Ex 6.1, 8 - A balloon is being inflated by pumping in 900 cubic - Finding rate of change

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.1,8 A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimeters of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm. Balloon is spherical Let r be the radius of spherical balloon . & v be the volume of spherical balloon. It is given that The balloon is inflated by pumping in in 900 cubic cm of gas per sec i.e. volume of balloon increasing at the rate of 9000 cm3/sec i.e. ๐‘‘๐‘ฃ๏ทฎ๐‘‘๐‘ก๏ทฏ = 900 cm3/sec We need to calculate the rate of change at which radius of Balloon increases when radius is 15 cm i.e. we need to calculate ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ when r = 15 cm we know that Volume of sphere = 4๏ทฎ3๏ทฏ ฯ€r3 i.e. V = 4๏ทฎ3๏ทฏ ฯ€r3 Differentiate w.r.t time(๐‘ก) ๐‘‘๐‘ฃ๏ทฎ๐‘‘๐‘ก๏ทฏ = ๐‘‘ 4๏ทฎ3๏ทฏ ๐œ‹๐‘Ÿ3๏ทฏ๏ทฎ๐‘‘๐‘ก๏ทฏ ๐‘‘๐‘ฃ๏ทฎ๐‘‘๐‘ก๏ทฏ = 4๏ทฎ3๏ทฏ ฯ€ ๐‘‘ (๐‘Ÿ3)๏ทฎ๐‘‘๐‘ก๏ทฏ 900 = 4๏ทฎ3๏ทฏ ฯ€ . ๐‘‘(๐‘Ÿ3)๏ทฎ๐‘‘๐‘ก๏ทฏ ร— ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘Ÿ๏ทฏ 900 = 4๏ทฎ3๏ทฏ ฯ€ . ๐‘‘(๐‘Ÿ3)๏ทฎ๐‘‘๐‘ก๏ทฏ ร— ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ 900 = 4๏ทฎ3๏ทฏ ฯ€ . 3r2 . ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ 900 = 4 ร— ฯ€ ร— r2 . ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ 900๏ทฎ4 ร— ๐œ‹ ร—๐‘Ÿ2๏ทฏ = ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ = 900 ร— 7๏ทฎ2 ร—22 ร—๐‘Ÿ2๏ทฏ We need to find ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ at r = 15 cm ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ = 900๏ทฎ4 ร—๐œ‹ ร— 15๏ทฏ๏ทฎ2๏ทฏ๏ทฏ ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ = 1๏ทฎ๐œ‹๏ทฏ Since radius in cm & time in sec So, ๐‘‘๐‘Ÿ๏ทฎ๐‘‘๐‘ก๏ทฏ = ๐Ÿ ๐’„๐’Ž๏ทฎ๐… ๐’”๐’†๐’„๏ทฏ Hence ,the radius of the balloon is increasing at the rate of 1๏ทฎ๐œ‹๏ทฏ cm/sec when r = 15 cm.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.