Question

Each side of a square is increase at a rate of 6 cm/s. At what rate is the area of the square increasing when the area of the square is \(16\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\)?

Short answer

The required answer is \(48\,{\rm{c}}{{\rm{m}}^{\rm{2}}}{\rm{/s}}\).

Step-by-step solution

The area of the square

It is known that the area of the square whose side is \(s\) equals to \(A = {s^2}\).

The rate of change of area

The rate of change of area with respect to time is given by \(\frac{{dA}}{{dt}}\).

Here, \(A = {s^2}\). Differentiate \(A\) with respect to \(t\):

\(\begin{array}{l}\frac{{dA}}{{dt}} = \frac{d}{{dt}}\left( {{s^2}} \right)\\\frac{{dA}}{{dt}} = \left( {2s} \right) \cdot \frac{{ds}}{{dt}}\end{array}\)

Hence, \(\frac{{dA}}{{dt}} = 2s \cdot \frac{{ds}}{{dt}}\).

Increase in area

Here, given that each side of the square is increasing at the rate of \(6\,\,{\rm{cm/s}}\) when the area of the square is\(16\,{\rm{c}}{{\rm{m}}^{\rm{2}}}\), means \(\frac{{ds}}{{dt}} = 6\) and \(A = 16\).

Now, obtain the side of the square as follows:

\(\begin{array}{c}A = 16\\{s^2} = 16\\s = 4\end{array}\)

Substitute in \(\frac{{dA}}{{dt}} = 2s \cdot \frac{{ds}}{{dt}}\) :

\(\begin{array}{c}\frac{{dA}}{{dt}} = 2\left( 4 \right) \cdot 6\\ = 48\end{array}\)

Hence, the area of the square is increasing at \(48\,{\rm{c}}{{\rm{m}}^2}{\rm{/s}}\).