a spherical snowball is melting symmetrically at the rate of 4pi cubic cm per hr. how fast is the diameter changing when it is 20 cm?
回答 (3)
2020-08-15 3:36 pm
✔ 最佳答案
V = (4/3)πr³ dV/dt = 4πr².dr/dt
so, when the diameter is 20 cm we have:
-4π = 4πr².dr/dt
Hence, dr/dt = -1/r²
i.e. dr/dt = -1/10² => -1/100
Now, d = 2r
so, d(d)/dt = 2.dr/dt
=> d(d)/dt = -1/50 cm/hour
or, -0.2 mm/hour
:)>
2020-08-15 5:49 pm
r: radius and d: diammeter
Volume, V = (4/3)πr³ ⇒ V = (4/3)π(d/2)³ ⇒ V = (1/6)πd³
dV/dt = d[(1/6)πd³]/dt
dV/dt = (1/2)πd²(dd/dt)
Now, dV/dt = -4π cm³/hr and d = 20 cm:
-4π =(1/2)π(20)²(dd/dt)
dd/dt = -0.02 cm/hr
The diameter DECREASES at a rate of 0.02 cm/hr = 0.2 mm/hr
Volume, V = (4/3)πr³ ⇒ V = (4/3)π(d/2)³ ⇒ V = (1/6)πd³
dV/dt = d[(1/6)πd³]/dt
dV/dt = (1/2)πd²(dd/dt)
Now, dV/dt = -4π cm³/hr and d = 20 cm:
-4π =(1/2)π(20)²(dd/dt)
dd/dt = -0.02 cm/hr
The diameter DECREASES at a rate of 0.02 cm/hr = 0.2 mm/hr
2020-08-15 7:01 pm
V = (4/3)πr^3 = (π/6)D^3
dV/dt = dV/dD * dD/dt
-4π = (π/2)D^2* dD/dt, (where V is in cubic cm)
dD/dt = -8/D^2, and when D = 20
dD/dt = -8/400 = -1/50 cm/hour
dD/dt = - 0.2 mm/hour
dV/dt = dV/dD * dD/dt
-4π = (π/2)D^2* dD/dt, (where V is in cubic cm)
dD/dt = -8/D^2, and when D = 20
dD/dt = -8/400 = -1/50 cm/hour
dD/dt = - 0.2 mm/hour
收錄日期: 2021-04-24 07:59:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200815070726AAKL08d