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Math Help - Series and sequences.

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by christina View Post
    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.)
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  3. #3
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    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})
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  4. #4
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    Quote Originally Posted by christina View Post
    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 .
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