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(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.Integrating both sides with respect to $t$:$A = \int \frac{2}{\sqrt{t}} dt = 2 \int

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$100$ minutes

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(C) Given the rate of digging $\frac{dA}{dt} = \frac{2}{\sqrt{t}}$.Integrating both sides with respect to $t$:$A = \int \frac{2}{\sqrt{t}} dt = 2 \int t^{-1/2} dt = 2 \cdot \frac{t^{1/2}}{1/2} + C = 4\sqrt{t} + C$.At $t = 0$,the area dug $A = 0$,so $C = 0$.Thus,$A = 4\sqrt{t}$.To find the time $t$ to dig $40$ square metres,set $A = 40$:$40 = 4\sqrt{t} \Rightarrow \sqrt{t} = 10$.Squaring both sides,$t = 10^2 = 100$ minutes.

$A$ gardener is digging a plot of land. As he gets tired,he works more slowly. After $t$ minutes,he is digging at a rate of $\frac{2}{\sqrt{t}}$ square metres per minute. How long will it take him to dig an area of $40$ square metres?

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