Score: 50% (4 of 8 pts) 5 of 8 complete Ex. Score: 0 of 1 pt A trough is shaped

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Score: 50% (4 of 8 pts) 5 of 8 complete Ex. Score: 0 of 1 pt A trough is shaped like a half cylinder with length 5 m and radius 1 m. The trough is full of water when a valve is opened and water flows out of the bottom of the trough at a rate of 4.5m/hr. (Hint: The area of a sector of a circle of radius r subtended by an angle is Complete parts (a) and (b) below. aHow fast is the water level changing when the water r2a 2 -5m 5m, level is 0.5 m from the bottom of the trough? dh dt Type an exact answer using radicals as needed.) hr Outflow 4.5m/ Enter your answer in the answer box, then clck Check Answer 1 part remaining Clear A Check Answer Save

Explanation / Answer

The cross-section of the liquid in the tank is a spherical segment of height h (= the depth of the liquid)
(for a sketch see [1]

The total volume of liquid equals cross-sectional area time the length of the cylinder:
V = AL
The volume of decreases by a constant rate Q = 4.5 m³hr¹, i.e.
dV/dt = -Q
When you expand left hand side using chain rule, you get:
d(AL)/dt = L(dA/dt) = L(dA/dh)(dh/dt) = -Q
So the rate of change of the liquid level is given by:
dh/dt = -Q / (L(dA/dh)
So when you know dA/dh as function of h you can compute the change of the liquid level at depth h.

The area of the segment is:
A = R²cos¹((R-h)/R) - (R - h)(2Rh - h²)
= R²cos¹((1 -(h/R)) - (R - h)(2Rh - h²)
(R is the radius of the circle)
=>
(dA/dh) = [R²(-1/R)(1/(1 - ((1 -(h/R))²)] + (2Rh - h²) - [(R - h)(1/2)(2R - 2h)/(2Rh - h²)]
= [R/(1 - 1 + 2(h/R) - (h/R)²)] + (2Rh - h²) - [(R - h)²/(2Rh - h²)]
= [R/(2(h/R) - (h/R)²)] + (2Rh - h²) - [(R² - 2Rh + h²) /(2Rh - h²)]
= (R² + (2Rh - h²) - (R² - 2Rh + h²)) / (2Rh - h²)
= 2(2Rh - h²) / (2Rh - h²)
= 2(2Rh - h²)

Hence,
dh/dt = -Q / (2L(2Rh - h²))
= - 4.5 m³hr¹ / (2 5m (2 1m 0.5m - (0.5m)²))
= -3.173 mhr¹


B.
The area of the liquid surface equals the length of the chord, which bounds the circular segment times the length of the cylinder:
S = aL
=>
dS/dt = L(da/dt) = L(da/dh)(dh/dt)

with
a = 2(2Rh - h²)
=>
da/dh = (2R - 2h)/(2Rh - h²) = 2(R - h)/(2Rh - h²)
=>
dS/dt = L [2(R - h)/(2Rh - h²)] [- Q/(2L(2Rh - h²))]
= - Q (R - h)/(2Rh - h²)
= - 4.5 m³hr¹ (1m - 0.5m)/(21m0.5m - (0.5m)²)
= - 3 m²hr¹

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