Cane sugar, when placed in water, converts to dextrose at a rate proportional to the amount of unconverted material remaining. That is, if M grams is the amount of material converted after t minutes, then dM = k(S – M) where S grams is the initial amount of cane dT sugar and k is a constant. (i) Show that M = S + Ae¬kt satisfies the equation, where A is constant. (ii) If a certain amount of cane sugar is placed in water at time t = 0 and loges 40% of it has been converted after 10 minutes, show k =÷loge 10 3'
Cane sugar, when placed in water, converts to dextrose at a rate proportional to the amount of unconverted material remaining. That is, if M grams is the amount of material converted after t minutes, then dM = k(S – M) where S grams is the initial amount of cane dT sugar and k is a constant. (i) Show that M = S + Ae¬kt satisfies the equation, where A is constant. (ii) If a certain amount of cane sugar is placed in water at time t = 0 and loges 40% of it has been converted after 10 minutes, show k =÷loge 10 3'
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.7: Applications
Problem 13EQ
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Cane sugar, when placed in water, converts to dextrose at a rate proportional
to the amount of unconverted material remaining. That is, if M grams is the
amount of material converted after t minutes, then
dM
= k(S – M) where S grams is the initial amount of cane
dT
sugar and k is a constant.
(i)
Show that M = S + Ae¬kt satisfies the equation, where A is constant.
(ii) If a certain amount of cane sugar is placed in water at time t = 0 and
40% of it has been converted after 10 minutes, show k =-
5
loge
10
3
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