Water flows from the base of a rectangular ta | vedclass

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(B) Let $x$ be the depth of water at time $t$. Given that the rate of flow is proportional to the square root of the depth,we have $\frac{dx}{dt} = -k

Vedclass · Accepted Answer

$0.25$

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(B) Let $x$ be the depth of water at time $t$. Given that the rate of flow is proportional to the square root of the depth,we have $\frac{dx}{dt} = -k\sqrt{x}$,where $k > 0$ is a constant (negative sign indicates decreasing depth).Separating the variables,we get $\frac{dx}{\sqrt{x}} = -k dt$.Integrating both sides,we get $\int x^{-1/2} dx = \int -k dt$,which gives $2\sqrt{x} = -kt + C$.At $t = 0$,the initial depth $x = 16 \ m$. Substituting these values,$2\sqrt{16} = -k(0) + C \Rightarrow C = 8$.So,the equation becomes $2\sqrt{x} = -kt + 8$.At $t = 2 \ hours$,the depth $x = 4 \ m$. Substituting these values,$2\sqrt{4} = -k(2) + 8 \Rightarrow 4 = -2k + 8 \Rightarrow 2k = 4 \Rightarrow k = 2$.Thus,the equation is $2\sqrt{x} = -2t + 8$,or $\sqrt{x} = 4 - t$.For $t = 3.5 \ hours$,$\sqrt{x} = 4 - 3.5 = 0.5$.Squaring both sides,$x = (0.5)^2 = 0.25 \ m$.

Water flows from the base of a rectangular tank of depth $16 \ m$. The rate of flow of the water is proportional to the square root of the depth at any time $t$. If the depth is $4 \ m$ when $t = 2 \ hours$,then after $3.5 \ hours$,the depth (in meters) is:

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