



Finding rate of change
Ex 6.1, 1 Deleted for CBSE Board 2022 Exams
Ex 6.1,17 (MCQ) Deleted for CBSE Board 2022 Exams
Example 5 Deleted for CBSE Board 2022 Exams
Ex 6.1,15 Important Deleted for CBSE Board 2022 Exams
Example 6 Deleted for CBSE Board 2022 Exams
Ex 6.1,16 Deleted for CBSE Board 2022 Exams
Ex 6.1, 18 (MCQ) Important Deleted for CBSE Board 2022 Exams
Example 2 Deleted for CBSE Board 2022 Exams
Ex 6.1,2 Deleted for CBSE Board 2022 Exams
Example 49 Deleted for CBSE Board 2022 Exams
Example 3 Deleted for CBSE Board 2022 Exams
Ex 6.1,5 Important Deleted for CBSE Board 2022 Exams
Ex 6.1,3 Deleted for CBSE Board 2022 Exams
Ex 6.1,6 Deleted for CBSE Board 2022 Exams
Ex 6.1,12 Deleted for CBSE Board 2022 Exams
Ex 6.1,13 Important Deleted for CBSE Board 2022 Exams
Misc. 19 (MCQ) Deleted for CBSE Board 2022 Exams
Ex 6.1,14 Deleted for CBSE Board 2022 Exams
Example 43 Important Deleted for CBSE Board 2022 Exams You are here
Ex 6.1,4 Important Deleted for CBSE Board 2022 Exams
Example 42 Important Deleted for CBSE Board 2022 Exams
Example 4 Important Deleted for CBSE Board 2022 Exams
Ex 6.1,7 Deleted for CBSE Board 2022 Exams
Ex 6.1,8 Deleted for CBSE Board 2022 Exams
Ex 6.1,9 Deleted for CBSE Board 2022 Exams
Ex 6.1,11 Important Deleted for CBSE Board 2022 Exams
Misc 3 Important
Ex 6.1,10 Important Deleted for CBSE Board 2022 Exams
Example 44 Important Deleted for CBSE Board 2022 Exams
Finding rate of change
Last updated at April 19, 2021 by Teachoo
Example 43 A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tanβ1 (0.5). Water is poured into it at a constant rate of 5 cubic meter per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4 m.Water tank is in shape of cone Let r be the radius of cone, h be the height of cone, & πΆ be the semiβvertical angle Given Semi-vertical angle is tan^(β1)β‘(0.5) Ξ±=tan^(β1)β‘(0.5) tan Ξ± =(0.5) π/β = 0.5 π/β = 1/2 π = π/π Also, Water is poured at a constant rate of 5 cubic meter per hour π π½/π π=π π^π/ππ Where V is volume of cone Now, π=1/3 ππ^2 β π=1/3 π(π/π)^2 β π=1/3 π(γh/4γ^2 )β π½=(π π^π)/ππ Differentiate w.r.t. t ππ/ππ‘=π((πβ^3)/12)/ππ‘ ππ/ππ‘=π/12 . (πβ^3)/ππ‘ ππ/ππ‘=π/12 . (πβ^3)/πβ .πβ/ππ‘ ππ/ππ‘=π/12. 3β^2 . πβ/ππ‘ ππ/ππ‘=(πβ^2)/4 . πβ/ππ‘ Putting π π½/π π=π 5 =(πβ^2)/4 . πβ/ππ‘ π π/π π=ππ/(π π^π ) We need to find, Rate at which level of water is rising when depth is 4 m i.e. β π π/π πβ€|_(π = π π) Putting h = 4 m in (3) β πβ/ππ‘β€|_(β = 4π)=20/(π(β)^2 )=20/16 Γ 1/π=5/4 Γ 1/(22/7)=5/4 Γ 7/22=35/88 Hence, rate of change of water level is ππ/ππ m/hr.