Given:
Inverted square pyramid filled at constant rate of 75 cm^3/s
Sides of square base (at top) are 8 cm.
Height is 15 cm.
Find: Rate the water level is rising when water level is 13 cm.
Solution:
Let s = side length and r = s/2
As water level rises, both the height and side increase.
The height and side are proportional to the structure.
15/4 = h/r
h = (15/4)*r
h = (15/4)*(s/2)
h = (15/8)*s Relates height and side.
s = (8/15)*h
V = (1/3)*h*s^2 Vol. of square pyramid
V = (1/3)*h*[ (8/15)*h ]*2 Vol. in terms of h only
V = (1/3)*(64/225)*h^3
V = (64/675)*h^3
V' = (64/675)*(3)*(h^2)*(h') Take derivative of both sides
75 = (64/675)*(3)*(13^2)*(h') Use given information
Solve for h'
h' = (75)*(675/64) / [ (3)*(13^2) ]
h' = (50625/64) / (507)
h' = 16875 / 10816
h' = 1.56019 cm/s