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(C) Let $S = 4\pi r^2$ be the surface area. Given $\frac{dS}{dt} = kS$.Integrating, $\ln S = kt + C$, so $S(t) = S_0 e^{kt}$.At $t = 0$, $r = 2 	ext{

Vedclass · Accepted Answer

$2464$

Vedclass · Answer

(C) Let $S = 4\pi r^2$ be the surface area. Given $\frac{dS}{dt} = kS$.Integrating, $\ln S = kt + C$, so $S(t) = S_0 e^{kt}$.At $t = 0$, $r = 2 	ext{ cm}$, so $S_0 = 4\pi(2)^2 = 16\pi 	ext{ cm}^2$.At $t = 5$, $r = 7 	ext{ cm}$, so $S(5) = 4\pi(7)^2 = 196\pi 	ext{ cm}^2$.$196\pi = 16\pi e^{5k} \implies e^{5k} = \frac{196}{16} = 12.25$.$e^k = (12.25)^{1/5}$.At $t = 12$, $S(12) = 16\pi (e^k)^{12} = 16\pi (12.25)^{12/5}$.$(12.25)^{2.4} \approx 12.25^{2} 	imes 12.25^{0.4} \approx 150.0625 	imes 3.24 \approx 49$. More precisely, $S(12) = 16\pi (12.25)^{2.4} \approx 16 	imes 3.14159 	imes 49 \approx 2463.01 	ext{ cm}^2$.Rounding to the nearest option, $S(12) = 2464 	ext{ cm}^2$.

$A$ spherical balloon expands at a rate proportional to its surface area. Initially, its radius is $2 \text{ cm}$ and $5$ minutes later it increases to $7 \text{ cm}$. What will be the surface area of the spherical balloon after $12$ minutes (in $\text{ cm}^2$)?

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