Question

A bacteria culture initially contains 250 cells and grows at a rate proportional to its size. After 2 hours, the population has increased to 600 cells.

A) Find a function for the number of bacteria after t hours (FINISHED).
B) Find the number of bacteria after 8 hours (FINISHED).
C) Find the rate of growth after 8 hours.
D) When will the population reach 100,000?

My questions are: Does (C) involve derivatives? And is (D) related to (C), or can I use the normal exponential growth/decay model?

Answer 1

So (c) is asking you to find the rate of growth, I guess you knew that, but..
So, the formula for exponential growth \[y = a(1 + r)^{x}\]
where a = initial amount
r = rate of growth
x = number of time intervals that have gone by

Answer 2

wot

Answer 3

So then, this scenario would be \(y=250(1+r)^{8}\)?

Answer 4

Yep :)

Answer 5

Do you have a graphing calculator?

Answer 6

Er, I'm not sure the professor wants me to use one. But I do have a graphing calculator

Answer 7

Answer to your first question;
No it does not involve derivatives.

Answer 8

If I solved for \(r\) would that work as well?

Answer 9

o:
Eh... but the entire chapter section talks about derivatives. :(

Answer 10

Do you know what a derivative is?

Answer 11

And yes, if you solved for 'r' it should work, if done correctly

Answer 12

http://prntscr.com/cxyijf This is the first page of the section.

Answer 13

Immediately started off with (implicit?) differentiation

Answer 14

Oh, oops, I meant "differentiation," not "derivatives"
But yeah -

Answer 15

Um.. I'm not sure how to address differentiation

Answer 16

Yeah... I don't know either. But thanks for trying.

Answer 17

The first sentence can be translated into a differential equation. If the size of the culture at time \(t\) is \(y(t)\), then the growth rate is \(\dfrac{\mathrm dy}{\mathrm dt}=y'(t)\). You're told that the growth rate is proportional to the culture size, so there is some constant \(k\) for which
\[y'(t)=ky(t)\]This is a separable ODE, so you can write
\[\frac{\mathrm dy}{y}=k\,\mathrm dt\]then integrate and solve for the arbitrary constant of integration using the fact that the culture starts with \(250\) cells, so \(y(0)=250\). Meanwhile, you're also told that at \(t=2\), the culture has \(600\) cells, so \(y(2)=600\), and this is enough info to solve for \(k\).

I assume this is what you did for parts (a) and (b).

Part (c) is asking you to find the (instantaneous) growth rate at \(t=8\), which is a matter of computing
\[y'(8)=ky(8)=\cdots\]

Answer 18

Thank you @HolsterEmission :)
I actually found an explanation online at the same time as your reply haha