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6) The volume of a spherical balloon is increasing at the rate of 20cm / sec. Find the rate of change of its surface area at the instant when its radius is 8 cm. |
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Answer» At any instant t, let r be the radius, V the volume and S the surface area of the balloon. Then, `(dV)/(dt) = 20 cm^(3)//sec` (given ) ...(i) Now, `V = (4)/(3) pi r^(3) rArr (dV)/(dt) = (dV)/(dr).(dr)/(dt)` `rArr 20 = (d)/(dr) ((4)/(3) pi r^(3)).(dr)/(dt)` `rArr 20 = (4)/(3) pi xx 3r^(2) xx (dr)/(dt) = 4 pi r^(2). (dr)/(dt)` `rArr (dr)/(dt) = (5)/(pir^(2))`...(ii) `:. S = 5pir^(2) rArr (dS)/(dt) = (dS)/(dr) .(dr)/(dt)` `= (d)/(dr) (4pir^(2)).(5)/(pir^(2))` `= (8 pi rxx (5)/(pir^(2))) = (40)/(r)` `rArr [(dS)/(dt)]_(r =8 cm) = ((40)/(8)) cm^(2)//sec = 5 cm^(2)//sec` Hence, the rate of change of surface area at the instant when `r = 8 cm" is " 5 cm^(2)//sec` |
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