Question #5475

Water is flowing at the rate of
50 m3min from a concrete conical reservoir (vertex down) of
base radius 45 m and height 6 m. (a) How fast is the water level
falling when the water is 5 m deep? (b) How fast is the radius of
the water’s surface changing at that moment? Give your answer
in cmmin.

Expert's answer

Water is flowing at the rate of 50 m^3/min from a concrete conical reservoir (vertex down) of base radius 45 m and height 6 m.

Let's make following denotations:

R = 45 m

H = 6 m

F = -50 m^3/min.

tgα = (45/2)/6 = 3.75.

(a) How fast is the water level falling when the water is 5 m deep?

Let's mark V0 the volume of a reservoir.

V0 - F*t = pi*R^2*H/3 ==> H = (V0 - F*t)*3/(pi*R^2)

dH/dt = - F*3/(pi*R^2) - velocity of H changing

When H = 5m, the R = 5*tgα = 5*3.75 = 18.75.

dH/dt = 50*3/(pi*45^2) = 0.02358 m/min.

(b) How fast is the radius of the water’s surface changing at that moment? Give your answer in cm/min.

V0 - F*t = pi*R^2*H/3 ==> R = 3*(V0 - F*t)/(pi*H)

dR/dt = - 3*F/(pi*H) - velocity of R changing

dR/dt = 3*50/(pi*5) = 9.5493 m/min = 954.93 cm/min.

Let's make following denotations:

R = 45 m

H = 6 m

F = -50 m^3/min.

tgα = (45/2)/6 = 3.75.

(a) How fast is the water level falling when the water is 5 m deep?

Let's mark V0 the volume of a reservoir.

V0 - F*t = pi*R^2*H/3 ==> H = (V0 - F*t)*3/(pi*R^2)

dH/dt = - F*3/(pi*R^2) - velocity of H changing

When H = 5m, the R = 5*tgα = 5*3.75 = 18.75.

dH/dt = 50*3/(pi*45^2) = 0.02358 m/min.

(b) How fast is the radius of the water’s surface changing at that moment? Give your answer in cm/min.

V0 - F*t = pi*R^2*H/3 ==> R = 3*(V0 - F*t)/(pi*H)

dR/dt = - 3*F/(pi*H) - velocity of R changing

dR/dt = 3*50/(pi*5) = 9.5493 m/min = 954.93 cm/min.

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