A trough is 8 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 13 ft3/min, how fast is the water level rising when the water is 7 inches deep?
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Volume of the trough is V = (1/2)bh8 = 4bh
implicitly differentiate with respect to time
dV/dt = 4(h db/dt + b dh/dt)
dV/dt = 13 cubic feet/min, when the height is 7 inches, by similar triangles b/(7/12) = 4/1 --> b = 7/3
Note that b/h = 4/1 --> b = 4h and db/dt = 4dh/dt
13 = 4((7/12) 4dh/dt + (7/3)(dh/dt) --> 13 = 4( 14/3 dh/dt) --> 39/56 ft/min = dh/ht --> about 8.357 in/min
implicitly differentiate with respect to time
dV/dt = 4(h db/dt + b dh/dt)
dV/dt = 13 cubic feet/min, when the height is 7 inches, by similar triangles b/(7/12) = 4/1 --> b = 7/3
Note that b/h = 4/1 --> b = 4h and db/dt = 4dh/dt
13 = 4((7/12) 4dh/dt + (7/3)(dh/dt) --> 13 = 4( 14/3 dh/dt) --> 39/56 ft/min = dh/ht --> about 8.357 in/min