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The radius of a balloon is increasing at the rateof 10 cm/sec. At what rate is the surface area of the balloon increasing whenthe radius is 15 cm? |
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Answer» Correct Answer - `1200pi cm^(2)//s` `S = 4pi r^(2) rArr (dS)/(dt) = 8pir.(dr)/(dt) = (8pi xx 15 xx 10) cm^(2)//s` |
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