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

Asked by Sakshi | 1 year ago |  117

##### Solution :-

From the question it is given that,

Initial radius of balloon = 1 unit

After 3 seconds radius of balloon = 2 units

So, let us assume A be the surface area of balloon,

($$\dfrac{dA}{dt}$$) ∝ t

Then,

$$\dfrac{dA}{dt}$$ = λt

$$\dfrac{d(4\pi r^2)}{dt}= λt$$

$$8πr (\dfrac{dr}{dt}) = λt$$

Integrating on both side we get,

$$8π ∫rdr = λ ∫t dt$$

$$8π (\dfrac{r^2}{2}) = (\dfrac{\lambda t ^2}{2}) + c$$

4πr2 = ($$\dfrac{\lambda t ^2}{2}$$) + c …. … [equation (i)]

From question, r = 1 unit when t = 0,

4π (1)2 = 0 + c

4π = c

From equation (i) c = 4πr2 – ($$\dfrac{\lambda t ^2}{2}$$)

Then, 4πr2 = ($$\dfrac{\lambda t ^2}{2}$$) + 4π … [equation (ii)]

And also form question, given r = 2 units when t = 3 sec

4π(2)2 =$$\dfrac{\lambda(3)2}{2} + 4π$$

16π = $$( \dfrac{9}{2})λ + 4π$$

$$( \dfrac{9}{2})λ$$ = 12π

By cross multiplication we get,

λ = ($$\dfrac{24}{9}$$) π

λ = ($$\dfrac{8}{3}$$) π

So, now equation (ii) becomes,

4πr2 = ($$\dfrac{8\pi}{6}$$) t2 + 4π

4π (r2 – 1) = ($$\dfrac{4\pi}{3}$$) πt2

By cross multiplication,

r2 – 1 =$$(\dfrac{1}{3}) t^2$$

r2 = 1 +$$(\dfrac{1}{3}) t^2$$

r = $$\sqrt{(1 + (\dfrac{1}{3})t^2)}$$

Answered by Aaryan | 1 year ago

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