IBHM 446 472 .pdf
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16 Integration 3 – Applications
1k
y
1
1
ln 2
4
2
1
x2
x2
1
1
ln11 x2 2 ln x
ln 2
2
2
4
4
2
˛
˛
˛
.
This is the particular
solution.
Many questions involving differential equations are set in a real-world context as many
natural situations can be modelled using differential equations.
Example
The rate of change of the volume (V) of a cone as it is filled with water is directly
proportional to the natural logarithm of the time (t) it takes to fill. Given that
dV
1
cm3>s when t 5 seconds and that V 25 cm3 when t 8 seconds,
dt
2
find the formula for the volume.
dV
r ln t.
dt
To turn a proportion sign into an equals sign we include a constant of proportionality, say k, which then needs to be evaluated.
dV
k ln t
Hence
dt
1
dV
when t 5 we get:
Given that
dt
2
1
k ln 5
2
1
1k
0.310 p
2 ln 5
dV
So
0.311 ln t
dt
We start with
1V
冮 0.311 ln t dt
1 V 0.311
冮 ln t dt
Hence V 0.311Bt ln t
冮 t a t b dtR using integration by parts
1
1 V 0.3113t ln t t c4
Given that V 25 when t 8
25 0.31138 ln 8 8 c4
1 c 71.8
Hence V 0.3113t ln t t 71.84
or V 0.311t ln t 0.311t 22.3
The constant can be included
within the brackets or it can
be outside. It will evaluate to
the same number finally.
Given that the question is
dealing with volume and
time, this formula is only
valid for t 7 0.
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