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

Last updated at April 16, 2024 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 𝒓 be the radius
& 𝒉 be height of the sand cone
& V be the volume of cone
Also, Sand is pouring from a pipe at the rate of 12𝑐𝑚^3/sec
i.e. Rate of volume of a cone w.r.t time is 12𝑐𝑚^3/sec
i.e. 𝒅𝑽/𝒅𝒕 ="12" 𝒄𝒎^𝟑 "/sec"
We need to find
how fast height of the Cone is increasing when height is 4cm
i.e. find 𝒅𝒉/𝒅𝒕 when 𝒉=𝟒𝒄𝒎
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. ℎ=1/6 𝑟
6ℎ=𝑟
𝒓=𝟔𝒉
We know that
Volume of a cone = 1/3 𝜋(𝑟^2 )ℎ
V = 1/3 π〖 𝒓〗^2 ℎ
V = 1/3 π (𝟔𝒉)^2 ℎ
V = 1/3 π × 36ℎ^2 ×ℎ
V = 1/3 π × 36 ℎ^3
V = 12π 𝒉^𝟑
(█("From (2)" : 𝑟" = 6" ℎ))
Differentiating w.r.t 𝑡
𝑑𝑉/𝑑𝑡=𝑑(12𝜋ℎ^3 )/𝑑𝑡
𝑑𝑉/𝑑𝑡=12𝜋 𝑑(ℎ^3 )/𝑑𝑡
𝑑𝑉/𝑑𝑡=12𝜋 ×𝑑(ℎ^3 )/𝑑𝑡 × 𝑑ℎ/𝑑ℎ
𝑑𝑉/𝑑𝑡=12𝜋 ×𝒅(𝒉^𝟑 )/𝒅𝒉 × 𝑑ℎ/𝑑𝑡
𝒅𝑽/𝒅𝒕=12𝜋 × 3ℎ^2 × 𝑑ℎ/𝑑𝑡
𝟏𝟐=12𝜋 × 3ℎ^2 × 𝑑ℎ/𝑑𝑡
12/(12𝜋 × 3ℎ^2 )=𝑑ℎ/𝑑𝑡
𝒅𝒉/𝒅𝒕=𝟏/(𝟑𝝅𝒉^𝟐 )
(From (1): 𝒅𝑽/𝒅𝒕 ="12" )
Putting ℎ = 4 cm
𝑑ℎ/𝑑𝑡= 1/(3𝜋 × (4)^2 )
𝑑ℎ/𝑑𝑡 =1/48𝜋
Since height is in cm & time is in sec
∴ 𝒅𝒉/𝒅𝒕=𝟏/𝟒𝟖𝝅 cm/s
Hence, Height of the sand cone is increasing at the rate of 𝟏/𝟒𝟖𝝅 cm/s

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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