Asked by Jane
Hi guys can you help me! Please teach me step by step.. I really need it. pls! Thank you.........
A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at ¼ cubic meter per minute, and there is 1 meter of water at the deep end.
a.) What percent of the pool is filled?
b.) At what rate is the water level rising?
A swimming pool is 12 meters long, 6 meters wide, 1 meter deep at the shallow end, and 3 meters deep at the deep end. Water is being pumped into the pool at ¼ cubic meter per minute, and there is 1 meter of water at the deep end.
a.) What percent of the pool is filled?
b.) At what rate is the water level rising?
Answers
Answered by
Steve
The cross-section is a trapezoid of area
(1+3)/2 * 6 = 12 m^2
So, the volume is 12 * 12 = 144 m^3
with only 1 m of water, the cross-section of the water is a triangle of area 3/2 m^2, so its volume of water is 18 m^3
Thus the pool is 18/144 = 1/8 full
At depth y of water, the surface has width y/2, so
v = (1/2)(y)(y/2)(12) = 3y^2
dv/dt = 6y dy/dt
At y=1, then
1/4 = 6 dy/dt
and the height is rising at 1/24 m/s
As usual, check my arithmetic.
(1+3)/2 * 6 = 12 m^2
So, the volume is 12 * 12 = 144 m^3
with only 1 m of water, the cross-section of the water is a triangle of area 3/2 m^2, so its volume of water is 18 m^3
Thus the pool is 18/144 = 1/8 full
At depth y of water, the surface has width y/2, so
v = (1/2)(y)(y/2)(12) = 3y^2
dv/dt = 6y dy/dt
At y=1, then
1/4 = 6 dy/dt
and the height is rising at 1/24 m/s
As usual, check my arithmetic.
Answered by
Braeden
Nope not a thing
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