作业帮一道英文的微积分题目?If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)(a) What is the average temperatur
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一道英文的微积分题目?If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)(a) What is the average temperatur
一道英文的微积分题目?
If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)
(a) What is the average temperature of the tea during the first 25 minutes?
(b) At what time does the tea reach the average temperature that you found in (a)?
温度一个是98度，一个是18度
一道英文的微积分题目?If a cup of tea has temperature 980C in a room where the temperature is C,then according to Newton's Law of Cooling the temperature of the tea after t minutes is,T(t) = 18 + 80e^(-t/50)(a) What is the average temperatur
平均值=曲边梯形面积/底边长
曲边梯形面积=T积分(t) = 18t + 4000(1-e^(-t/50))
(a):
平均值=曲边梯形面积/底边长=T积分(25)/25=(18*25 + 4000(1-e^(-25/50)))/25=……（用计算器算出）
(b):
令已求出的平均值=T(t) = 18 + 80e^(-t/50),反解出t=……即可
水杯的温度为98℃，环境温度是C，根据牛顿定律，杯中的水温度与时间的关系T(t) = 18 + 80e^(-t/50)
a、前25分钟水杯的平均温度？
平均温度*1500=
从0到1500得定积分T(t) = 18t - 80*50e^(-t/50)
=18*1500-[400*exp(-1500/50)-(0-80*50*exp^0]=31000
平均温度...
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水杯的温度为98℃，环境温度是C，根据牛顿定律，杯中的水温度与时间的关系T(t) = 18 + 80e^(-t/50)
a、前25分钟水杯的平均温度？
平均温度*1500=
从0到1500得定积分T(t) = 18t - 80*50e^(-t/50)
=18*1500-[400*exp(-1500/50)-(0-80*50*exp^0]=31000
平均温度=31000/1500=20.66666667
t1=0
t2=2500
b、什么时候水杯降低到平均温度？
20.66666667=18 + 80e^(-t/50)
t=-50*ln[(20.667-18)/80]=170s
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