Question

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

Answer 1

differentiate it

Answer 2

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

Answer 3

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

Answer 4

do you by chance have the answer?

Answer 5

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

Answer 6

but the question says the first five hour

Answer 7

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

Answer 8

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

Answer 9

heh you're smarter than me brah

Answer 10

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

Answer 11

so i did y2-y1/x2-x1

Answer 12

which is 1250-500/5-0

Answer 13

yeah 750; the answer

Answer 14

how do you work it out but?

Answer 15

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

Answer 16

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?

Answer 17

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

Answer 18

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

Answer 19

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!

Answer 20

ahhh i get you

Answer 21

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

Answer 22

ahhh bingo, got it. Thankyou

Answer 23

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?

Answer 24

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

Answer 25

No worries - just verify your answer!

Answer 26

thankyou!!!!

Answer 27

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

Answer 28

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.

Answer 29

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!

Answer 30

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

Answer 31

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

Answer 32

haha, anytime mate ;)