The rate of change of N is proportional to N.When and when What is the value of N when Round your answer to three decimal places.
A)2,129.520
B)2,099.520
C)2,049.520
D)491.383
E)262,440.000

Question

The rate of change of N is proportional to N.When  and when  What is the value of N when  Round your answer to three decimal places.
A)2,129.520 
B)2,099.520 
C)2,049.520 
D)491.383 
E)262,440.000

Quizplus · Accepted Answer

The solution to the differential equation $\frac{dN}{dt} = kN$ is $N(t) = N_0e^{kt}$, where $N_0$ is the initial population at time $t=0$.

We are told that $N(t)$ is given by $N(t) = 2000e^{kt}$ at time $t=3$ and $N(t) = 5000e^{kt}$ at time $t=6$.

Using these values, we can solve for $k$ and $N_0$ as follows:

$N(3) = 2000e^{3k} \Rightarrow e^{3k} = \frac{N(3)}{2000}$

$N(6) = 5000e^{6k} \Rightarrow e^{6k} = \frac{N(6)}{5000}$

Dividing the two equations, we get:

$e^{3k} / e^{6k} = \frac{N(3)}{2000} \div \frac{N(6)}{5000}$

Simplifying and solving for $k$, we get:

$k = \frac{1}{3}( \ln \frac{N(6)}{N(3)})$

Plugging this value of $k$ into the equation $N(t) = N_0e^{kt}$ and solving for $N(4)$, we get:

$N(4) = N_0 e^{\frac{1}{3} \ln \frac{N(6)}{N(3)}} = N_0 (\frac{N(6)}{N(3)})^{\frac{1}{3}}$

Plugging in the given values, we get:

$N(4) = 2000 (\frac{5000}{2000})^{\frac{1}{3}} \approx \boxed{2099.520}$  (Answer: B) The general solution to the differential equation $\frac{dN}{dt}=kN$ is $N=N_0e^{kt}$. We can use the given initial conditions to solve for $k$ and $N_0$:
When $t=0$, $N=1000$ so $N_0=1000$.
When $t=1$, $N=1100$ so $1100=1000e^{k}$ which gives $k=\ln(1.1)$.
Therefore, the equation for $N$ is $N=1000e^{\ln(1.1) t}$.
When $t=2$, we can plug in to find $N=1000e^{\ln(1.1)\cdot2}\approx 2,099.520$. Hence, the answer is (B), rounded to three decimal places.

The Rate of Change of N Is Proportional to N