What rate is the water level decreasing when the water is 2m deep?
A water tank has the shape of an inverted circulur cone with base radius 3m and height 10m.
(a) If water is leaking out of the tank at a rate of 1m^3/min, at what rate is the water level decreasing when the water is 2m deep?
(b) If the empty tank is being filled with water at a rate of 1.5m^3/min find the rate at which the water level is rising when the water is 7m deep.
Volume of cone = V = (1/3)(pi)(r^2h)
where
r = radius of cone
h = height of cone
By similar triangles,
(3/10) = r/h
hence, r = 0.6h
Substituting r = 0.6h in the volume formula for the cone,
V = (1/3)(pi)(0.6h)^2(h)
V = (1/3)(pi)(0.36h^3)
V = 0.12(pi)h^3
Differentiating,
dV/dT = 0.12(pi)(3h^2)(dh/dT)
Solving for (dh/dT),
(dh/dT) = (dV/dT)/(0.12 * pi * 3h^2)
and at h = 2,
(dh/dT) = (-1)/(0.12 * 3.1416 * 3(2^2))
(dh/dT) = -1/4.5239
(dh/dT) = -0.221 m^3/min.
b) When the tank is being filled, the same original function and its derivative are still the same. Substituting
dV/dt = 1.5 m^3/min
h = 7 m
then
(dh/dT) = 1.5/(.12 * 3.1416 * 3(7)^2)
(dh/dT) = 1.5/55.418
(dh/dT) = 0.027 m^3/min.
Volume of cone = V = (1/3)(pi)(r^2h)
where
r = radius of cone
h = height of cone
By similar triangles,
(3/10) = r/h
hence, r = 0.6h
Substituting r = 0.6h in the volume formula for the cone,
V = (1/3)(pi)(0.6h)^2(h)
V = (1/3)(pi)(0.36h^3)
V = 0.12(pi)h^3
Differentiating,
dV/dT = 0.12(pi)(3h^2)(dh/dT)
Solving for (dh/dT),
(dh/dT) = (dV/dT)/(0.12 * pi * 3h^2)
and at h = 2,
(dh/dT) = (-1)/(0.12 * 3.1416 * 3(2^2))
(dh/dT) = -1/4.5239
(dh/dT) = -0.221 m^3/min.
b) When the tank is being filled, the same original function and its derivative are still the same. Substituting
dV/dt = 1.5 m^3/min
h = 7 m
then
(dh/dT) = 1.5/(.12 * 3.1416 * 3(7)^2)
(dh/dT) = 1.5/55.418
(dh/dT) = 0.027 m^3/min.
References :
its volume
v=1/3 ∏r^2h
v=1/3 ∏ 3^2 10=30∏ m^3
a) dv/dt =1m^3 ……………given
dh/dt =to be detemined
v=1/3 ∏r^2h
dv/dt=3 ∏h(dh/dt)
1m^3=3∏2m(dh/dt)………h=2m is given
1m=6∏dh/dt
dh/dt=1/6∏ m/min……the first ans
b) dv/dt=1.5m^/min
dh/dt =?
dv/dt=3∏h (dh/dt)
1.5m^/min=3∏ 7 (dh/dt)
dh/dt=1/14∏ m^3/min
References :