## Lachlan1996 3 years ago Calculus. A colony of bacteria is increasing in such a way that the number of bacteria present after t hours is given by N where N=120+500t +10t^3. Find the average rate of increase in bacteria/hour in the first 5 hours # please show working

1. Omniscience

differentiate it

2. Lachlan1996

i have, i answered 2 of the questions so far, and sort of this one, but im having trouble with it. Differentiated it is 30t^2 + 500

3. Omniscience

then plug in \(t=5\) i think

4. Omniscience

do you by chance have the answer?

5. Lachlan1996

yes the book says the answer is 750 bacteria per hour. but i disagree

6. Omniscience

but the question says the first five hour

7. Lachlan1996

it wants the average rate of change for the first 5 hours

8. Omniscience

@lgbasallote: what do you think? I forgot how to do this; or i will hahve to look in my book lol

9. lgbasallote

heh you're smarter than me brah

10. Lachlan1996

well differentiated for t= 0 and t=5 you get 500 and 1250 respectively

11. Lachlan1996

so i did y2-y1/x2-x1

12. Lachlan1996

which is 1250-500/5-0

13. Omniscience

14. Lachlan1996

how do you work it out but?

15. Omniscience

sorry im no use; i forgot how to do these hopefully someone else can help

16. apoorvk

Okay so you need to find the *average* rate of growth, for the first 5 hours. So, you need to find the net growth in the given time, and divide it by the time or no. of hours. So, what was the bacteria-count at the onset, that is at t=0?

17. Lachlan1996

500 as t=0 but differentiating the equation we get 30t^2 + 500

18. Lachlan1996

so plugging f'(0) into the differential we get 500 for initial

19. apoorvk

No no, do not differentiate - you have that equation defining 'N', which is the bacteria population - differentiating it would give you the expression for the instantaneous rate of population growth! I just need the average!

20. Lachlan1996

ahhh i get you

21. Lachlan1996

uhhh the equation is 10t^3 +500t +120 thus at t=0 bacteria = 120

22. Lachlan1996

ahhh bingo, got it. Thankyou

23. apoorvk

Just plug in t=0, and t=5 in the original expression for 'N' - and find the increase in population, which the difference of the to 'N' values. yeah 120 for t=0.. annnd?

24. Lachlan1996

we would go 10(5)^3 + 500t + 120 = 3870. then 3870 - 120/5 = 750, the answer

25. apoorvk

26. Lachlan1996

thankyou!!!!

27. apoorvk

Right, 750 per hour would be the average rate of growth - very well!! :)

28. Lachlan1996

thankyou very much. so in truth i just would have taken the y2-y1/x2-x1 like i had before, but without first differentiating the equation.

29. Lachlan1996

my teacher taught me wrong then, he said whenever you see the word rate, you differentiate, yet effectively this does not apply here. Nonetheless thankyou for your help mate!

30. apoorvk

yeah right - for average rate you do need the instantaneous rate equation!

31. Lachlan1996

okie doke, will keep that in mind, thanks again :) have a good one mate, and cheers for lending a hand.

32. apoorvk

haha, anytime mate ;)