# Derivative to find rate of change/population.

• October 18th 2010, 08:12 AM
Goose
Derivative to find rate of change/population.
I'm just looking for someone to point me in the right direction for this question. It's easy in theory but I'm stuck on it for some reason!

8. The fish population in a lake can be modeled by the function p(t) = 15(t^2 + 30)(t + 8), where t is time in years from now.
a) Determine the rate of change of the fish population when there are 5000 fish in the lake.
b) When will the fish population double from its current level? What is the rate of change in the fish population at this time?

So p'(t)= 45t^2 + 240t + 450

a) 0= 15t^3 + 120t^2 + 450t - 1400
b) 0= 15t^3 + 120t^2 + 450t - 3600

From there I just can't seem to factor and work out the answer. Thanks.
• October 18th 2010, 08:26 AM
Unknown008
a) look for:

5000 = 15t^3 + 120t^2 + 450t + 3600
Solve for t. Then use this value in p'(t)

b) Find the number of fish at the current time, that is when t = 0. Let that amount by Y

Double it, and solve for Y = 15t^3 + 120t^2 + 450t + 3600

Then, use that time in p'(t).
• October 18th 2010, 08:34 AM
Goose
"a) 0= 15t^3 + 120t^2 + 450t - 1400
b) 0= 15t^3 + 120t^2 + 450t - 3600

From there I just can't seem to factor and work out the answer. "

I've set up my equations to solve but I just can't seem to factor them.
• October 18th 2010, 09:19 AM
Unknown008
Factor 5.

0= 3t^3 + 24t^2 + 90t - 280

Ok... Hm... Maybe use an iteration?

$t_{n+1} = \sqrt{\dfrac{280-90t}{3t +24}}$

I get:

t = 1.91

Starting from 1.9

(I used wolframalpha to get an idea of the first point to use)

b) 0 = 15t^3 + 120t^2 + 450t - 3600

Well, I use nearly the same iteration.

$t_{n+1} = \sqrt{\dfrac{240-30t}{t +8}}$

I get t = 3.45

You can now find the rate of growth with the two values of t now.
• October 18th 2010, 09:35 AM
Goose
So far we've learned to use the factor theorem for higher degree polynomials, which is apparently why I couldn't work this out! Thanks for your help!
• October 18th 2010, 09:49 AM
Unknown008
I can give an example to iteration if you like :)

It's a way to find roots that are otherwise not easy to get.

If you have $0 = x^3 -27$

The usualy way is to make it like this: $x^3 = 27$

Hence, $x = ^3\sqrt{27} = 3$

The same principle is used, but with x.

$0 = 15t^3 + 120t^2 + 450t - 3600$

$0 = t^3 + 8t^2 + 30t - 240$

By preference, I find this way better:

$t^3 + 8t^2 = 240 - 30t$

Factorise t^2;

$t^2(t + 8) = 240 - 30t$

Divide by (t+8) then take the square root.

$t = \sqrt{\dfrac{240 - 30t}{t+8}}$

Now, put a random number not too far fetched. The previous number had 5000 and t was 1.9. I take 3 for 7200.

$t_{n+1} = \sqrt{\dfrac{240 - 30t_n}{t_n+8}}$

Take t_n = 3

$t_{1} = \sqrt{\dfrac{240 - 30(3)}{(3)+8}} = 3.69$

It's quite close to 3 itself... take this number into the equation now.

$t_{2} = \sqrt{\dfrac{240 - 30(3.69)}{(3.69)+8}} = 3.32$

Again, do the same thing.

$t_{3} = \sqrt{\dfrac{240 - 30(3.32)}{(3.32)+8}} = 3.51$

As you can see, the values oscillate about a number. This number is the solution we are looking for. Continue this process until the desired accuracy is obtained and we get t = 3.45 at 3 significant figures