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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

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

Grade:Upto college level

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
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
askIITians Faculty

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