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Math Help - Related Rates

  1. #1
    Junior Member
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    Related Rates

    Hello!

    I have a question in regards below, and have answered the question but do not understand number line 5, could you please explain the reason for this.

    A water tank has the shape of an inverted circular cone with base radius 3m and height 5m. If water is pumped into the tank at a rate of 1.5m3/min, find the rate at which the water level is rising when the water is 3.5m deep.

    Let t represent time (in minutes),
    h(t) represent height (in metres) of water at time t,
    r(t) represent radius (in metres) at time t,
    V(t) represent volume (in cubic metres) of water at time t.

    dV/dt = 1.5m3/min h=5m r=3m r/h=3/5 so, r=3h/5

    Formula: V(t) = 1/3 \pi [r(t)] h(t)

    1. d/dt{V(t)} = d/dt{1/3 \pi [r(t)] h(t)}

    ok i am going to skip to a few steps to get to the problem i am having.

    4. dV/dt= d/dt { 3 \pi h(t)3 / 25}
    5. d/dv = 9 \pi h(t) / 25 dh/dt
    ok, so where did the "dh/dt" come from.. is it because we are looking for
    the rate at which the water level is rising over time.. if so please advise
    if that is the reason.

    So why is "dh/dt" inserted so far down in the steps? is there a better method where it has "dh/dt" from the beginning since thats what we are looking for...

    in the end the answer is dh/dt \approx 0.108


    Thanks heaps!!!

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  2. #2
    MHF Contributor

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    Quote Originally Posted by Jon123 View Post
    Hello!

    I have a question in regards below, and have answered the question but do not understand number line 5, could you please explain the reason for this.

    A water tank has the shape of an inverted circular cone with base radius 3m and height 5m. If water is pumped into the tank at a rate of 1.5m3/min, find the rate at which the water level is rising when the water is 3.5m deep.

    Let t represent time (in minutes),
    h(t) represent height (in metres) of water at time t,
    r(t) represent radius (in metres) at time t,
    V(t) represent volume (in cubic metres) of water at time t.

    dV/dt = 1.5m3/min h=5m r=3m r/h=3/5 so, r=3h/5

    Formula: V(t) = 1/3 \pi [r(t)] h(t)

    1. d/dt{V(t)} = d/dt{1/3 \pi [r(t)] h(t)}

    ok i am going to skip to a few steps to get to the problem i am having.

    4. dV/dt= d/dt { 3 \pi h(t)3 / 25}
    5. d/dv = 9 \pi h(t) / 25 dh/dt
    ok, so where did the "dh/dt" come from.. is it because we are looking for
    the rate at which the water level is rising over time.. if so please advise
    if that is the reason.

    Yes, that is the reason.

    So why is "dh/dt" inserted so far down in the steps? is there a better method where it has "dh/dt" from the beginning since thats what we are looking for...

    in the end the answer is dh/dt \approx 0.108

    Thanks heaps!!!
    It is there "from the beginning". Your first equation is V= \frac{\pi}{3}r^2(t)h(t) and then the very next equation is \frac{dV}{dt}= \frac{\pi}{3}\frac{d}{dt}(r^2(t)(h(t)). So you already have the derivative with respect to t in there. Now use the relationship between h and r to write as a function of h only: \frac{dV}{dt}= \frac{\pi}{3}\frac{d}{dt}(h^3(t)/25) = \frac{\pi}{75} \frac{dh^3}{dt}
    (You have an error- the "1/3" accidently became "3")

    Now use the chain rule to differentiate h^3 with respect to t.
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