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    A water tank has rectangular base 1.5m by .2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45 cubed metres per minute. At what rate is the depth in the tank increasing?

    I don't know if i'm being simple but, wouldnt you just divide the volume by the area of the base to get 0.25 m per minute.

    but it is a 4 mark question so i don't know what its asking for?
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    (Original post by Custardcream000)
    A water tank has rectangular base 1.5m by .2m. The sides are vertical and water is being added to the tank at a constant rate of 0.45 cubed metres per minute. At what rate is the depth in the tank increasing?

    I don't know if i'm being simple but, wouldnt you just divide the volume by the area of the base to get 0.25 m per minute.

    but it is a 4 mark question so i don't know what its asking for?
    Well no. Dividing volume (which you dont know by the way) by the area of base would give you the height of the tank.

    The question essentially wants you to find \frac{dH}{dt} and it gives you \frac{dV}{dt} where H is the height of the water and V is the volume of the water.
 
 
 
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