Question

The volume \[V\]and depth \[x\] of water in a vessel are connected by the relation \[V = 5x - \left( {\dfrac{{{x^2}}}{6}} \right)\] and the volume of water is increasing at the rate of \[5{\text{ c}}{{\text{m}}^3}/\sec \], when \[x = 2{\text{ cm}}\]. The rate at which the depth of water is increasing is
A. \[\dfrac{5}{{18}}{\text{ cm/sec}}\]
B. \[\dfrac{1}{4}{\text{ cm/sec}}\]
C. \[\dfrac{5}{{16}}{\text{ cm/sec}}\]
D. None of these

Answer 1

Hint: In this question, we need to find the rate at which the depth of water is increasing. For this, we have to differentiate the given equation of \[V\] with respect to \[t\]. After that by putting \[x = 2{\text{ cm}}\] in the result of differentiation, we will get the desired result.

Formula used: we will use the following rule of differentiation.
\[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Here, \[n\] is an integer.

Complete step-by-step solution:
We know that the volume \[V\] and depth \[x\] of water in a vessel are connected by the relation
\[V = 5x - \left( {\dfrac{{{x^2}}}{6}} \right)\]
Differentiate the above equation of \[V\] with respect to \[t\].
Thus, we get
\[\dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {5x - \left( {\dfrac{{{x^2}}}{6}} \right)} \right)\]
\[\dfrac{{dV}}{{dt}} = \dfrac{d}{{dt}}\left( {5x} \right) - \dfrac{d}{{dt}}\left( {\dfrac{{{x^2}}}{6}} \right)\]
Let us find the derivative.
\[\dfrac{{dV}}{{dt}} = 5\dfrac{d}{{dt}}\left( x \right) - \dfrac{1}{6}\dfrac{d}{{dt}}\left( {{x^2}} \right)\]
But we know that \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
So, we get
\[\dfrac{{dV}}{{dt}} = 5\dfrac{{dx}}{{dt}} - \dfrac{1}{6}\left( {2x} \right)\dfrac{{dx}}{{dt}}\]
\[\dfrac{{dV}}{{dt}} = 5\dfrac{{dx}}{{dt}} - \dfrac{x}{3}\dfrac{{dx}}{{dt}}\]
Now, put \[x = 2{\text{ cm}}\] in the above equation.
Hence, we get
\[\dfrac{{dV}}{{dt}} = 5\dfrac{{dx}}{{dt}} - \dfrac{2}{3}\dfrac{{dx}}{{dt}}\]
By simplifying, we get
\[\dfrac{{dV}}{{dt}} = \dfrac{{dx}}{{dt}}\left( {5 - \dfrac{2}{3}} \right)\]
\[\dfrac{{\dfrac{{dV}}{{dt}}}}{{\left( {5 - \dfrac{2}{3}} \right)}} = \dfrac{{dx}}{{dt}}\]
But we know that the volume of water is increasing at the rate of \[5{\text{ c}}{{\text{m}}^3}/\sec \]
That means \[\dfrac{{dV}}{{dt}} = 5{\text{ c}}{{\text{m}}^3}/\sec \]
By putting \[\dfrac{{dV}}{{dt}} = 5{\text{ c}}{{\text{m}}^3}/\sec \] in the above equation, we get
\[\dfrac{5}{{\left( {\dfrac{{13}}{3}} \right)}} = \dfrac{{dx}}{{dt}}\]
\[\dfrac{{15}}{{13}} = \dfrac{{dx}}{{dt}}\]
Hence, the rate at which the depth of water is increasing is \[\dfrac{{15}}{{13}}{\text{ cm/sec}}\].

Therefore, the correct option is (D).

Note: Many students generally make mistakes in finding the derivative. Rather than using the generalized method of differentiation, the differentiation rules enable us to determine the derivatives of specific functions. It helps to find the value of the rate at which the depth of water is increasing.