The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :
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Official solution
We know,
Surface area of balloon (s) = 4\[\pi\]r2
\[\therefore\] \[{{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})\]
\[ \Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}\]
\[ \Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}\]
Given that, surface area of balloon is increasing in constant rate.
\[\therefore\] \[{{ds} \over {dt}}\] = constant = k (Assume)
\[ \Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}\]
\[ \Rightarrow \int {k\,dt = \int {8\pi r\,dr} } \]
\[ \Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c\]
\[ \Rightarrow kt = 4\pi {r^2} + c\] ..... (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
\[ \Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi \]
\[ \Rightarrow 8 \times 9 = {r^2} - 9\]
\[ \Rightarrow {r^2} = 81 \Rightarrow r = 9\]