Slide11.JPG

Slide12.JPG
Slide13.JPG Slide14.JPG Slide15.JPG

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Ex 6.1, 14 Sand is pouring from a pipe at the rate of 12 cm3/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm? Given that sand is pouring from a pipe & falling sand forms a cone Let ๐’“ be the radius & ๐’‰ be height of the sand cone & V be the volume of cone Also, Sand is pouring from a pipe at the rate of 12๐‘๐‘š^3/sec i.e. Rate of volume of a cone w.r.t time is 12๐‘๐‘š^3/sec i.e. ๐’…๐‘ฝ/๐’…๐’• ="12" ๐’„๐’Ž^๐Ÿ‘ "/sec" We need to find how fast height of the Cone is increasing when height is 4cm i.e. find ๐’…๐’‰/๐’…๐’• when ๐’‰=๐Ÿ’๐’„๐’Ž Given that sand forms cone on the ground in such a way that the height of the cone is always one sixth of the radius i.e. โ„Ž=1/6 ๐‘Ÿ 6โ„Ž=๐‘Ÿ ๐’“=๐Ÿ”๐’‰ We know that Volume of a cone = 1/3 ๐œ‹(๐‘Ÿ^2 )โ„Ž V = 1/3 ฯ€ใ€– ๐’“ใ€—^2 โ„Ž V = 1/3 ฯ€ (๐Ÿ”๐’‰)^2 โ„Ž V = 1/3 ฯ€ ร— 36โ„Ž^2 ร—โ„Ž V = 1/3 ฯ€ ร— 36 โ„Ž^3 V = 12ฯ€ ๐’‰^๐Ÿ‘ (โ–ˆ("From (2)" : ๐‘Ÿ" = 6" โ„Ž)) Differentiating w.r.t ๐‘ก ๐‘‘๐‘‰/๐‘‘๐‘ก=๐‘‘(12๐œ‹โ„Ž^3 )/๐‘‘๐‘ก ๐‘‘๐‘‰/๐‘‘๐‘ก=12๐œ‹ ๐‘‘(โ„Ž^3 )/๐‘‘๐‘ก ๐‘‘๐‘‰/๐‘‘๐‘ก=12๐œ‹ ร—๐‘‘(โ„Ž^3 )/๐‘‘๐‘ก ร— ๐‘‘โ„Ž/๐‘‘โ„Ž ๐‘‘๐‘‰/๐‘‘๐‘ก=12๐œ‹ ร—๐’…(๐’‰^๐Ÿ‘ )/๐’…๐’‰ ร— ๐‘‘โ„Ž/๐‘‘๐‘ก ๐’…๐‘ฝ/๐’…๐’•=12๐œ‹ ร— 3โ„Ž^2 ร— ๐‘‘โ„Ž/๐‘‘๐‘ก ๐Ÿ๐Ÿ=12๐œ‹ ร— 3โ„Ž^2 ร— ๐‘‘โ„Ž/๐‘‘๐‘ก 12/(12๐œ‹ ร— 3โ„Ž^2 )=๐‘‘โ„Ž/๐‘‘๐‘ก ๐’…๐’‰/๐’…๐’•=๐Ÿ/(๐Ÿ‘๐…๐’‰^๐Ÿ ) (From (1): ๐’…๐‘ฝ/๐’…๐’• ="12" ) Putting โ„Ž = 4 cm ๐‘‘โ„Ž/๐‘‘๐‘ก= 1/(3๐œ‹ ร— (4)^2 ) ๐‘‘โ„Ž/๐‘‘๐‘ก =1/48๐œ‹ Since height is in cm & time is in sec โˆด ๐’…๐’‰/๐’…๐’•=๐Ÿ/๐Ÿ’๐Ÿ–๐… cm/s Hence, Height of the sand cone is increasing at the rate of ๐Ÿ/๐Ÿ’๐Ÿ–๐… cm/s

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.