so we have P(0) = 500.
Next, we are told that after 1 day, the population has grown to 600. This gives us the population size at t = 1, so we have P(1) = 600.
Now, let's solve the differential equation to find the value of k. We have:
dp/dt = k√(t)
To integrate this, we can rewrite it as dp = k√(t)dt.
Now, integrate both sides with respect to their respective variables:
∫dp = ∫k√(t)dt
Integrating the left side gives us P, and integrating the right side gives us (2/3)k(t^(3/2)) + C, where C is the constant of integration.
So we have:
P = (2/3)k(t^(3/2)) + C
Since we know the initial condition P(0) = 500, we can substitute this into the equation:
500 = (2/3)k(0^(3/2)) + C
Simplifying, we find that C = 500.
Now, we can rewrite our equation as:
P = (2/3)k(t^(3/2)) + 500
To find the value of k, we can use the information given that P(1) = 600:
600 = (2/3)k(1^(3/2)) + 500
Substituting t=1 gives us:
600 = (2/3)k + 500
Simplifying, we find:
(2/3)k = 100
Multiplying both sides by 3/2, we get:
k = 150
Now that we have the value of k, we can substitute it back into our equation for P:
P = (2/3)(150)(t^(3/2)) + 500
To estimate the population after 7 days, substitute t = 7 into the equation:
P(7) = (2/3)(150)(7^(3/2)) + 500
Calculate this expression to find the estimated population after 7 days.