1. ## Confirmation / Help

Hi there,

Could I get some confirmation that I'm using the correct method to answer the question?
I'm also having some trouble with one of the questions, it's been annotated.
I've attached two images and the .doc file

Help is much appreciated,
Thanks!

edit: couldn't attach .doc, so attached .pdf

Bump

3. ## Re: Confirmation / Help

Most forums try to discourage the bumping of topics (esp. after less than 2 hours), unless the OP has additional progress to post.

You will be more likely to get help if you take the time to type out your questions/work, preferably using $\LaTeX$ as many who help simply don't want to download files. Just a bit of unsolicited advice.

4. ## Re: Confirmation / Help

Sorry about that, just getting a little impatient, spent a good 30 - 45 minutes typing everything up. Is there a tutorial page on LATEX somewhere?

5. ## Re: Confirmation / Help

In (b), the reduction would be a curve (not a line) $f(t)=2-\sqrt{t-4}$. I replaced t with t-4 because the reduction period starts at t=4 hours.

For (c), they want to know the rate of increase of the area of the disc. The rate of increase of the radius of the disc is 0.5 cm/hour. So it's a related rates problem.

Everything else looked good.

- Hollywood

6. ## Re: Confirmation / Help

Just google or do a search here for lots of information on using LaTeX. Once you get started, you'll find its fairly straightforward, and once you know most of the commands it is a breeze.

In looking over you work, I find:

a) You have correctly determined the growth period of 4 hours and the reduction period of 4 hours to get a total time of 8 hours for the experiment.

b) You want to graph:

$r(t)=\begin{cases}0.5t & 0 \le t \le4 \\ 2-\sqrt{t-4} & 4\le t\le8 \\ \end{cases}$

c) Here you are asked how fast the area is growing just before the antibiotic is introduced, so begin with:

$A=\pi r^2=\pi\left(0.5t \right)^2=\frac{\pi}{4}t^2$

Now differentiate with respect to $t$, then plug in $t=4\text{ hr}$ to get the growth of the disk in $\frac{\text{cm}^2}{\text{hr}}$.

d) Correct.

e) This isn't correct.This should be done similarly to part c).

7. ## Re: Confirmation / Help

Here's what I came back with

(b) Table of values
t |r(t)
0|0
1|0.5
2|1
3|1.5
4|2
5|1
6|0.586
7|0.268
8|0

Where I got a graph looking like this

Attachment 28305

(c)

$\\A=\pi r^2\\A=\frac {\pi}{4}t^2\\\A= \pi-\frac{2\pi}{\sqrt{4}}= \pi -\frac {2\pi}{2}$

Sub 4 for t
$=\frac {\pi(4)}{2}\\ \\=6.283 cm^2/hr$

(e)I don't think this one is right

$A=\pi r^2\\=\pi (2-\sqrt{t-4})^2\\=\frac {d}{dt}\pi (2-\sqrt{t-4})^2\\=\pi-\frac {2\pi}{\sqrt{(8)-4}}\\=\pi -\frac {2\pi}{\sqrt{4}}=0 cm^2/hr$

Thanks for the help, I know my LaTeX isn't the cleanest but I'll try to clean it up in the future. I think I'm starting to appreciate the time people spend typing up each function/equation using LaTeX. It really goes to show how generous they are.

Another BIG thanks!

edit: The graph didn't come out right, had to fix it, can't get rid of the 'Attachment 28305' though

8. ## Re: Confirmation / Help

Yes, part e) is correct.