Answer to Question #134076 in Calculus for Higgins

Question

Determine the average rate of change of the surface area when the surface area decreases from 2827.43 cm^2 to 1256.64 cm^2.

Answer 1

$\textsf{Rate of change of the surface area} \hspace{0.1cm} = \frac{\mathrm{d}A}{A}\\ \mathrm{d}A = 1256.64 - 2827.43 \\ \frac{\mathrm{d}A}{A} = \frac{1256.64 - 2827.43}{2827.43} \approx -0.5556 = -55.56\%\\ \therefore \hspace{0.1cm} \textsf{The average rate of change}\\\textsf{of the surface area when the}\\\textsf{surface area decreases from}\\\hspace{0.1cm} 2827.43 \hspace{0.1cm}\textsf{cm}^2\hspace{0.1cm} \textsf{to} \hspace{0.1cm}1256.64 \hspace{0.1cm}\textsf{cm}^2 \hspace{0.1cm} \textsf{is} \\-55.56\%$