The volume, V cm3 , of a cube at time t seconds is given by
V=(4 + 1/3t)^3
find
the rate at which its volume is increasing at the instant when t = 2.
To find the rate at which the volume of the cube is increasing at the instant when t = 2, we need to find the derivative of the volume function with respect to time (t) and then evaluate it at t = 2.
First, let's find the derivative of the volume function V(t):
V(t) = (4 + (1/3)t)^3
To take the derivative, we can use the chain rule:
dV/dt = 3(4 + (1/3)t)^2 * (1/3)
Now, let's evaluate this derivative at t = 2:
dV/dt = 3(4 + (1/3)(2))^2 * (1/3)
= 3(4 + (2/3))^2 * (1/3)
= 3(10/3)^2 * (1/3)
= (30/9) * (1/3)
= 10/9
Therefore, the rate at which the volume of the cube is increasing at the instant when t = 2 is 10/9 cm^3/s.