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

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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

Mathematics
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differentiate it
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
then plug in \(t=5\) i think

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do you by chance have the answer?
yes the book says the answer is 750 bacteria per hour. but i disagree
but the question says the first five hour
it wants the average rate of change for the first 5 hours
@lgbasallote: what do you think? I forgot how to do this; or i will hahve to look in my book lol
heh you're smarter than me brah
well differentiated for t= 0 and t=5 you get 500 and 1250 respectively
so i did y2-y1/x2-x1
which is 1250-500/5-0
yeah 750; the answer
how do you work it out but?
sorry im no use; i forgot how to do these hopefully someone else can help
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?
500 as t=0 but differentiating the equation we get 30t^2 + 500
so plugging f'(0) into the differential we get 500 for initial
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!
ahhh i get you
uhhh the equation is 10t^3 +500t +120 thus at t=0 bacteria = 120
ahhh bingo, got it. Thankyou
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?
we would go 10(5)^3 + 500t + 120 = 3870. then 3870 - 120/5 = 750, the answer
No worries - just verify your answer!
thankyou!!!!
Right, 750 per hour would be the average rate of growth - very well!! :)
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.
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!
yeah right - for average rate you do need the instantaneous rate equation!
okie doke, will keep that in mind, thanks again :) have a good one mate, and cheers for lending a hand.
haha, anytime mate ;)

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