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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

Answers (1)

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.

Posted by

infoexpert26

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