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ODE-The-rate-at-which-the-ice-melt-is-proportional-to-the-amount-of-ice-present-at-the-instant-Find-the-amount-of-ice-left-after-2-hours-if-half-the-quantity-melt-in-30-minute-

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Question Number 15800 by tawa tawa last updated on 14/Jun/17
ODE The rate at which the ice melt is proportional to the amount of ice present at the instant. Find the amount of ice left after 2 hours if half the quantity melt in 30 minute.
$$\mathrm{ODE} \\ $$$$\mathrm{The}\:\mathrm{rate}\:\mathrm{at}\:\mathrm{which}\:\mathrm{the}\:\mathrm{ice}\:\mathrm{melt}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{amount}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{present} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{instant}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{amount}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{left}\:\mathrm{after}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{if}\:\mathrm{half}\:\mathrm{the}\:\mathrm{quantity} \\ $$$$\mathrm{melt}\:\mathrm{in}\:\mathrm{30}\:\mathrm{minute}. \\ $$
Answered by mrW1 last updated on 14/Jun/17
the ice melt in 0.5 hour by half of its amount, i.e t=0.5h: (1/2) ice left t=1h: (1/2)×(1/2) ice left t=1.5h: (1/2)×(1/2)×(1/2) ice left t=2h: (1/2)×(1/2)×(1/2)×(1/2) ice left ⇒after 2 hours the amount of ice is (1/(16)) general solution: M=amout of ice (dM/dt)=−kM with k=some constant (dM/M)=−kdt ln M=−kt+C M=e^(−kt+C) =M_0 e^(−kt) with M_0 =amout at t=0 let m=(M/M_0 ) ⇒m=e^(−kt) at t=t_1 =0.5h: m_1 =e^(−0.5k) =(1/2) ⇒e^(−0.5k) =(1/2) at t=t_2 =2h: m_2 =e^(−2k) =(e^(−0.5k) )^4 =((1/2))^4 =(1/(16)) further examples: after t=10 minutes (t/t_1 )=(1/3) m=e^(−kt) =(e^(−kt_1 ) )^(t/t_1 ) =((1/2))^(1/3) =0.79 (79%) after t=45 minutes (t/t_1 )=1.5 m=((1/2))^(1.5) =0.35 (35%)
$$\mathrm{the}\:\mathrm{ice}\:\mathrm{melt}\:\mathrm{in}\:\mathrm{0}.\mathrm{5}\:\mathrm{hour}\:\mathrm{by}\:\mathrm{half}\:\mathrm{of}\:\mathrm{its}\:\mathrm{amount},\:\mathrm{i}.\mathrm{e} \\ $$$$\mathrm{t}=\mathrm{0}.\mathrm{5h}:\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ice}\:\mathrm{left} \\ $$$$\mathrm{t}=\mathrm{1h}:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ice}\:\mathrm{left} \\ $$$$\mathrm{t}=\mathrm{1}.\mathrm{5h}:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ice}\:\mathrm{left} \\ $$$$\mathrm{t}=\mathrm{2h}:\:\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{ice}\:\mathrm{left} \\ $$$$\Rightarrow\mathrm{after}\:\mathrm{2}\:\mathrm{hours}\:\mathrm{the}\:\mathrm{amount}\:\mathrm{of}\:\mathrm{ice}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$ \\ $$$$\mathrm{general}\:\mathrm{solution}: \\ $$$$ \\ $$$$\mathrm{M}=\mathrm{amout}\:\mathrm{of}\:\mathrm{ice} \\ $$$$\frac{\mathrm{dM}}{\mathrm{dt}}=−\mathrm{kM}\:\mathrm{with}\:\mathrm{k}=\mathrm{some}\:\mathrm{constant} \\ $$$$\frac{\mathrm{dM}}{\mathrm{M}}=−\mathrm{kdt} \\ $$$$\mathrm{ln}\:\mathrm{M}=−\mathrm{kt}+\mathrm{C} \\ $$$$\mathrm{M}=\mathrm{e}^{−\mathrm{kt}+\mathrm{C}} =\mathrm{M}_{\mathrm{0}} \mathrm{e}^{−\mathrm{kt}} \:\mathrm{with}\:\mathrm{M}_{\mathrm{0}} =\mathrm{amout}\:\mathrm{at}\:\mathrm{t}=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{m}=\frac{\mathrm{M}}{\mathrm{M}_{\mathrm{0}} } \\ $$$$\Rightarrow\mathrm{m}=\mathrm{e}^{−\mathrm{kt}} \\ $$$$\mathrm{at}\:\mathrm{t}=\mathrm{t}_{\mathrm{1}} =\mathrm{0}.\mathrm{5h}:\:\mathrm{m}_{\mathrm{1}} =\mathrm{e}^{−\mathrm{0}.\mathrm{5k}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{e}^{−\mathrm{0}.\mathrm{5k}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{at}\:\mathrm{t}=\mathrm{t}_{\mathrm{2}} =\mathrm{2h}:\:\mathrm{m}_{\mathrm{2}} =\mathrm{e}^{−\mathrm{2k}} =\left(\mathrm{e}^{−\mathrm{0}.\mathrm{5k}} \right)^{\mathrm{4}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$ \\ $$$$\mathrm{further}\:\mathrm{examples}: \\ $$$$\mathrm{after}\:\mathrm{t}=\mathrm{10}\:\mathrm{minutes} \\ $$$$\frac{\mathrm{t}}{\mathrm{t}_{\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{m}=\mathrm{e}^{−\mathrm{kt}} =\left(\mathrm{e}^{−\mathrm{kt}_{\mathrm{1}} } \right)^{\frac{\mathrm{t}}{\mathrm{t}_{\mathrm{1}} }} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} =\mathrm{0}.\mathrm{79}\:\left(\mathrm{79\%}\right) \\ $$$$ \\ $$$$\mathrm{after}\:\mathrm{t}=\mathrm{45}\:\mathrm{minutes} \\ $$$$\frac{\mathrm{t}}{\mathrm{t}_{\mathrm{1}} }=\mathrm{1}.\mathrm{5} \\ $$$$\mathrm{m}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}.\mathrm{5}} =\mathrm{0}.\mathrm{35}\:\left(\mathrm{35\%}\right) \\ $$
Commented by tawa tawa last updated on 14/Jun/17
God bless you sir. i really appreciate.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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