Help maths question calculus
this isn't the crazy hard calculus... but I still can't get the correct answer.
The number of microbes (N) in a culture dish after t hours is given by the rule
N(t) = 1/3 (10t^2 - 1/9 t^3), 0 < t < 100
a Find the average rate at which the population is growing in the first 9 hours.
I think I might be overthinking it....
anyway, the answer is supposed to be 27
Help is much appreciated
October 16, 2011
6 Comments • Newest first
[quote=AnasF]Average rate of change is [N(9) - N(0)]/9[/quote]
Oh wow.. it was that easy...
<best answer here>
Thanks Anasf
Average rate of change is [N(9) - N(0)]/9.
Wow at the guy using integrals. That is the most roundabout way of getting N(9) - N(0) possible. (But useful in vector calculus).
[quote=samlee95]You both are wrong Average =/= plug in 4.5
average = [1/(9-0)] (integral of N'(x) from 0 to 9)
= (1/9) (N(9)-N(0))
= (1/9) (243)
= 27[/quote]
Umm is there a simpler way to do it without the use of integrals? .. because i haven't learnt that yet haha
[quote=skylifter]N(t) = 1/3 (10t^2 - 1/9 t^3)
Yep, first post is right.
One method is to use the Power Rule to find the derivative; it should be the easiest as well.
Note: Don't plug in NINE, it asks for the AVERAGE. therefore, plug in 4.5
(Average of start - finish = 0 - 9)
First, distribute the 1/3 so you get
10/3t^2 - 1/27 t^3
20/3t - 1/9t^2
20/3 (4.5) - 1/9 (4.5^2)
=27.75[/quote]
Oh right THANK YOU VERY MUCH
Really appreciate it
[quote=DrHye]Take the derivative and then plug in 9 for t? (someone verify please)
And for a note: Derivative=rate of change (or slope/velocity), 2nd derivative=acceleration[/quote]
Yeh I tried that, but I get 51 using that, which is the answer to question b) "Find the rate at which the population is growing after 9 hours."
keep in mind question a) says "[b]average[/b] rate of change",
Take the derivative and then plug in 9 for t? (someone verify please)
And for a note: Derivative=rate of change (or slope/velocity), 2nd derivative=acceleration
EDIT: Ok, been corrected