Question:
The volume V and depth x of water in a vessl are connected by the relation $V = 5x - \frac{x^2}{6}$ and the volume of water is increasing , at the rate of $5 \, cm^3/sec$, when x = 2 cm. The rate at which the depth of water is increasing, is
The volume V and depth x of water in a vessl are connected by the relation $V = 5x - \frac{x^2}{6}$ and the volume of water is increasing , at the rate of $5 \, cm^3/sec$, when x = 2 cm. The rate at which the depth of water is increasing, is
Updated On: Jul 7, 2022
- $\frac{5}{18} cm / \sec $
- $\frac{1}{4} cm / \sec $
- $\frac{5}{16} cm / \sec $
- None of these
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The Correct Option is D
Solution and Explanation
$V = 5x - \frac{x^{2}}{6} \Rightarrow \frac{dV}{dt} = 5 \frac{dx}{dt} - \frac{x}{3}. \frac{dx}{dt} $
$\Rightarrow \frac{dx}{dt} = \frac{\frac{dV}{dt}}{\left(5 - \frac{x}{3}\right)}$
$ \Rightarrow \left(\frac{dx}{dt}\right)_{x=2} = \frac{5}{5- \frac{2}{3}} = \frac{15}{13} cm / \sec$
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Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
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