hi! just needed help on an FRQ for ap calculus ab. let me know if you have any questions for me. I'm just really confused as far as what I am meant to do. If you could walk me through it that would be amazing. THANKS!!

A population is modeled by a function P that satisfies the logistic differential equation. dP/dt = P/5(1-(P/10)).
A. If P(0) = 3, what is lim (as t approaches infinity)P(t)?
B. If P(0) = 20, what is lim (as t approaches infinity)P(t)?
C. A different population is modeled by a function Y that satisfies the seperable differential equation:
dY/dt=(Y/5)(1-(t/10)). Find Y(t) if Y(0) =3.
D. For the function Y found in part (c), what is lim(as t approaches infinity) Y(t)?

well, you should start out by solving the DE:

P = 10/(1+c*e^(-t/5))

(A) P(0) = 3, so
10/(1+c) = 3
c = 7/3

P(t) = 10/(1 + 7/3 e^-t/5)

Now it's clear that P->10

Now try the rest.

Hi! I'd be happy to help you with your AP Calculus AB FRQ. Let's go through each part of the question step by step.

A. The first part asks you to find the limit of P(t) as t approaches infinity, given that P(0) = 3. To solve this, you need to solve the logistic differential equation dP/dt = P/5(1 - (P/10)).

To solve this differential equation, you can separate the variables. Start by multiplying both sides by dt and then rearrange the equation to have all P terms on one side and all t terms on the other side.

You'll end up with (1/(P(1 - P/10))) dP = dt/5.

Next, integrate both sides. The integral of (1/(P(1 - P/10))) dP is the natural logarithm of the absolute value of the expression (1 - P/10). The integral of dt/5 is (1/5)t + C, where C is the constant of integration.

Now, you can solve for P. Take the natural logarithm of both sides to eliminate the exponential and absolute value. You'll get ln|1 - P/10| = (1/5)t + C.

To find the specific solution, you'll need to use the initial condition P(0) = 3. Substitute t = 0 and P = 3 into the equation obtained above.

ln|1 - 3/10| = (1/5)(0) + C.

Simplify and solve for C.

ln|7/10| = C.

Now you have the particular solution. Substitute t = infinity into the equation above. The limit as t approaches infinity will be the value of P(t) at infinity. ln|1 - P(infinity)/10| = (1/5)infinity + ln|7/10|.

Since exponential growth cannot exceed 10, P(infinity) must be 10. Thus, ln|1 - 1/10| = ln|9/10| = ln|7/10| + ∞/5.

Therefore, the limit of P(t) as t approaches infinity, given P(0) = 3, is 10.

B. The process for solving part B is similar to part A, but with P(0) = 20. Repeat the steps above, substituting P(0) = 20 into the equation and solve for P(infinity).

You'll find that ln|1 - P(infinity)/10| = ln|9/10| + ∞/5.

Since exponential growth cannot exceed 10, P(infinity) must be 10. Thus, ln|1 - 1/10| = ln|9/10| = ln|9/10| + ∞/5.

Therefore, the limit of P(t) as t approaches infinity, given P(0) = 20, is 10.

C. For this part, you are given a different differential equation: dY/dt = (Y/5)(1 - (t/10)). To find Y(t) with the initial condition Y(0) = 3, you can solve this separable differential equation.

Similar to the previous parts, separate the variables by multiplying both sides by dt and rearranging to have Y terms on one side and t terms on the other.

You'll get (1/Y) dY = (1/5)(1 - (t/10)) dt.

Now, integrate both sides. The integral of (1/Y) dY is the natural logarithm of the absolute value of Y, and the integral of (1/5)(1 - (t/10)) dt is (1/5)t - (1/50)t^2/2 + K, where K is the constant of integration.

Combine the integrals to get ln|Y| = (1/5)t - (1/50)t^2/2 + K.

Using the initial condition Y(0) = 3, substitute t = 0 and Y = 3 into the equation.

ln|3| = (1/5)(0) - (1/50)(0)^2/2 + K.

Simplify and solve for K.

ln|3| = K.

Now that you have the particular solution, substitute t = infinity into the equation above. The limit as t approaches infinity will be the value of Y(t) at infinity. ln|Y(infinity)| = (1/5)(∞) - (1/50)(∞)^2/2 + ln|3|.

The term (1/50)(∞)^2/2 will approach infinity. Since Y(t) cannot be negative and the population cannot exceed 10, you can conclude that Y(infinity) will approach 10.

Therefore, the limit of Y(t) as t approaches infinity, given Y(0) = 3, is 10.

D. For this part, you already have Y(t) from part C. To find the limit of Y(t) as t approaches infinity for the function Y found in part (C), you can directly substitute t = infinity into the equation ln|Y| = (1/5)t - (1/50)t^2/2 + ln|3|.

Similar to part C, the term (1/50)(∞)^2/2 will approach infinity. Since Y(t) cannot be negative and the population cannot exceed 10, you can conclude that Y(infinity) will approach 10.

Therefore, the limit of Y(t) as t approaches infinity, given Y(0) = 3, is 10 as well.

I hope this explanation helps! Let me know if you have any further questions.