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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 1 maths textbook solution

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Answer: r=\sqrt{1+\frac{1}{3} t^{2}}

Given: It is given that initial radius of balloon=1 unit. After 3 seconds, radius of balloon=2 units.

To find: We have to find the radius after time t.

Hint: Use surface area of a balloon =4\pi r^{2}. Then differentiate it to find the radius.

Solution: We have radius of the balloon=1 unit

After 3 seconds, radius of balloon=2 units

Let r be the radius and A be the surface area of the balloon at any time t.

        \begin{aligned} &=\frac{d A}{d t} \propto t \\\\ &=\frac{d A}{d t}=k t \end{aligned}        where k is constant.

        \begin{aligned} &=\frac{d\left(4 \pi r^{2}\right)}{d t}=k t \\\\ &=8 \pi r \frac{d r}{d t}=k t \end{aligned}

Integrating on both sides

        \begin{aligned} &=8 \pi \frac{r^{2}}{2}=k \frac{t^{2}}{2}+C \\\\ &=4 \pi r^{2}=k \frac{t^{2}}{2}+C \ldots(i) \end{aligned}

We are given that unit at t=0, r=1 unit

        \begin{aligned} &=4 \pi(1)^{2}=k \times 0+C \\\\ &=C=4 \pi \end{aligned}

And t=3, r=2 units

        \begin{aligned} &=4 \pi(2)^{2}=k \frac{(3)^{2}}{2}+C \quad \text { [From (i)] }\\ \\&=16 \pi=\frac{9}{2} k+4 \pi\\\\ &=\frac{9}{2} k=12 \pi \end{aligned}

By cross multiplication we get

        \begin{aligned} &9 k=24 \pi \\\\ &k=\frac{24}{9} \pi \\\\ &k=\frac{8}{3} \pi \end{aligned}

Now substitute C=4 \pi and   k=\frac{8}{3} \pi  in equation (i)


\begin{aligned} &4 \pi r^{2}=\frac{8}{3} \pi \frac{t^{2}}{2}+4 \pi \\\\ &4 \pi r^{2}-4 \pi=\frac{8}{6} \pi t^{2} \\\\ &4 \pi\left(r^{2}-1\right)=\frac{8}{6} \pi t^{2} \\\\ &\pi\left(r^{2}-1\right)=\frac{1}{4}\cdot \frac{8}{6} \pi t^{2} \end{aligned}

        \begin{aligned} &\left(r^{2}-1\right)=\frac{1}{\pi} \frac{1}{3} \pi t^{2} \\\\ &r^{2}-1=\frac{1}{3} t^{2} \\\\ &r^{2}=1+\frac{1}{3} t^{2}=\sqrt{1+\frac{1}{3} t^{2}} \end{aligned}




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