An inverted pyramid is being filled with water at a constant rate of 25 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 6 cm, and the height is 8 cm. Find the rate at which the water level is rising when the water level is 6 cm.

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2014-10-21T05:12:51+05:30

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   Let  a = side of the base, b = base area, h = altitude 
    
V = \frac{1}{3}bh = \frac{1}{3} a^2 h\\ \\\frac{dV}{dt}=a^2\frac{dh}{dt},\ \ \ \frac{dh}{dt}=\frac{1}{a^2}\frac{dV}{dt}\\ \\h_0=8cm, a_0=6cm,h=water-level=6cm\\ \\\frac{h}{h_0}=\frac{a}{a_0},\ \ \ \ a=\frac{a_0h}{h_0}=\frac{6*6}{8}=4.5cm\\ \\\frac{dh}{dt}=\frac{1}{4.5^2}25 = 1.23456790\ cm/s

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Detailed explanation:

Let us assume it is a regular pyramid. The (altitude) height "h" of apex from base square, slanting height and the side "a" of the base square rise uniformly and proportionally when we move from the apex point to the base.

If height (altitude) h becomes x times, then side "a" of the base square becomes x times, the slanting height also becomes x times.  Thus the area of the base square rises x² times.

     Δh / h = Δa / a    =>     Δb / b  = 2 Δa / a = 2 Δh / h    or      h1/h2 = a1/a2

     V = 1/3 b h = 1/3 a² h
 
   d V / dt = 1/3 a² dh/dt  + 1/3 h 2 a da/dt = 1/3 a² dh /dt + 2/3 h a  (a/h) (dh/dt)
               = 1/3 a² dh / dt + 2/3 a² dh/dt

     dV/dt  = a² dh/dt         =>   dh/dt = (1/a²) dV/dt

    h_0 = 8 cm     a_0 = 6 cm    at the base
    When h = 6cm,     h / h_0 = a/a_0   => a = a_0 * h / h_0 = 6*6/8 = 4.5 cm
 
     dh/dt = 1/4.5² * 25 = 100/81 cm/sec = 1.23456790 cm/sec


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