2007
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Apr
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Small Increments & Approximations – A Familiar-Looking Inverted Cone of Water

(9)
Posted at 7:07 pm by Miss Loi in A-Maths Questions

After thwarting the Persians with the three Spartans, Miss Loi will be busily lighting up many many joss sticks to ‘celebrate’ this long weekend!

To spice up your holidays, she will leave you with a commonly-asked question starring an enigmatic inverted cone of water that has been sighted many many times …

Diagram of a very familiar inverted cone of waterThe diagram shows a vertical cross-section of an inverted cone of height 10 cm and base radius 6 cm. Water is poured in to a height of x cm.

  1. Show that the volume, V cm3, of water in the container is (3π/25)x3 cm3.
  2. Find the approximate increase in V when x increases from 3 to 3.02 cm.

Next time, instead of daydreaming at the water dispensor, why not think about how familiar the little cone you’re drinking from is going to be in your exam? :)

がんばって!!!

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Comments & Reactions

9 Comments



  1. 2007
    Apr
    5
    Thu
    8:31pm
     
    1

    woah, that was challenging... been like almost 6 years since i last attempted such quesion.

    let the unknown radius of the cone with x height be rx

    by porportion,

    rx = 6x/10

    Therefore, by the cone formula

    Vx = 1/3π(6x/10)^2*x
    = (3π/25)x3

    correct?

    the 2nd part, Vx3.02-Vx3.00 or

    3/25π[(x a)^3 - x^3)
    = 3/25π[x^2a xa^2 a^3]

    so, let x = 3 and a = 2, and u can solve for the differences.

    correct?

    hahaha... really been long since i ever did this man!!!



  2. 2007
    Apr
    5
    Thu
    11:47pm
     
    2

    Mossie you sure sound like an over-aged student!

    Part 1 is correct as you rightly used similar triangles for the proving.

    But Part 2 is WRONG! Wrong approach! Wrong formula! WRONG WRONG WRONG!!! *jumps up and down like a siow char bo*

    So *tidies hair* I think your 6 years of rustiness is showing :)

    Students, this question is still open - keep trying!



  3. 2007
    Apr
    6
    Fri
    8:27am
     
    3

    hahaha... i can't imagine if my teacher ever jumps up and down, left and right like a siaw char boh...

  4. Kiroii's Avatar
    Kiroii says


    2007
    May
    5
    Sat
    2:22am
     
    4

    r = 3/5x
    v= 1/3 π r^2 h
    v= 1/3 π 9x^2/25 x
    v= 3πx^3/ 25
    v=3π/25)x3 (shown)

    dv/dx = 9/3625π x^2 (thro quotion rule)
    delta v = dv/dx times 0.2 (with x as 3)
    delta v = 0.648 π
    correcto?



  5. 2007
    May
    5
    Sat
    10:31pm
     
    5

    Kiroii, answer correcto but working a little incorrecto which Miss Loi already corrected in red for you. Judging by the time of your answer, think you're a bit sleepy lah.

    Mossie see how the model student did it??? You can't just anyhow substitute the x values directly into the above volume formula!

    You are finding the change in V (i.e. dV) with respect to a small change in x (i.e. dx) so you die die have to differentiate first!

  6. 123's Avatar
    123 says


    2007
    Sep
    30
    Sun
    1:45pm
     
    6

    Familiar TYS qn....



  7. 2007
    Oct
    1
    Mon
    12:26am
     
    7

    Now that it's O-Level crunch time, more familiar 'must-do' questions will be coming your way.

    Do ask your fellow sufferers friends to come here to get an idea of the commonly recurring questions and also burn some joss sticks! :D

  8. stupid question's Avatar
    stupid question says


    2007
    Oct
    23
    Tue
    12:52am
     
    8

    ok a stupid question. must memorise figure formulas anot ar. like cone. ok probably too late to ask.



  9. 2007
    Oct
    23
    Tue
    1:50am
     
    9

    But of course! How many times have you seen 1/3 πr2h served to you on a platter?

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