An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft3/min, find the rate of change of the depth of the water when the water is 10 feet deep.
3 answers
Do you know what kind of formula I could use to solve this?
sure
change in volume = surface area * change in depth
or
dV = pi r^2 dh
dV/dt = pi r^2 dh/dt = 12 ft^3/min
so
dh/dt = 6/(pi r^2)
if r = 7, h = 24
or
r = (7/24)h = (7/24)(10) = 70/24
so
dh/dt = 6 / [ pi (70/24)^2 ]
change in volume = surface area * change in depth
or
dV = pi r^2 dh
dV/dt = pi r^2 dh/dt = 12 ft^3/min
so
dh/dt = 6/(pi r^2)
if r = 7, h = 24
or
r = (7/24)h = (7/24)(10) = 70/24
so
dh/dt = 6 / [ pi (70/24)^2 ]
Hmmm. I get
r = 7/24 h
so, dr = 7/24 dh
r = 35/12 when h=10
v = 1/3 pi r^2 h
using the product rule,
dv = 1/3 pi (2rh dr + r^2 dh)
12 = pi/3 (2*(35/12)(10)(7/24 dh)+(35/12)^2 dh)
12 = 1225/144 pi dh
dh = 1728/1225 pi
= 12^3/35^2 pi
Maybe you can figure out whether one of us is correct.
r = 7/24 h
so, dr = 7/24 dh
r = 35/12 when h=10
v = 1/3 pi r^2 h
using the product rule,
dv = 1/3 pi (2rh dr + r^2 dh)
12 = pi/3 (2*(35/12)(10)(7/24 dh)+(35/12)^2 dh)
12 = 1225/144 pi dh
dh = 1728/1225 pi
= 12^3/35^2 pi
Maybe you can figure out whether one of us is correct.
2 answers