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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
 one year ago
 one year 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
 one year ago
 one year ago

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Lachlan1996Best ResponseYou've already chosen the best response.0
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
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
then plug in \(t=5\) i think
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
do you by chance have the answer?
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
yes the book says the answer is 750 bacteria per hour. but i disagree
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
but the question says the first five hour
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
it wants the average rate of change for the first 5 hours
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
@lgbasallote: what do you think? I forgot how to do this; or i will hahve to look in my book lol
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
heh you're smarter than me brah
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
well differentiated for t= 0 and t=5 you get 500 and 1250 respectively
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
so i did y2y1/x2x1
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
which is 1250500/50
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
yeah 750; the answer
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
how do you work it out but?
 one year ago

OmniscienceBest ResponseYou've already chosen the best response.1
sorry im no use; i forgot how to do these hopefully someone else can help
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
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 bacteriacount at the onset, that is at t=0?
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
500 as t=0 but differentiating the equation we get 30t^2 + 500
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
so plugging f'(0) into the differential we get 500 for initial
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
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!
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
uhhh the equation is 10t^3 +500t +120 thus at t=0 bacteria = 120
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
ahhh bingo, got it. Thankyou
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
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?
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
we would go 10(5)^3 + 500t + 120 = 3870. then 3870  120/5 = 750, the answer
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
No worries  just verify your answer!
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
Right, 750 per hour would be the average rate of growth  very well!! :)
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
thankyou very much. so in truth i just would have taken the y2y1/x2x1 like i had before, but without first differentiating the equation.
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
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!
 one year ago

apoorvkBest ResponseYou've already chosen the best response.2
yeah right  for average rate you do need the instantaneous rate equation!
 one year ago

Lachlan1996Best ResponseYou've already chosen the best response.0
okie doke, will keep that in mind, thanks again :) have a good one mate, and cheers for lending a hand.
 one year ago
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