At a certain instant of time, the sides of the cube are 3 in. long and increasing at the rate of 0.2 in./sec. How fast is the volume of the cube changing at that instant of time?
To find how fast the volume of the cube is changing at a certain instant of time, we need to use the concept of related rates. We know that the volume of a cube is given by V = s^3, where V is the volume and s is the length of the side of the cube.
First, let's find the derivative of the volume with respect to time, dV/dt. We can rewrite the volume formula as V = (s(t))^3, where s(t) represents the side length of the cube at time t.
Now, we can take the derivative of both sides of the equation with respect to time:
dV/dt = 3(s(t))^2 * ds/dt.
In this equation, ds/dt represents the rate at which the side length is changing with respect to time. Given that the side length is increasing at a rate of 0.2 in./sec, we have ds/dt = 0.2 in./sec.
Now, we can substitute the given values into the equation:
dV/dt = 3(3 in.)^2 * (0.2 in./sec)
Simplifying this expression, we have:
dV/dt = 27 * 0.2 in.^3/sec
So, the volume of the cube is changing at a rate of 5.4 in.^3/sec at that instant of time.