Question:
The volume V and depth x of water in a vessel 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 vessel 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: Apr 19, 2024
- $\frac{5}{18} cm/sec$
- $\frac{1}{4} cm/sec$
- $\frac{5}{16} cm/sec$
- None of these
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The Correct Option is D
Solution and Explanation
$V=5x-\frac{x^{2}}{6} \Rightarrow \frac{d v}{d t}=5\frac{d x}{d t}\frac{x}{3}. \frac{d x}{d t}$
$\Rightarrow \frac{d x}{d t}=\frac{\frac{d v}{d t}}{\left(5-\frac{x}{3}\right)}$
$\Rightarrow\, \left(\frac{d x}{d t}\right)_{x=2} =\frac{5}{5-\frac{2}{3}}=\frac{15}{13} cm/sec .$
$\Rightarrow \frac{d x}{d t}=\frac{\frac{d v}{d t}}{\left(5-\frac{x}{3}\right)}$
$\Rightarrow\, \left(\frac{d x}{d t}\right)_{x=2} =\frac{5}{5-\frac{2}{3}}=\frac{15}{13} cm/sec .$
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
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This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
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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