# Thread: What rate is the water level decreasing when the water is 2m deep?

1. ## What rate is the water level decreasing when the water is 2m deep? var addthis_config = {"data_track_clickback":false};

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.

2. 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.

3. 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

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