Answer:
a) 3.32 mm/s
b) 1.36 mm/s
c) 8.53 mm/s
Step-by-step explanation:
a) <em>Relation</em>: V(t) = πr(t)²h
<em>Derivative</em>: V'(t) = 2πr(t)h·r'(t)
<em>With values substituted</em>: 5 dm³/s = 2π·(120 dm)·(0.2 dm)·r'(t) . . . . 1 L = 1 dm³
<em>Final answer</em>:
... r'(t) = (5 dm³/s)/(2π·24 dm²) = 5/(48π) dm/s ≈ 0.0331573 dm/s ≈ 3.32 mm/s
b) <em>Relation</em>: In addition to the above relation between volume and radius, we also have the relation between volume and time:
... V(t) = ∫V'(t)·dt = V'(t)·t . . . . . since V'(t) is a constant
... r(t) = √(V(t)/(πh)) = √(V'(t)t/(πh))
<em>Derivative</em>:
... Let r(t) = k√t . . . . k = √(V'(t)/(πh))
Then
... r'(t) = k/(2√t)
<em>With values substituted</em>: k = ((5 dm³/s)/(π·0.2 dm)) = √((25/π dm²/s)
... k = (5/√π) (dm/√s)
... r'(t) = (5/√π) (dm/√s) / (2√(3·3600 s)) = 1/(24√(3π)) dm/s ≈ 0.0135722 dm/s
<em>Final answer</em>:
... r'(t) ≈ 1.36 mm/s . . . . . after 3 hours
c) <em>Relation</em>: C(t) = 2πr(t)
<em>Derivative</em>: C'(t) = 2πr'(t)
<em>With values substituted</em>: C'(t) = 2π/(24√(3π)) dm/s = 1/12·√(π/3) dm/s
<em>Final answer</em>: C'(t) ≈ 0.0852772 dm/s ≈ 8.53 mm/s
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<em>Comment on liters in volume calculations</em>
A liter is one cubic decimeter (1 dm³). Often, it is convenient to use that measure, rather than m³ or cm³ in calculations involving liters. The decimeter is readily interconverted with meters, cm, or mm, so it is no great handicap to use this somewhat unusual measure.
Here, we're given a radius in meters. The units of radius rate seem best presented in mm/s. Messing with meters or centimeters or millimeters would add powers of 10 that would only confuse the calculations.
