The volume V and depth x of water in a vessel | vedclass

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(D) Given the relation between volume $V$ and depth $x$ is $V = 5x - \frac{x^2}{6}$.Differentiating both sides with respect to time $t$,we get:$\frac{

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(D) Given the relation between volume $V$ and depth $x$ is $V = 5x - \frac{x^2}{6}$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \frac{d}{dt}(5x - \frac{x^2}{6}) = 5\frac{dx}{dt} - \frac{2x}{6}\frac{dx}{dt} = (5 - \frac{x}{3})\frac{dx}{dt}$.We are given that $\frac{dV}{dt} = 5 \, cm^3/sec$ and we need to find $\frac{dx}{dt}$ when $x = 2 \, cm$.Substituting the values into the derivative equation:$5 = (5 - \frac{2}{3})\frac{dx}{dt}$$5 = (\frac{15 - 2}{3})\frac{dx}{dt}$$5 = \frac{13}{3}\frac{dx}{dt}$Solving for $\frac{dx}{dt}$:$\frac{dx}{dt} = 5 	imes \frac{3}{13} = \frac{15}{13} \, cm/sec$.Since $\frac{15}{13} \, cm/sec$ is not among the given options $(A)$,$(B)$,or $(C)$,the correct answer is $(D)$.

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V = 5x - \frac{x^2}{6}$. If the volume of water is increasing at the rate of $5 \, cm^3/sec$,find the rate at which the depth of water is increasing when $x = 2 \, cm$.

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