# Ex 6.1,14 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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 form a cone Let & be the radius & height of the sand cone respectively & V be the volume of cone Sand form cone on the ground in such a way that the height of the cone is always one sixth of the radius i.e. = 1 6 6 = =6 & Sand is pouring from a pipe at the rate of 12 3 /sec i.e. Rate of volume of a cone w.r.t is 12 3 /sec i.e. =12 3 /sec And we need to find how fast height of the Cone is increasing when height is 4cm i.e. find when =4 We have = 12 3 /sec We know that Volume of a cone = 1 3 2 i.e. V = 1 3 2 V = 1 3 6 2 V = 1 3 36 2 V = 1 3 36 3 V = 12 3 Diff w.r.t = 12 3 =12 3 =12 3 =12 3 2 12 = 2 3 2 12 2 3 2 = = 12 12 3 2 We need when =4 When = 4 , =4 = 12 12 3 4 2 = 12 12 3 4 4 = 1 48 Since height is in cm & time is in sec So, = / Hence height of the sand cone is increasing at the rate of 1 48 /

Chapter 6 Class 12 Application of Derivatives

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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.