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Ex 6.1, 14 - Sand is pouring from a pipe at rate of 12 cm3/s - Ex 6.1

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  1. Chapter 6 Class 12 Application of Derivatives
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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= 1﷮6﷯𝑟 6 ℎ=𝑟 𝑟=6ℎ & Sand is pouring from a pipe at the rate of 12𝑐 𝑚﷮3﷯/sec i.e. Rate of volume of a cone w.r.t 𝑡 is 12𝑐 𝑚﷮3﷯/sec i.e. 𝑑𝑣﷮𝑑𝑡﷯=12𝑐 𝑚﷮3﷯/sec And we need to find how fast height of the Cone is increasing when height is 4cm i.e. find 𝑑ℎ﷮𝑑𝑡﷯ when ℎ=4𝑐𝑚 We have 𝑑𝑣﷮𝑑𝑡﷯= 12𝑐 𝑚﷮3﷯/sec We know that Volume of a cone = 1﷮3﷯𝜋 𝑟﷮2﷯﷯ℎ i.e. V = 1﷮3﷯ π 𝑟﷮2﷯ℎ V = 1﷮3﷯ π 6ℎ﷯﷮2﷯ℎ V = 1﷮3﷯ π × 36 ℎ﷮2﷯ ×ℎ V = 1﷮3﷯ π × 36 ℎ﷮3﷯ V = 12π ℎ﷮3﷯ Diff w.r.t 𝑡 𝑑𝑣﷮𝑑𝑡﷯= 𝑑 12𝜋 ℎ﷮3﷯﷯﷮𝑑𝑡﷯ 𝑑𝑣﷮𝑑𝑡﷯=12𝜋 𝑑 ℎ﷮3﷯﷯﷮𝑑𝑡﷯ 𝑑𝑣﷮𝑑𝑡﷯=12𝜋 × 𝑑 ℎ﷮3﷯﷯﷮𝑑ℎ﷯ × 𝑑ℎ﷮𝑑𝑡﷯ 𝑑𝑣﷮𝑑𝑡﷯=12𝜋 ×3 ℎ﷮2﷯ × 𝑑ℎ﷮𝑑𝑡﷯ 12 = 2𝜋 ×3 ℎ﷮2﷯ × 𝑑ℎ﷮𝑑𝑡﷯ 12﷮2𝜋 ×3 ℎ﷮2﷯﷯= 𝑑ℎ﷮𝑑𝑡﷯ 𝑑ℎ﷮𝑑𝑡﷯= 12﷮12𝜋 ×3 ℎ﷮2﷯﷯ We need 𝑑ℎ﷮𝑑𝑡﷯ when ℎ=4𝑐𝑚 When ℎ = 4 , 𝑑ℎ﷮𝑑𝑡﷯│﷮ℎ=4﷯= 12﷮12𝜋 ×3 4﷯﷮2﷯﷯ 𝑑ℎ﷮𝑑𝑡﷯ = 12﷮12𝜋 ×3 ×4 ×4﷯ 𝑑ℎ﷮𝑑𝑡﷯ = 1﷮48 𝜋﷯ Since height is in cm & time is in sec So, 𝑑ℎ﷮𝑑𝑡﷯= 𝟏﷮𝟒𝟖 𝝅﷯ 𝒄𝒎/𝒔𝒆𝒄 Hence height of the sand cone is increasing at the rate of 1﷮48 𝜋﷯ 𝑐𝑚/𝑠𝑒𝑐

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