Note that
∆MSN ~ ∆ASB
or
MS
AS  = MN
AB
or
AS = 3s (as MN =
2 and AB = 6 (given))
Thus
AM = 3s – s = 2s But AM = l
So
l = 2s
Therefore
dl
dt  = 2 ds
dt
Since 
5
dl
dt =
km/h Hence, the length of the shadow increases at the rate 5
2 km/h Example 33 Find intervals in which the function given by
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
is (a) increasing (b) decreasing