an inverted cylindrical cone, 44 ft deep and 22 ft across at the top, is being filled with water at a rate of 11 ft3/min. at what rate is the water rising in the tank when the depth of the water is:

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The rate of the  water rising  in the tank when the depth of the water is 1 ft is 56.05 ft/min.  Explain the term rate of change?   The term ' rate of change ' (ROC) describes the rate at which something changes over time.     The rate of change of volume is-  dV/dt = 11 ft3/min.    Volume of  cylindrical cone  = (1/3)πr²h  As we're trying to find dh/dt, we need to calculate the volume within terms of just one variable, h.  Diameter =  22 ft  Thus, radius is 11 ft     By using height and radius of a water in the tank, we may calculate h using similar ratios.  11/44 = r/h   1/4 = r/h   r= h/4    Put into volume;  V = (1/3)π(h/4)h²   V = (1/3)(π)h³ / (16)  dV/dt = [(1/3)(3)(π)(h)² / 16](dh/dt)   V = [(π)(h)² / 16](dh/dt)    Solve for (dh/dt) ;  dh/dt = (dV/dt)(16) / [((π)(h)²]    Now, the change in volume is dV/dt = 11 ft³/min  dh/dt = (11 ft³/min)(16) / [((π)(h)²]  For h = 1 ft,   dh/dt = (11 ft³/min)(16) / [((π)(1 ft)²]   dh/dt ≈ 56.05 ft/min    Thus, the rate of the water rising in the tank when the depth of the water is 1 ft is 56.05 ft/min.    To know more about the  rate of change , here   https://brainly.com/question/25184007   #SPJ4

Explain the term rate of change?

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