1. anonymous

Hospital officials estimate that approximately N(p)=p^2+5p+900 people will seek treatment in an emergency room each year if the population of the community is thousand. The population is currently 20,000 and is growing at the rate of 1,200 per year. At what rate is the number of people seeking emergency room treatment increasing?

2. anonymous

Did you forget something between the words "is" and "thousand"?

3. anonymous

I think you're looking for $\frac{dN}{dt}$

4. anonymous

which can be found by looking at $\frac{dN}{dt}=\frac{dN}{dp}\frac{dp}{dt}$

5. anonymous

First, start with your function$N(p)=p ^{2}+5p+900$ Let's take a derivative of both sides to unlock the rates of change that are related$\frac{d}{dt}N(p)=\frac{d}{dt}(p ^{2}+5p+900)$ Then we can simplify to this:$\frac{dN}{dt}=2p \frac{dp}{dt}+5$ From here, let's plug in what we know:$1,200=2(20,000) \frac{dp}{dt}+5$ From here, you can solve for dp/dt

6. anonymous

so why did dp/dt only come up with the p^2 term and not the 5p term

7. anonymous

I'm wondering about that myself actually...

8. anonymous

and I'm actually looking for dN/dt according to the way this is making me input my answer

9. anonymous

Good gawd - then I really messed that one up... lol

10. anonymous

haha its all good I've been messing this one up for about an hour

11. anonymous

Let's try that one again... From the top - take two!

12. anonymous

Here is our function:$N(p)=p^2+5p+900$ Let's first identify some things what we are given, and we'll identify what they're asking us for (whenever I skip that, I screw things up).

13. anonymous

right

14. anonymous

that was right to the function not to you screwing up by the way haha

15. anonymous

The population is currently 20,000; so p = 20,000 It is growing at a rate of 1,200 per year; so dp/dt = 1,200

16. anonymous

They're asking for the the rate at which the number of people seeking medical attention is increasing; so dN/dt = ?

17. anonymous

That's what we're trying to find. :)

18. anonymous

ok well what if we take what you had a second ago 2(20)(dp/dt) +5 and plug in 1200 for dp/dt and yes thats what we are trying to find hha

19. anonymous

well actually thats just going to give us a ridiculously huge number

20. anonymous

But, will the number make sense?

21. anonymous

What did you get?

22. anonymous

no it was like 48 million

23. anonymous

$\frac{d}{dt}N(p)=\frac{d}{dt}(p^2+5p+900)$Lets plug these things into the right places this time... $\frac{d}{dt}N(p)=2p \frac{dp}{dt}+5\frac{dp}{dt}$ $\frac{dN}{dt}=2(20,000)(1,200)+5(1,200)$

24. anonymous

ya thats what i did

25. anonymous

Oh man... I got nothing... And nothing was left out of the question?

26. anonymous

ya i don't get it either man thanks anyway

27. anonymous

I'm going to take a look at this one on the calculator really quick, just to see if that will shine a little light on this one.

28. anonymous

Wait a sec...

29. anonymous

You originally wrote: "Hospital officials estimate that approximately N(p)=p^2+5p+900 people will seek treatment in an emergency room each year if the population of the community is thousand. The population is currently 20,000 and is growing at the rate of 1,200 per year. At what rate is the number of people seeking emergency room treatment increasing?"

30. anonymous

Did you instead mean: "Hospital officials estimate that approximately N(p)=p^2+5p+900 people will seek treatment in an emergency room each year if the population of the community IN thousands." ?

31. anonymous

Because if you did, then we should be plugging in 20 instead of 20,000. Actually, that may fix the problem. :)

32. anonymous

If the function is set up to inherently measure in thousands, then by typing in 20,000 - we're accidentally making the population 20,000,000.

33. anonymous

@TheFigure you already reached the answer: 48,006,000 people seek for emergency care per year!

34. anonymous

But he would be entering the answer in wrong if he had those extra zeros attached to the back of it.

35. anonymous

You've been right since 30 min before :)