A metal cube contracts when it is cooled. if the edge of the cube is decreasing at a rate of 0.2 cm/hr, how fast is the volume changing when the edge is 60 centimeters?

2160 cm³/hour

Step-by-step explanation:

By default, we know that the volume of a cube is given as s³

Thus, the Volume function, V = s³

When we differentiate with respect to time we have

dV/dt = 3s² (ds/dt), where ds/dt = 0.2

Then we go ahead and substitute all the given parameters

dV/dt = 3 x 60 x 60 x 0.2

dV/dt = 10800 * 0.2

dV/dt = 2160 cm³/hour

This means that the volume decreases by a rate of 2160 cm³/hour at the instant its edge is 60 cm

351°f

step-by-step explanation: