JEE Main 2022 — Application Of Derivatives Question with Solution

We know,

Surface area of balloon (s) = 4$$\pi$$r²

$$\therefore$$ $$\frac{ds}{dt}=\frac{d}{dt}(4\pi r^2)$$

$$\Rightarrow \frac{ds}{dt}=4\pi (2r)\times \frac{dr}{dt}$$

$$\Rightarrow \frac{ds}{dt}=8\pi r\times \frac{dr}{dt}$$

Given that, surface area of balloon is increasing in constant rate.

$$\therefore$$ $$\frac{ds}{dt}$$ = constant = k (Assume)

$$\Rightarrow k=8\pi r.\frac{dr}{dt}$$

$$\Rightarrow \int kdt=\int 8\pi rdr$$

$$\Rightarrow kt=8\pi \times \frac{r^2}{2}+c$$

$$\Rightarrow kt=4\pi r^2+c$$ ..... (1)

Given at t = 0, radius r = 3

So, $$0=4\pi (3^2)+c$$

$$\Rightarrow c=-36\pi$$

$$\therefore$$ Equation (1) becomes

$$kt=4\pi r^2-36\pi$$

Also given, at t = 5, radius r = 7

$$\therefore$$ $$k(5)=4\pi (7{)}^2-36$$

$$\Rightarrow k=32\pi$$

$$\therefore$$ Equation (1) is

$$32\pi t=4\pi r^2-36\pi$$

Now at $$t=9$$

$$\Rightarrow 32\pi (9)=4\pi r^2-36\pi$$

$$\Rightarrow 8\times 9=r^2-9$$

$$\Rightarrow r^2=81\Rightarrow r=9$$