Finding rate of change

Chapter 6 Class 12 Application of Derivatives (Term 1)
Concept wise

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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 forms a cone Let π be the radius & π be height of the sand cone & V be the volume of cone Also, Sand is pouring from a pipe at the rate of 12ππ^3/sec i.e. Rate of volume of a cone w.r.t time is 12ππ^3/sec i.e. ππ½/ππ ="12" ππ^π "/sec" We need to find how fast height of the Cone is increasing when height is 4cm i.e. find ππ/ππ when π=πππ 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. β=1/6 π 6β=π π=ππ We know that Volume of a cone = 1/3 π(π^2 )β V = 1/3 Ογ πγ^2 β V = 1/3 Ο (ππ)^2 β V = 1/3 Ο Γ 36β^2 Γβ V = 1/3 Ο Γ 36 β^3 V = 12Ο π^π (β("From (2)" : π" = 6" β)) Differentiating w.r.t π‘ ππ/ππ‘=π(12πβ^3 )/ππ‘ ππ/ππ‘=12π π(β^3 )/ππ‘ ππ/ππ‘=12π Γπ(β^3 )/ππ‘ Γ πβ/πβ ππ/ππ‘=12π Γπ(π^π )/ππ Γ πβ/ππ‘ ππ½/ππ=12π Γ 3β^2 Γ πβ/ππ‘ ππ=12π Γ 3β^2 Γ πβ/ππ‘ 12/(12π Γ 3β^2 )=πβ/ππ‘ ππ/ππ=π/(πππ^π ) (From (1): ππ½/ππ ="12" ) Putting β = 4 cm πβ/ππ‘= 1/(3π Γ (4)^2 ) πβ/ππ‘ =1/48π Since height is in cm & time is in sec β΄ ππ/ππ=π/πππ cm/s Hence, Height of the sand cone is increasing at the rate of π/πππ cm/s