Page 33 - Pra U STPM 2022 Penggal 2 - Mathematics

Mathematics Semester 2 STPM Chapter 4 Differential Equations

Example 14

The rate at which a substance evaporates is k times the amount of the substance that has not been evaporated.
If the amount of the substance at the beginning is A and the amount that has been evaporated at time t is
x, write down a differential equation involving x, A, k and t.
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(a) Solve this differential equation and sketch a graph of x against t.
(b) Show that the time taken for half the amount to evaporate is 1 ln 2.
k 1
(c) Find the percentage of the amount of substance that has not been evaporated after time ln 2.
2k

Solution: (a) The differential equation is dx = k(A – x).
dt
1
∫
∫ (A – x) dx = k dt
– ln | A – x | = kt + C

When t = 0, x = 0, therefore C = –ln A.
Hence, – ln | A – x | = kt – ln A
x


In A – x = –kt A
A

 A – x = e –kt
A
t
0
–kt
x = A(1 – e )
(b) When half of the amount has evaporated, x = 1 A.
2
Thus, 4
–kt
A = A(1 – e )
2
1 – e –kt = 1
2
e = 1
–kt
2
–kt = ln 1 2
t = 1 ln 2
k
(c) When t = 1 ln 2,
2k
1
–— ln 2
x = A(1 – e )
2
1
–—
= A(1 – 2 )
2
= 0.2929A

Thus, the amount that has not evaporated is 0.7071A.
Hence, 70.71% of the substance has not evaporated.

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04 STPM Math(T) T2.indd 141 28/01/2022 5:44 PM