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

- anonymous

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- anonymous

differentiate it

- anonymous

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

- anonymous

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

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- anonymous

do you by chance have the answer?

- anonymous

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

- anonymous

but the question says the first five hour

- anonymous

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

- anonymous

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

- lgbasallote

heh you're smarter than me brah

- anonymous

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

- anonymous

so i did y2-y1/x2-x1

- anonymous

which is 1250-500/5-0

- anonymous

yeah 750; the answer

- anonymous

how do you work it out but?

- anonymous

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

- 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?

- anonymous

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

- anonymous

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

- 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!

- anonymous

ahhh i get you

- anonymous

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

- anonymous

ahhh bingo, got it. Thankyou

- 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?

- anonymous

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

- apoorvk

No worries - just verify your answer!

- anonymous

thankyou!!!!

- apoorvk

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

- anonymous

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.

- anonymous

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!

- apoorvk

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

- anonymous

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

- apoorvk

haha, anytime mate ;)

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