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Lachlan1996

  • 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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  1. Omniscience
    • 3 years ago
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    differentiate it

  2. Lachlan1996
    • 3 years ago
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    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

  3. Omniscience
    • 3 years ago
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    then plug in \(t=5\) i think

  4. Omniscience
    • 3 years ago
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    do you by chance have the answer?

  5. Lachlan1996
    • 3 years ago
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    yes the book says the answer is 750 bacteria per hour. but i disagree

  6. Omniscience
    • 3 years ago
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    but the question says the first five hour

  7. Lachlan1996
    • 3 years ago
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    it wants the average rate of change for the first 5 hours

  8. Omniscience
    • 3 years ago
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    @lgbasallote: what do you think? I forgot how to do this; or i will hahve to look in my book lol

  9. lgbasallote
    • 3 years ago
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    heh you're smarter than me brah

  10. Lachlan1996
    • 3 years ago
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    well differentiated for t= 0 and t=5 you get 500 and 1250 respectively

  11. Lachlan1996
    • 3 years ago
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    so i did y2-y1/x2-x1

  12. Lachlan1996
    • 3 years ago
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    which is 1250-500/5-0

  13. Omniscience
    • 3 years ago
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    yeah 750; the answer

  14. Lachlan1996
    • 3 years ago
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    how do you work it out but?

  15. Omniscience
    • 3 years ago
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    sorry im no use; i forgot how to do these hopefully someone else can help

  16. apoorvk
    • 3 years ago
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    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?

  17. Lachlan1996
    • 3 years ago
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    500 as t=0 but differentiating the equation we get 30t^2 + 500

  18. Lachlan1996
    • 3 years ago
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    so plugging f'(0) into the differential we get 500 for initial

  19. apoorvk
    • 3 years ago
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    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!

  20. Lachlan1996
    • 3 years ago
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    ahhh i get you

  21. Lachlan1996
    • 3 years ago
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    uhhh the equation is 10t^3 +500t +120 thus at t=0 bacteria = 120

  22. Lachlan1996
    • 3 years ago
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    ahhh bingo, got it. Thankyou

  23. apoorvk
    • 3 years ago
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    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?

  24. Lachlan1996
    • 3 years ago
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    we would go 10(5)^3 + 500t + 120 = 3870. then 3870 - 120/5 = 750, the answer

  25. apoorvk
    • 3 years ago
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    No worries - just verify your answer!

  26. Lachlan1996
    • 3 years ago
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    thankyou!!!!

  27. apoorvk
    • 3 years ago
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    Right, 750 per hour would be the average rate of growth - very well!! :)

  28. Lachlan1996
    • 3 years ago
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    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.

  29. Lachlan1996
    • 3 years ago
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    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!

  30. apoorvk
    • 3 years ago
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    yeah right - for average rate you do need the instantaneous rate equation!

  31. Lachlan1996
    • 3 years ago
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    okie doke, will keep that in mind, thanks again :) have a good one mate, and cheers for lending a hand.

  32. apoorvk
    • 3 years ago
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    haha, anytime mate ;)

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