Last updated at April 19, 2021 by Teachoo

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 12๐๐^3/sec i.e. Rate of volume of a cone w.r.t time is 12๐๐^3/sec i.e. ๐ ๐ฝ/๐ ๐ ="12" ๐๐^๐ "/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=๐๐๐ 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=1/6 ๐ 6โ=๐ ๐=๐๐ We know that Volume of a cone = 1/3 ๐(๐^2 )โ V = 1/3 ฯใ ๐ใ^2 โ V = 1/3 ฯ (๐๐)^2 โ V = 1/3 ฯ ร 36โ^2 รโ V = 1/3 ฯ ร 36 โ^3 V = 12ฯ ๐^๐ (โ("From (2)" : ๐" = 6" โ)) Differentiating w.r.t ๐ก ๐๐/๐๐ก=๐(12๐โ^3 )/๐๐ก ๐๐/๐๐ก=12๐ ๐(โ^3 )/๐๐ก ๐๐/๐๐ก=12๐ ร๐(โ^3 )/๐๐ก ร ๐โ/๐โ ๐๐/๐๐ก=12๐ ร๐ (๐^๐ )/๐ ๐ ร ๐โ/๐๐ก ๐ ๐ฝ/๐ ๐=12๐ ร 3โ^2 ร ๐โ/๐๐ก ๐๐=12๐ ร 3โ^2 ร ๐โ/๐๐ก 12/(12๐ ร 3โ^2 )=๐โ/๐๐ก ๐ ๐/๐ ๐=๐/(๐๐ ๐^๐ ) (From (1): ๐ ๐ฝ/๐ ๐ ="12" ) Putting โ = 4 cm ๐โ/๐๐ก= 1/(3๐ ร (4)^2 ) ๐โ/๐๐ก =1/48๐ Since height is in cm & time is in sec โด ๐ ๐/๐ ๐=๐/๐๐๐ cm/s Hence, Height of the sand cone is increasing at the rate of ๐/๐๐๐ cm/s

Ex 6.1

Ex 6.1, 1
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About the Author

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.