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Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
The volume of spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.
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Answer

Let the rate of change of the volume of the balloon be k. (k is a constant)
$\therefore \frac{d y}{d t}=k$ 
Or,
$\frac{d}{d t}\left(\frac{4}{3} \pi r^{3}\right)=k$ {Volume of sphere = $\frac{4}{3} \pi r^{3}$ }
$\Rightarrow \frac{4}{3} \pi \cdot 3 r^{2} \frac{d r}{d t}=k$ 
$\Rightarrow 4 \pi r^{2} d r=k d t$ 
Integrating both sides,
$\Rightarrow 4 \pi \int r^{2} d r=k \int d t$ 
$\Rightarrow \frac{4 \pi r^{3}}{3}=k t+c$ .....(i) 
Now, given that
At t = 0, r = 3 units,
$\Rightarrow \frac{4 \pi(3)^{3}}{3}=k(0)+C$ 
$\Rightarrow 4 \pi(3)^{2}=C$ 
$\Rightarrow C=36 \pi$ 
Now, at t = 3, r = 6 units:
$\Rightarrow 4 \pi \times (6)^{3}=3(\mathrm{k} \times 3+\mathrm{c})$ 
$\Rightarrow$ k = $84 \pi$
Substituting the values of k and c in (i)
$\Rightarrow 4 \pi r^{3}=3(84 \pi t+36 \pi)$ 
$\Rightarrow 4 \pi r^{3}=4 \pi(63 t+27)$ 
$\Rightarrow r^{3}=63 t+27$ 
$\Rightarrow r=\sqrt[3]{63 t+27}$ 
$\therefore$ Radius of balloon after t seconds is $\sqrt[3]{63 t+27}$ units.

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