Ex. A triangular pyramid has base of area A = x cm2 and height h which is increasing at a constant
1
rate of 3 cm/s. Given that the volume of the pyramid is V = Ah cm3 /s find the rate at which the
3
volume is changing per second, when x = 4.
A sketch is usually helpful:
dV
dV
dV
but V is defined in terms of A and h, so only
or
may be found
dt
dA
dh
dh
we can use the equality:
directly. Since we are given
dt
We are asked to find
dV dV dh
=
× .
dt
dh dt
Since
dV 1
dh
= A and
= 3 we get
dh 3
dt
dV
dt
=
1
3
A × 3 = A cm3 /s.
So the rate at which volume changes is exactly the base area; when x = 4,
dV
= 4 cm3 /s.
dt