Derivatives: Chain Rule - Melting Ice Block?

A cubical block of ice is melting in such a way that each edge decreses steadily by 5.2 cm every hour. At what rate is its volume decreasing when each edge is 5 meters long?



Solution: Let l=l(t) be the length of each edge at time t. Then the volume of the block is given by V = _______ .



Since the length of each edge is changing in time, we conclude that the volume V is also a function of time t. We note that the rate of change of l is constant and that is given as [(dl)/(dt)]= _______cm/h = ________m/h.



The question is to find the rate of change of V when

l= ____________m.



The chain rule gives



dV/dt = (dV/dl)*(dl/dt)



Therfore , when each edge is 5 m long, the rate of change of the volume of the ice block is ________m3/h.



I don't know how to approach this question..

Please answer..

Thank youDerivatives: Chain Rule - Melting Ice Block?
Volume of a cube with side l is: V = l^3



dl/dt = 5.2 cm/hr = 0.052 m/hr



Find rate of change of V when l = 5 m.



Ok, for this last part, you need to use the chain rule as you have it stated: dV/dt = (dV/dl)*(dl/dt)

Now dV/dl is simply the derivative of the first equation with respect to l. This gives:

dV/dl = 3l^2

Now plug this into the equation for dV/dt.

dV/dt = 3l^2*(dl/dt)

Now you can substitute in all the constants, l = 5 m, dl/dt = 0.052 m/hr, to get:

dV/dt = 3*(5 m)^2*(0.052 m/hr)

= 75*0.052 m^3/hr

= 3.9 m^3/hr

Hope this clears it up for you.Derivatives: Chain Rule - Melting Ice Block?
V = l^3 = 5*5*5 = 125 cubic meters



dl/dt = 5.2 cm/hour = 0.052 meters per hour



Since V = l^3, then dV/dl = 3l^2 (as long as each edge shrinks at the same rate).



Then dV/dt = (dV/dl)*(dl/dt) = 3l^2*(dl/dt) = (3*5*5*)(0.052) = 3.9 cubic meters per hour.Derivatives: Chain Rule - Melting Ice Block?
dl /dt = 5.2 cm / h

dl /dt = 0.052 m / h

V = l 鲁

dV/dl = 3 l 虏 m 虏

dV/dt = (dv / dl) x (dl / dt)

dV/dt = 3 x 5虏 x 0.052 m鲁 / h (when l = 5m)

dV/dt = 3.9 m鲁 / h