1. ## Series and sequences.

Daily measurements taken each evening at a lake indicate that the lake loses a constant volume of water equal to 0.25% of its remaining volume each day.
i) find the approximate % of lake that remains after 30 days
ii) Find the approximate number of days before the lake is at 75% of its original capacity

I hope this is the right thread to post this in, i'm really not sure ><
I would really appreciate some help with this question, thank you.

2. Originally Posted by christina
Daily measurements taken each evening at a lake indicate that the lake loses a constant volume of water equal to 0.25% of its remaining volume each day.
i) find the approximate % of lake that remains after 30 days
ii) Find the approximate number of days before the lake is at 75% of its original capacity

I hope this is the right thread to post this in, i'm really not sure ><
I would really appreciate some help with this question, thank you.
If it "loses .25% of its volume", it retains 1- 0.0025= .9975 or 99.75% of its volume. Let V be the initial volume. Then the second day its volume is .9975V. The third day it is .9975 times that: $.9975(.9975V)= (.9975)^2V$. The fourth day it is .9975 times that: $.9975(.9975^2V)= .9975^3V$.

Get the point? This is a "geometric series". After n days, its volume will be $.9975^{n-1} V$.

i) What is $.9975^{30-1}$?
ii) Solve [/tex].9975^{n-1}= .75 (You may need to use logarithms.)

3. Hi Christina,

This is very similar to compound interest questions
(one in which a percentage is being lost instead of gained).

Initial volume of water = V, a constant.
Volume of water at the beginning of any day is a variable.
Amount lost per day is W(0.0025) as 0.25% is 0.0025,
where W=water volume remaining at the beginning of any day.

If you observe the pattern from day to day,
you will understand the formulation.

Day 1 (after 1 day) Y=V-(0.0025)V=V(1-0.0025)
Day 2 (after 2 days) Y(1-0.0025)=Z=V(1-0.0025)(1-0.0025)

You will begin to notice the pattern.
At the end of x days, the water volume remaining is $V(1-0.0025)^x$

Then you only need write

$\%\ of\ the\ lake\ that\ remains\ after\ x\ days\ is\ \frac{day\ 'x'\ volume}{initial\ volume}(100\%)$

You can then answer the first question, as x is known.

For the 2nd question, x is the unknown, so we need x=?

$\%\ W=0.75=\frac{V(1-0.0025)^x}{V}$

$0.75V=V(1-0.0025)^x$

$\log({0.75V})=x\log({[1-0.0025]V})$

4. Originally Posted by christina
Daily measurements taken each evening at a lake indicate that the lake loses a constant volume of water equal to 0.25% of its remaining volume each day.
i) find the approximate % of lake that remains after 30 days
ii) Find the approximate number of days before the lake is at 75% of its original capacity

I hope this is the right thread to post this in, i'm really not sure ><
I would really appreciate some help with this question, thank you.
HI

if the lake loses 0.25 % (1/400) of its water every day , the leftover would be 399/400 of its remaining volume

Let the initial volume be V , so

V , (399/400)V , (399/400)^2 V , ...

(1) Using $T_n=ar^{n-1}\Rightarrow T_{30}=V(\frac{399}{400})^{29}$

(2) T_n=0.75 % V , when is this gonna happen ?

$V(\frac{399}{400})^{n-1}=\frac{3}{4}V$ , don worry bout the v , they will eventually cancel off .

Apply logarithms , $(n-1)\log (\frac{399}{400})=\log (\frac{3}{4})$

n=115.9 days .