v = s^3
dv/dt = 3s^2 ds/dt
now just plug in your numbers.
An ice cube is 3 by 3 by 3 inches is melting in such a way that the length of one of its side is decreasing at a rate of half an inch per minute. Find the rate at which its surface area is decreasing at the moment when the volume of the cube is 8 cubic inches
2 answers
s.a. = 6 s²
da/dt = 12 s ds/dt
v = s³ ... 8 = 2³
da/dt = 12 s ds/dt
v = s³ ... 8 = 2³
2 answers