Please someone get back to me! I really don't know how to do this, need as soon as possible. Please help!
A rabbit population satisfies the logistic equation
dy
dt = 2x(10to the exponent 6)y(10to the exponent6)- y;
where t is the time measured in months. The population is suddenly reduced to
40% of its steady state size by myxamatosis. If the myxamatosis then has no further effeect, how large is the population 8 months later? How long will it take
for the population to build up again to 90% of its steady state size?
thank you!
There is no t in your equation and it is incomprehensible anyway.
dy/dt = 2 * 10^6 y - y ??? or something
I also had several questions about this when you posted it before
http://www.jiskha.com/display.cgi?id=1338029875
You had clarified only one of the problems, I still find x and y undefined and no sign of any t in the equation.
I noticed that Damon interpreted your x as a multiplication.
yes that is what i am having trouble with. that is the question exactly and there is no t within the equation =/
I bet you mean
dy/dt = 2*10^6 y
dy/y = 2*10^6 dt
ln y = 2*10^6 t + c
y = C e^2*10^6 t where C is the population
at t = 0 just after the 40% reduction
at t = 8
y = C e^16*10^6 = impossibly big
please post the problem more carefully
To solve this problem, we need to understand the given logistic equation and its implications.
The given logistic equation is:
dy/dt = 2x(10^6)y(10^6) - y
where:
- dy/dt represents the rate of change of the rabbit population with respect to time.
- x represents the steady-state size of the rabbit population.
- y represents the current size of the rabbit population.
Let's break down the problem into two parts:
1. How large is the population 8 months later after being reduced to 40% of its steady state size?
To solve for this, we need to find the steady-state size of the population (x) and then calculate 40% of it.
Step 1: Set dy/dt = 0 to find the steady-state size (x):
0 = 2x(10^6)x(10^6) - x
0 = 2x^2(10^12) - x
Step 2: Solve for x:
2x^2(10^12) - x = 0
x(2x(10^12) - 1) = 0
Either x = 0 or 2x(10^12) - 1 = 0
Since the population cannot be zero, we solve for the second condition:
2x(10^12) - 1 = 0
2x(10^12) = 1
x(10^12) = 1/2
x = 1/2 * (10^(-12))
So, the steady-state size of the population is x = 1/2 * (10^(-12)).
Step 3: Calculate 40% of the steady-state size:
40% = 0.4 * (1/2 * (10^(-12)))
= (0.4/2) * (10^(-12))
= (0.2) * (10^(-12))
= 2 * (10^(-13))
Therefore, the population 8 months later is 2 * (10^(-13)).
2. How long will it take for the population to build up again to 90% of its steady-state size?
To solve this, we need to find the time it takes for the population to reach 90% of the steady-state size (0.9x).
Step 1: Set y = 0.9x in the logistic equation:
dy/dt = 2x(10^6)y(10^6) - y
dy/dt = 2x(10^6)(0.9x)(10^6) - 0.9x
Step 2: Solve the differential equation separating variables and integrating:
∫(1/y)(dy) = ∫(2x(10^6)(0.9x)(10^6) - 0.9x)(dt)
Integrating both sides gives:
ln|y| = ∫(1.8)(x^2(10^12)) - 0.9x*dt
ln|y| = 0.6(x^3(10^12)) - 0.45(x^2) + C
Taking the exponential of both sides gives:
|y| = e^(0.6(x^3(10^12)) - 0.45(x^2) + C)
Considering the initial condition, when t = 0, y = 0.9x, we can solve for C:
0.9x = e^(0.6(x^3(10^12)) - 0.45(x^2) + C)
C = ln(0.9x) - 0.6(x^3(10^12)) + 0.45(x^2)
Now, we can determine the time it takes for the population to reach 90% of its steady-state size by setting y = 0.9x in our equation and solving for t.
0.9x = e^(0.6(x^3(10^12)) - 0.45(x^2) + ln(0.9x) - 0.6(x^3(10^12)) + 0.45(x^2))
Simplifying the equation, we get:
0.9 = e^ln(0.9)
Since x = 1/2 * (10^(-12)), we can substitute this value into the equation and solve for t.
Now, we can't provide the exact value of t without the exact value of x. You'll need to calculate the exact value using a calculator or software.
I hope this explanation helps you!