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解答

設半徑為 r (cm) 時,融化速率為 vr² (cm/min)
則經過 ∆t (min) 後,新半徑 r' (cm) 滿足
4π(r')³/3 - 4πr³/3 = -vr²∆t
(r')³-r³ = -(3v/4π)r²∆t
∆[r³] = -(3v/4π)r²∆t
d[r³] = -(3v/4π)r²dt
3r²dr = -(3v/4π)r²dt
dr = -(v/4π)dt
∫ₜ₌₀ᵗ dr = -(v/4π) ∫ₜ₌₀ᵗ dt
r |ₜ₌₀ᵗ = -(v/4π) t |ₜ₌₀ᵗ
r(t) - 10 = -(v/4π)t
t=5 (min) 代入得 5-10=-(v/4π)·5 → v/4π = 1
r(t) = 10-t
故若 r(t)=0 則 t=10

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