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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 54

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 54

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Answer: (63 t+27)^{\frac{1}{3}}

Hint: Separate the terms of x and y and then integrate them.

Given: The volume of a spherical balloon being inflated changes at a constant rate. If initially it radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds

Solution: Let V be the volume of spherical balloon

          Now it is being inflated changes at a constant rate

        \because \frac{d v}{d t}=kwhere k is any constant                                                         ……………….(*)

        New volume of spherical balloon=\frac{4}{3} \pi r^{3}, r is radius

        \begin{aligned} &\frac{d v}{d t}=\frac{4}{3} \pi 3 r^{2} \frac{d r}{d t}=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow k=4 \pi r^{2} \frac{d r}{d t}\\\\ &\Rightarrow d t=\frac{4 \pi}{k} r^{2} d r \end{aligned}

          Integrating both sides

        \begin{aligned} &\int d t=\frac{4 \pi}{k} \int r^{2} d r\\\\ &\Rightarrow t=\frac{4 \pi}{k}\left[\frac{r^{3}}{3}\right]+c\\\\ &\Rightarrow \frac{r^{8}}{3}=\frac{k}{4 \pi} t+c \end{aligned}                ...........(1)

        

        Now given conditions are:

         Whent = 0; r = 3 and when t = 3, r = 6

        We have to find r at t = t

        Put in (1) we get

        \begin{aligned} &\text { At } t=0, r=3 \Rightarrow \frac{3^{8}}{3}=\frac{k}{4 \pi} 0+c \Rightarrow 9=c \\ \end{aligned}                ?By(1)

        \frac{r^{8}}{3}=\frac{k}{4 \pi} t+9                ..................(2)

        Now at  \mathrm{t}=3, \mathrm{r}=6 \Rightarrow \frac{6^{3}}{3}=\frac{k}{4 \pi} 3+9

                                            \Rightarrow \frac{6 \times 6 \times 6}{3}-9=\frac{3 k}{4 \pi}    

                                            \begin{aligned} &\Rightarrow 72-9=\frac{3 k}{4 \pi} \Rightarrow \frac{63 \times 4 \pi}{3}=k \\ &\Rightarrow k=84 \pi \end{aligned}

        Put in (2)

        \begin{aligned} &\frac{r^{3}}{3}=\frac{84 \pi}{4 \pi} t+9 \Rightarrow \frac{r^{3}}{3}=21 t+9 \\\\ &\Rightarrow r^{3}=63 t+27 \\\\ &\Rightarrow r=(63 t+27)^{\frac{1}{3}} \end{aligned}

        which is our required radius.

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