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    A tank is shaped as a cuboid with a square base 10 cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, h cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of 2cm^3s^-1. Find an expression for dh/dt. Please "EXPLAIN" THE METHOD.
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    (Original post by YasinZahra)
    A tank is shaped as a cuboid with a square base 10 cm. Water runs out through a hole in the base at a rate proportional to the square root of the height, h cm, of water in the tank. At the same time, water is pumped into the tank at a constant rate of 2cm^3s^-1. Find an expression for dh/dt. Please "EXPLAIN" THE METHOD.
    you have to figure out an expression for the derivatives. i.e dv/dt
    for dv/dt water leaves the tank at a rate proportional to sqrt(h) but enters at a rate of 2

    dv/dt= 2-ksqrt(h)
    the k sqrt(h) is negative as it leaves

    try and see if you can do it from there
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    Yeah actually I tried it, but I really don't understand why we have to subtract the value of two different speed to get the resultant speed. Like 2-ksqrt(h)
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    Oh okay. I actually got it. I kind of brainstormed!. Thanks!
 
 
 

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