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

Vedclass · Accepted Answer

$2464$

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(C) Let $S = 4\pi r^2$ be the surface area. The rate of expansion is $\frac{dS}{dt} = kS$.Integrating, $\ln S = kt + C$, so $S(t) = S_0 e^{kt}$.At $t = 0$, $r = 2 	ext{ cm}$, so $S_0 = 4\pi(2)^2 = 16\pi$.At $t = 5$, $r = 7 	ext{ cm}$, so $S(5) = 4\pi(7)^2 = 196\pi$.$196\pi = 16\pi e^{5k} \implies e^{5k} = \frac{196}{16} = \frac{49}{4} = 12.25$.$e^k = (12.25)^{1/5}$.At $t = 12$, $S(12) = S_0 e^{12k} = 16\pi (e^{5k})^{12/5} = 16\pi (12.25)^{12/5}$.Using $12.25^{2.4} \approx 49$, $S(12) = 16\pi 	imes 49 = 784\pi$.$784 	imes \frac{22}{7} = 112 	imes 22 = 2464 	ext{ cm}^2$.

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

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