Asked by Claire
A conical cistern is 10 ft. across the top and 12 ft. deep. If water is poured into the cistern at the rate of 1 cubic foot per second, how fast is the surface rising when the water is 8 ft. deep?
Answers
Answered by
Reiny
let the water level have a radius of r ft
let the height of the water be h ft
by ratio:
r/h = 5/12
12r = 5h ---> r = 5h/12
V = (1/3)π r^2 h
= (1/3)π(25h^2/144)(h)
= (25/432)π h^3
dV/dt = (25/144)π h^2 dh/dt
plug in our given stuff
1 = (25/144)π(64) dh/dt
dh/dt = 144/(25(64)π) = 9/(100π)
check my arithmetic
let the height of the water be h ft
by ratio:
r/h = 5/12
12r = 5h ---> r = 5h/12
V = (1/3)π r^2 h
= (1/3)π(25h^2/144)(h)
= (25/432)π h^3
dV/dt = (25/144)π h^2 dh/dt
plug in our given stuff
1 = (25/144)π(64) dh/dt
dh/dt = 144/(25(64)π) = 9/(100π)
check my arithmetic
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.