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Math Help - rate of change problem

  1. #1
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    rate of change problem

    Hi, I'm new here and I would be very grateful if you guys can help me with these 2 exercises :-)
    ---
    1. A container in the shape of a right circular cone of heighr 20 cm and radius 5cm is held vertex downward and filled with water which then drips out from the vertex at the rate of 5cm3/s. Find the rate of change of the height of water in the cone when it is half empty (measured by volume) <= not quite sure here why the change of height is measured by volume, but this one I take from 'The core course for A-level', so it's rather confusing.

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    2. A 5m ladder is held against the wall. The top of the ladder is sliding down the wall at the rate of 1m/s. Find the rate af change of the angle between the ladder and the ground when the vertical distance between the top of the ladder and the foot of the wall is 3m.

    ---
    I know how this type of exercise is normally solved, but I just fail to apply it into these 2
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  2. #2
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    Re: rate of change problem

    Hi koplan,

    For the first question, you will need to use similar triangle to help you solve the answer.
    Let's say r is the radius and h is the height. Then you will need to use similar triangle to write r in term of h (like comparing two fractions).
    Then you need to derive the volume equation in terms of h.
    Then you will probably need to plug in the h that you found when is half volume.

    Question 2:
    You will need to use sin function to derive.
    Since sin t = opp/hyp
    t is your angle
    opp is the height
    hyp is the ladder which is constant

    Hope this help
    Francis
    Tutorat A+ Tutoring | Go for A+ , A big step toward the future.
    Thanks from kopklan
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  3. #3
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    Re: rate of change problem

    For the first one draw the cone and label. mark height and corresponding radius also. By similar triangles we get 4h/r = 20/5=4 that gives r = (1/4 ) h
    Volume of cone = 1/3 pi r^2 h = 1/3 pi ( 1/4)^2 h^2 h = 1/48 pi h^3
    Now differentiate dV/dt = d [1/48 pi h^3] / dt = 1/48 pi 3 h^2 dh / dt. It is given that volume is decreasing by 5 cm^3 / s therefore we have
    - 5 = 1/48 pi 3 h^2 dh / dt Now I am sure you can proceed further.
    Thanks from kopklan
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  4. #4
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    Re: rate of change problem

    Thank guys, the 1st one was solved correctly :-)
    But I'm still stuck with the 2nd one. With sin function, I have z = sint = h/5, so dz/dh = 1/5, that means (dz/dh).(dh/dt) = (1/5).(-1) = -1/5 (constant), so regardless of the height (h), the rate of change in sint is always -1/5 per sec?. And by the way, how can I deduce the rate of change of the angle from the rate of change of its sin?
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  5. #5
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    Re: rate of change problem

    Quote Originally Posted by kopklan View Post

    A 5m ladder is held against the wall. The top of the ladder is sliding down the wall at the rate of 1m/s. Find the rate af change of the angle between the ladder and the ground when the vertical distance between the top of the ladder and the foot of the wall is 3m.
    if the distance from wall is x,
    x^2 + h^2 = 25
    x = √(25-h^2)
    when h=3, x = √(25-9) = √16
    sinθ = h/5

    2x dx/dt + 2h dh/dt = 0

    So, dh/dθ

    h/5 = sinθ
    h = 5sinθ
    dh/dθ = 5cosθ = 5(x/5) = 5(√16/5)= 4

    Let me know if that was helpful

    edit: No, it wasn't - I did a mistake
    Last edited by dokrbb; March 28th 2013 at 07:26 AM.
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  6. #6
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    Re: rate of change problem

    I think your use of "t" is confusing you! On the left side of the equation, sin(t), you mean t to be the angle but when you differentiate, you are using t as the time.
    Instead write the angle as " \theta" so that sin(\theta)= h/5 so that cos(\theta)\frac{d\theta}{dt}= -1/5.
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  7. #7
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    Re: rate of change problem

    with cos(θ).(dθ/dt) = -1/5, then dθ/dt = (-1/5).(1/cosθ). When the vertical distance between top of the ladder and the foot of the wall is 3m, then cosθ = 4/5, so dθ/dt = (-1/5).(5/4) = -1/4. Am I right? if I am, how can I put this identity (dθ/dt = -1/4) into words?
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  8. #8
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    Re: rate of change problem

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  9. #9
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    Re: rate of change problem

    Quote Originally Posted by kopklan View Post
    with cos(θ).(dθ/dt) = -1/5, then dθ/dt = (-1/5).(1/cosθ). When the vertical distance between top of the ladder and the foot of the wall is 3m, then cosθ = 4/5, so dθ/dt = (-1/5).(5/4) = -1/4. Am I right? if I am, how can I put this identity (dθ/dt = -1/4) into words?
    do you mean "(dθ/dt = -1/4) into words" ? it's the rate of change of the angle with respect to time, and it's negative 1/4 m per seconds because it descends,

    hope this helps
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  10. #10
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    Re: rate of change problem

    Quote Originally Posted by dokrbb View Post
    do you mean "(dθ/dt = -1/4) into words" ? it's the rate of change of the angle with respect to time, and it's negative 1/4 m per seconds because it descends,

    hope this helps
    Not quite. Because it is an angle it is -1/4 radians per second, not meters.
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  11. #11
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    Re: rate of change problem

    Quote Originally Posted by HallsofIvy View Post
    Not quite. Because it is an angle it is -1/4 radians per second, not meters.
    thanks HallsofIvy, always I have problems with the radians vs all other
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