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Math Help - Word problem involving differentiation..

  1. #1
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    Word problem involving differentiation..

    A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

    So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

    I do know that the volume would be 1.5 x 1.2 x height.

    But I cannot put things together..!

    p.s. I have to use the chain rule.
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    Hello !
    Quote Originally Posted by struck View Post
    A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

    So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

    I do know that the volume would be 1.5 x 1.2 x height.

    But I cannot put things together..!

    p.s. I have to use the chain rule.
    Let h(t) denote the depth of the water at time t.

    Like you said, the volume would be : V(t)=1.5 \times 1.2 \times \text{height}=1.8 \times h(t)

    Now if you differentiate both sides, not by using the chain rule but by using : (a u(x))'=au'(x), where a is a constant, it'll give :

    \frac{dV}{dt}(t)=1.8 h'(t)

    0.45=1.8 h'(t)


    and h'(t) is the rate of change of the depth !
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  3. #3
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    Quote Originally Posted by struck View Post
    A water tank has a rectangular base 1.5m by 1.2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45m^3. At what rate is the depth of water in the tank increasing?

    So far, I can say that dV/dt = 0.45, i.e. the rate of change in volume with respect to time is 0.45m^3 / minute.

    I do know that the volume would be 1.5 x 1.2 x height.

    But I cannot put things together..!

    p.s. I have to use the chain rule.
    What is 0.45m^3/minute it is the volume that gets in one minute inside the tank, right? By how much would this volume increase the depth ? Volume/(Area of the base) or 0.45/(1.5*1.2) meters/minute
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