Question

A hypothetical square grows at a rate of 32 m²/min. How fast are the diagonals of the square increasing when the diagonals are 12 m each?

Answer 1

Let 's' be the length side of a square

A is its area

From pythagorean theorem, we know the diagonal, y, is:

`\begin{gathered} y^2=s^2+s^2 \\ y^2=2s^2 \end{gathered}`

Also, we know the area of a square is:

`A=s^2`

We can substitute this equation in previous one to get:

`\begin{gathered} y^2=2s^2 \\ y^2=2A \end{gathered}`

We can implicitly differentiate with respect to time (t) to get:

`\begin{gathered} y^2=2A \\ 2y((dy)/(dt))=2((dA)/(dt)) \end{gathered}`

dA/dt is rate of change of Area with time, which is 32.

dy/dt is rate of change of diagonal [what we want to find]

At the instant diagonals are 12, y = 12

Thus, we have:

`\begin{gathered} 2y((dy)/(dt))=2((dA)/(dt)) \\ 2(12)((dy)/(dt))=2(32) \\ (dy)/(dt)=(2(32))/(2(12)) \\ (dy)/(dt)=2.67\text{ m/s} \end{gathered}`