Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 251: 3

Area is increasing at $48 \frac{cm^{2}}{s}$

The area of the square is $A(x) = x^{2}$ $\frac{dA}{dt} = \frac{d(x^{2})}{dt}$ $\frac{dA}{dt} = 2x \times \frac{dx}{dt}$ $\frac{dA}{dt} = 2x \times 6$ Using the radius of the area $x = 4$ $\frac{dA}{dt} = 2(4) \times 6$ $\frac{dA}{dt} = 8 \times 6$ $\frac{dA}{dt} = 48 \frac{cm^{2}}{s}$