when an ice cube is put in warm water, it starts to melt.\nin a particular case, the expression\n\n v(t) =…

Explanation:

Step1: Simplify the function

First, rewrite (V(t)=8(1 - t)\sqrt{1 - t}=8(1 - t)^{\frac{3}{2}}).

Step2: Differentiate using the chain rule

Let (u = 1 - t), then (V = 8u^{\frac{3}{2}}). The derivative of (V) with respect to (u) is (V^\prime_u=8\times\frac{3}{2}u^{\frac{1}{2}} = 12u^{\frac{1}{2}}), and the derivative of (u) with respect to (t) is (u^\prime_t=-1). By the chain rule (\frac{dV}{dt}=\frac{dV}{du}\times\frac{du}{dt}), so (\frac{dV}{dt}=12(1 - t)^{\frac{1}{2}}\times(-1)=- 12\sqrt{1 - t}).

Step3: Find the initial rate of change

The initial time is (t = 0). Substitute (t = 0) into (\frac{dV}{dt}): (\frac{dV}{dt}\big|_{t = 0}=-12\sqrt{1-0}=-12).

Answer:

(-12)