diameter /depth = 8/16 = 1/2
so when deprh h = 12 diameter of water surface = 6 and r = 3 cm
area of water surface = pi r^2 = 9 pi cm^2
so pi r^2 dh/dt = 9 pi (1/3) cm^3/min = 3 pi cm^3/min
Water at the rate of 10 cm3/ min is pouring into the leaky cistern whose shape is a cone 16 cm. deep and 8 cm. in diameter at the top. At the time the water is 12 cm. deep, the water level is observed to be rising 1/3 cm/min. how fast is the water leaking away?
2 answers
At a time of t min, let the height of the water level be h cm, and the radius
of its surface be r cm
Vol of water = (1/3)π r^2 h
but by simple ratios:
r/h = 4/16 = 1/4
h = 4r or r = h/4
since we need dh/dt, let's keep the h's
V = (1/3)π (h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16) π h^2 dh/dt
let the leakage be x cm^3/min
so dV/dt = (10-x) cm^3/min
when h = 12
V = (1/48)π (1728) = 36π cm^3
"At the time the water is 12 cm. deep, the water level is observed to be rising 1/3 cm/min"
----> dV/dt = (1/16) π h^2 dh/dt
10-x = (1/16)π(144)(1/3)
10-x = 3π
x = 10-3π = appr .575
So the leak is appr .575 cm^3/min
check my arithmetic, I did not write this out first
of its surface be r cm
Vol of water = (1/3)π r^2 h
but by simple ratios:
r/h = 4/16 = 1/4
h = 4r or r = h/4
since we need dh/dt, let's keep the h's
V = (1/3)π (h^2/16)(h)
= (1/48)π h^3
dV/dt = (1/16) π h^2 dh/dt
let the leakage be x cm^3/min
so dV/dt = (10-x) cm^3/min
when h = 12
V = (1/48)π (1728) = 36π cm^3
"At the time the water is 12 cm. deep, the water level is observed to be rising 1/3 cm/min"
----> dV/dt = (1/16) π h^2 dh/dt
10-x = (1/16)π(144)(1/3)
10-x = 3π
x = 10-3π = appr .575
So the leak is appr .575 cm^3/min
check my arithmetic, I did not write this out first
2 answers