InterviewSolution
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InterviewSolution
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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» Here, Volume is changing at a constant rate. So, `(dV)/dt = k`, where `k` is a constant. Now, Volume of a sphere, `V = 4/3pir^3` So, `(dV)/dt = 4/3**3pir^2(dr)/dt` `=>(dV)/dt = 4pir^2(dr)/dt` As,`(dV)/dt = k`.So, `=>kdt = 4pir^2dr` Now, integrating both sides, `=>kt+c = 4/3pir^3+c_1`->(1) `=>kt+C = 4/3pir^3`, where `C= c-c_1` So, at `(r,t) = (3,0)` `=>C = 4/3pi**(3)^3 => C= 36pi` At`(r,t) = (6,3)` `=>k(3) + 36pi = 4/3pi(6)^3=>3k = 288pi-36pi=>k = 84pi` Putting values of `C` and `k` in (1), `84pit+36pi = 4/3pir^3=>84t+36 = 4/3r^3` `=>r^3 = (63t+27) => r = (63t+27)^(1/3)` So, at any time `t` radius of the baloon will be `(63t+27)^(1/3)`. |
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