Answer
(a) $$V=\frac{10,000}{\sqrt{1+t}}$$
(b) Rate of depreciation, 1767.77
(c) Rate of depreciation, 625
Work Step by Step
Step-1: It is given that $V \propto \frac{1}{\sqrt{1+t}}$
$$\therefore V=\frac{k}{\sqrt{1+t}}$$
where $k$ is the constant of proportionality.
Step-2: It is also given that the initial value of $V$ is $10,000$. Thus, at $t=0, V=10,000\implies k=10,000$. Thus,
$$V=\frac{10,000}{\sqrt{1+t}}$$
Step-3: Rate of depreciation,
$$\frac{dV}{dt}=(10,000)\cdot \bigg(-\frac{1}{2}\bigg)\cdot(1+t)^{-3/2}$$
$$\frac{dV}{dt} = -\frac{5000}{(1+t)^{3/2}}$$
Step-4: At $t=1$,
$$\frac{dV}{dt}=-\frac{5000}{(1+1)^{3/2}}=-1767.77$$
Step-5: At $t=3$,
$$\frac{dV}{dt} = -\frac{5000}{(1+3)^{3/2}}=-625$$