if a ice cube (sides of length 5) is dropped in water, the sides dec at 0.2cm/min,how fast is vol dec at 8cm^3

Question

i hate these darn related rates Q's, ive tried many attempts but i still cant get it, its a previous final, and the answer is either 8, .008,0.8, 2.4, 5.4, 38.4
thanks

Answer 1

the key is to recognize the relationship between the rates and then smack it into an algebraic relationship

If I read your question right, the cube is assumed to be melting in a way that keeps it a perfect cube (this is not realistic but is fine for a math problem).

initially the cube has surfaces that are squares, 5 cm on a side.

initially then, the volume is 125 cm^3

The rate that the cube volume changes (knowing the rate the length changes) is (I'm sure you already know) the key to the problem.

Before we do the calculus, lets do an estimate with arithmetic.

A cube with volume 8 cm^3 has sides that are 2 cm long.
So, at the time when each side is 2.1 cm it should take 1 minute for the sides to drop .2 cm each to 1.9 cm. This is right around the point we are looking at and should provide a good estimate.

The cube volume started at 2.1*2.1*2.1=9.26 cm^3 and in one minute went to 1.9*1.9*1.9=6.86.

That means that the cube volume changed 9.26-6.86 or 2.4 cm in that minute or approximately 2.4 cm/min. Now we know the answer but it is not exact because we calculated the average loss for the whole minute, rather than the instantaneous loss rate right at the 8 cm volume point. If we had carried more significant digits we would have got an answer a little different that 2.4.

This little excersize just helps us understand whats going on. Now the math.

V=s^3

dV/t = 3s^2 ds/dt

ds/dt, the change in s with time, or rate of change of the sides, is given as .2 cm/min

so, dV/dt=3s^2 * (.2)

at V=8 cm^3, s=2 cm

at that point then, the rate of change of V, dV/dt=3*(2)^2*.2

dV/dt=2.4

Answer 2

Try:
dV/dt = (dV/ds)(ds/dt
V = s^3
dV/ds = 3s^2
ds/dt = 0.2

dV/dt = 0.2*3s^2 cc/min
@ V = 6 cc, s = 2 cm, so
dV/dt = 0.2*3*2^2 cc/min
dV/dt = 2.4 cc/min