anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
differentiate it
anonymous
  • 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
  • anonymous
then plug in \(t=5\) i think

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

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