Was your answer to the first part y = 70e^[e^(-3t) - 1] ?Can anybody help with its (ii) part? It is from M/J 2012 v33 and the answer is 100/e.
View attachment 64388
As t goes to infinity,
e^-3t tends to zero
thus, y tends to 70e^[0 - 1] = 70e^-1
so we have:
yFinal = 70e^-1
yStart = 70 (from main question)
p = (yFinal/yStart)*100
p = 100e^-1
yStart = 70 (from main question)
p = (yFinal/yStart)*100
p = 100e^-1
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