Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 3

How fast is the area of the square increasing when the radius is 30m?

Given: $\displaystyle \frac{dx}{dt} = 6 cm/s$

Required: $\displaystyle \frac{dA}{dt}$ when $\displaystyle \frac{dr}{dt}$

Solution: Let $A = x^2$ be the area of square where $x$ = side of the square

$\displaystyle \frac{dA}{dt} = \frac{dA}{dx} \left( \frac{dx}{dt} \right) = 2 x \frac{dx}{dt} $

$\displaystyle \frac{dA}{dt} = 2x \frac{dx}{dt}$

To get the value of $x$, we use the given area to get $x$


$
\begin{equation}
\begin{aligned}

A =& x^2
\\
\\
x =& \sqrt{A} ; A = 16
\\
\\
\frac{dA}{dt} =& 2 (4)(6)
\\
\\
\end{aligned}
\end{equation}
$


$\fbox{$\large \frac{dA}{dt} = 48 cm^2 /s $}$