a population grows according to p(t)=10e^t-e^(3t/2) for t=0
a. at what rate is the population changing at t=2 and t=4
b. When is the population maximuma dn what is the maximum population
c. What is the initial population? When is the population zero?

Question

a population grows according to p(t)=10e^t-e^(3t/2) for t>=0
a.  at what rate is the population changing at t=2 and t=4
b.  When is the population maximuma dn what is the maximum population
c.  What is the initial population?  When is the population zero?

anonymous · Answer

Since the population is modeled by $p(t)$, the rate of change of the population is given by $p'(t)$.
$$p(t)=10e^t-e^{3t/2}\
p'(t)=10e^t-\frac{3}{2}e^{3t/2}$$
For the first question, find $p'(2)$ and $p'(4)$.

For the second, use the first derivative test to find the maxima (if any) of $p(t)$.

For the last, the initial population is the population at time $t=0$, so find $p(0)$. The population is zero when $p(t)=0$, so find the $t$ that makes this happen (if it does).

anonymous · Answer

I am unclear on the second part?  I did the first derivitave test.  Am i finding the local maxs?  Also how do I find when adn what they are?

anonymous · Answer

Local max's occur between intervals on which $p'(t)$ is positive and negative.

So, for example, the function $y=\dfrac{1}{1+x^2}$ has a max at $x=0$ because its derivative is positive (and thus $y$ is increasing) over $(-\infty,0)$, and negative (and thus decreasing) over $(0,\infty)$.