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anonymous
 3 years 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
anonymous
 3 years 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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0then plug in \(t=5\) i think

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you by chance have the answer?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes the book says the answer is 750 bacteria per hour. but i disagree

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but the question says the first five hour

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it wants the average rate of change for the first 5 hours

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@lgbasallote: what do you think? I forgot how to do this; or i will hahve to look in my book lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0heh you're smarter than me brah

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0well differentiated for t= 0 and t=5 you get 500 and 1250 respectively

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0which is 1250500/50

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0how do you work it out but?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry im no use; i forgot how to do these hopefully someone else can help

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Okay 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0500 as t=0 but differentiating the equation we get 30t^2 + 500

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so plugging f'(0) into the differential we get 500 for initial

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No 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
 3 years ago
Best ResponseYou've already chosen the best response.0uhhh the equation is 10t^3 +500t +120 thus at t=0 bacteria = 120

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ahhh bingo, got it. Thankyou

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0No worries  just verify your answer!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Right, 750 per hour would be the average rate of growth  very well!! :)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0my 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah right  for average rate you do need the instantaneous rate equation!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0haha, anytime mate ;)
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