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# The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t

Jitender Singh IIT Delhi
7 years ago
Ans:
Surface Area ‘S’:
$S = 4\pi r^{2}$
$\frac{\partial S}{\partial t} = 8\pi r.\frac{\partial r}{\partial t}$
Given:
$8\pi r.\frac{\partial r}{\partial t} = at, a = constant$
$r.dr = \frac{a}{8\pi }t.dt$
$\int_{1}^{r} r.dr = \frac{a}{8\pi }\int_{0}^{t}t.dt$
$r^{2}-1 = \frac{a}{8\pi }t^{2}$
$t = 3 \rightarrow r = 2$
$\Rightarrow a = \frac{8\pi }{3}$
$r = \sqrt{1 + \frac{t^{2}}{3}}$
Thanks & Regards
Jitender Singh
IIT Delhi