工程數學/工數/應用問題///20點

圖片參考：https://s.yimg.com/rk/AC09299269/o/801387009.png


Q : 冰塊為球體還是立方體?
Ans : 題目是說 ice "ball", 所以是指"球體".

Sol :
設時間 t 時,此球半徑為R(t),表面積為S(t),體積為V(t),
由題意知融化速率跟表面積成比例,
則存在一負值常數K,使得下式成立:
dV(t)/dt = K S(t)
再利用球體體積與表面積公式可得:
d[ (4/3)πR(t)^3 ] / dt = K 4πR(t)^2
(4/3)π d[ R(t)^3 ] / dt = 4πK R(t)^2
d[ R(t)^3 ] / dt = 3K R(t)^2
3 R(t)^2 R'(t) = 3K R(t)^2
R'(t) = K

圖片參考：https://s.yimg.com/rk/AC09299269/o/959960219.png

R(T) - R(0) = KT .....(1式)

又 V(t) = (4/3)πR(t)^3
R(t)^3 = 3 V(t) / (4π)
R(t) = [ 3 V(t) / (4π) ] ^ (1/3)

R(0) = [ 3*1000 / (4π) ] ^ (1/3) ≒ 6.204
R(1) = [ 3*729 / (4π) ] ^ (1/3) ≒ 5.583
設 t = Te 時,體積為125 cm^3 ,
R(Te) = [ 3*125 / (4π) ] ^ (1/3) ≒ 3.102

當 T = 1, 由(1式)得:
K = R(1) - R(0) = 5.583 - 6.204 = - 0.621

當 T = Te, 由(1式)得:
R(Te) - R(0) = KTe
3.102 - 6.204 = -0.621*Te
Te = (3.102-6.204)/(-0.621) ≒ 4.995 ≒ 5
Ans: 若以體積為1000 cm^3時為第0分鐘,
則5分鐘時,體積為125 cm^3.
( 1000 cm^3 → 1分鐘 → 729 cm^3 → 4分鐘 → 125 cm^3 )

參考看看他的答案
TS777.CC

(1) Ball融化速率跟表面積成比例: dV/dt = -k*AV=Volume=4πr^3/3A=Surface=4πr^2dV/dt=4πr^2*dr/dt=A*dr/dt=-k*A=> dr/dt = -k0 = dr + kdtC = r + k*tInitial condition: C = ro + 0= ro= (3Vo/4π)^1/3= a*Vo^1/3.....a=(3/4π)^1/3= a*1000^1/3= 10a=> 10a = r + k*t
Boundary condition: t=110a = r1 + k*1k = 10a - r1= 10a - a*V1^(1/3)= 10a - a*729^1/3= 10a - 9a= a=> r(t) = 10a - at=> t = [10a - r(t)]/a = (10a - a*125^1/3)/a= (10a - 5a)/a= 5 min= ans.

(2) CubeA=6a^2, V=a^3dV/dt=3a^2*(da/dt)=0.5A*(da/dt)=-kA=> da/dt = -2k0 = da + 2k*dtC = a + 2k*t
Initial condition: C = ao + 0= Vo^1/3= 1000^1/3= 10=> a(t) = 10 - 2k*t
Boundary condition: t=12k = 10 - a1= 10 - V1^1/3= 10 - 729^1/3= 10 - 9= 1=> k = 1/2=> a(t) = 10 - t=> t = 10 - a(t)= 10 - 125^1/3= 10 - 5= 5 min= ans.


2015-02-07 04:03:55 補充：
補充:

V(t) = (4π/3)*r^3

= (1/a^3) * a^3*(10-t)^3

= (10 - t)^3

V(0) = 10^3 = 1000 cc

V(1) = 9^3 = 729 cc

V(5) = 125 cc