Equation problem

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lamaclass

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Oct 18, 2009
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I've partially solved this problem but am not sure how to continue from where I've left off:

A lake is filled with 500 fish. Their population increases according to the equation

P(t) = 10,000/1 + 19e[sup:17qm3e9l]-.02t[/sup:17qm3e9l]

1) After how many months is the population increasing most rapidly? (When is the rate of growth at a maximum?)

2) What is the point in part 1 called with respect to the graph of P(t)?

So far I've found the derivative of P(t) and set it equal to zero but am not sure what to do after that.

P'(t) = -10,000(19*-0.2)*e[sup:17qm3e9l]-0.2t[/sup:17qm3e9l]/1+19e[sup:17qm3e9l]-0.2[/sup:17qm3e9l]

=38,000e[sup:17qm3e9l]-0.2t[/sup:17qm3e9l]/1+19e[sup:17qm3e9l]-0.2t[/sup:17qm3e9l]=0 ??
 
I disagree with your derivative. The denominator should be squared.
P(t)=10000/[1+19e^(-.02t)]
dP(t)/dt = [1+19e^(-.02t)] d/dt [10000] - 10000[19e^(-.02t)[-.02] all over [1+19e^(-.02t)]^2
dP(t)=[3800e^-(.02t)] / [1+19e^(-.02t)]^2

we want to know when the derivative of the rate of change[dP/dt] is equal to zero

take the second derivative of P(t) and set it = to 0 to determine the max or minimun of the rate of change

Arthur
 
In your question you have two different values, e^(-.2t) and e^(-.02t), so you'll need to clarify that first.

Also, your derivative is slightly wrong; it should read 38,000e^(-.2t)/(1+19e^(-.2t))^2 - you just forgot to square the bottom.

The first part of the question is asking for when the rate of change is at a maximum, not the population. If it were asking for the population's maximum you would be correct in setting the derivative equal to zero, however since it want's the rate of change's (derivative's) maximum, you would need to take the SECOND derivative and then go ahead and set that equal to zero. Another way to do this is by taking the derivative once (shown above) and then graphing it on a graphing calculator to find the maximum.

The second part of the question is asking what this point is called-- an Inflection Point. This is the point on the original graph (of population) where it changes concavity.

Hope this helps some!
 
Oh whoops! It should as -0.2 sorry about that!

Okay I see that now! However after reading your post, I am unable to find the second derivative for it and keep getting different answers for it. :?
 
dP/dt = 38,000e^(-.2t)/(1+19e^(-.2t))^2

Take the derivative of this using the quotient rule, d/dx f(x)/g(x) = (f'(x)*g(x)-f(x)g'(x))/(g(x))^2.

Remember that when taking the derivative of the bottom (g(x)) you will have to use the chain rule:

EX) d/dx (3x+2)^2 = 2 * (3x+2) * 3
 
I'm still having a hard time trying to find the second derivative of this one. :(
 
The second derivative of the given equation is \(\displaystyle \frac{76e^{\frac{t}{50}}\left(19-e^{\frac{t}{50}}\right)}{\left(e^{\frac{t}{50}}+19\right)^{3}}\)

Set to 0 and solve for t. This is easier than it looks. Look at the numerator.

What t value makes \(\displaystyle 19-e^{\frac{t}{50}}=0\)
 
lamaclass wrote:

Is there a way to do it on a calculator?
Click to expand...

Yes, remember in my original post I mentioned that you could graph the FIRST derivative, then use the calculator to solve for the maximum value on this graph (where the second derivative equals zero).

Hope this helps.
 
\(\displaystyle lamaclass;\)

\(\displaystyle P(t) \ = \ 10,000/1+19e^{-.02t} \ = \ 10,000+19e^{-.02t}, \ your \ notation.\)

\(\displaystyle P(t) \ = \ 10,000/(1+19e^{-.02t}) \ = \ \frac{10,000}{1+19e^{-.02t}}, \ I \ assume \ this \ is \ the \ equation.\)

\(\displaystyle Unless \ you \ marry \ the \ bosses' \ daughter, \ you'll \ have \ a \ hard \ road \ to \ follow \ in \ the \ real \ world.\)
 
Thank you everyone for your help and patience in helping me to understand problemes of these types! It's greatly appreciated! :)
 
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