# Ex 6.1,14

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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 𝑟 & ℎ 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. ℎ= 16𝑟 6 ℎ=𝑟 𝑟=6ℎ & Sand is pouring from a pipe at the rate of 12𝑐 𝑚3/sec i.e. Rate of volume of a cone w.r.t 𝑡 is 12𝑐 𝑚3/sec i.e. 𝑑𝑣𝑑𝑡=12𝑐 𝑚3/sec And we need to find how fast height of the Cone is increasing when height is 4cm i.e. find 𝑑ℎ𝑑𝑡 when ℎ=4𝑐𝑚 We have 𝑑𝑣𝑑𝑡= 12𝑐 𝑚3/sec We know that Volume of a cone = 13𝜋 𝑟2ℎ i.e. V = 13 π 𝑟2ℎ V = 13 π 6ℎ2ℎ V = 13 π × 36 ℎ2 ×ℎ V = 13 π × 36 ℎ3 V = 12π ℎ3 Diff w.r.t 𝑡 𝑑𝑣𝑑𝑡= 𝑑 12𝜋 ℎ3𝑑𝑡 𝑑𝑣𝑑𝑡=12𝜋 𝑑 ℎ3𝑑𝑡 𝑑𝑣𝑑𝑡=12𝜋 × 𝑑 ℎ3𝑑ℎ × 𝑑ℎ𝑑𝑡 𝑑𝑣𝑑𝑡=12𝜋 ×3 ℎ2 × 𝑑ℎ𝑑𝑡 12 = 2𝜋 ×3 ℎ2 × 𝑑ℎ𝑑𝑡 122𝜋 ×3 ℎ2= 𝑑ℎ𝑑𝑡 𝑑ℎ𝑑𝑡= 1212𝜋 ×3 ℎ2 We need 𝑑ℎ𝑑𝑡 when ℎ=4𝑐𝑚 When ℎ = 4 , 𝑑ℎ𝑑𝑡│ℎ=4= 1212𝜋 ×3 42 𝑑ℎ𝑑𝑡 = 1212𝜋 ×3 ×4 ×4 𝑑ℎ𝑑𝑡 = 148 𝜋 Since height is in cm & time is in sec So, 𝑑ℎ𝑑𝑡= 𝟏𝟒𝟖 𝝅 𝒄𝒎/𝒔𝒆𝒄 Hence height of the sand cone is increasing at the rate of 148 𝜋 𝑐𝑚/𝑠𝑒𝑐

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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