

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
Ex 6.1,4 Important Deleted for CBSE Board 2022 Exams You are here
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
Ex 6.1, 4 An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of cube increasing when the edge is 10 cm long?Let π be the edge of cube. & V be the volume of cube. Given that Edge of cube is increasing at the rate of 3 cm/ sec β΄ π π/π π = 3 cm/sec We need to calculate how fast volume of cube increasing when edge is 10 cm i.e. we need to find π π½/π π when π₯ = 10 cm We know that Volume of cube = (Edge)3 V = π₯3 Differentiate w.r.t time π π½/π π = (π (ππ))/π π ππ/ππ‘ = (π(π₯3))/ππ‘ Γ ππ₯/ππ₯ ππ/ππ‘ = (π(π₯3))/ππ₯ Γ ππ₯/ππ‘ ππ/ππ‘ = 3π₯2 . π π/π π ππ/ππ‘ = 3π₯2 Γ 3 ππ/ππ‘ = 9π₯2 When π₯ = 10 β ππ/ππ‘β€|_(π₯ =10) = 9(10)2 β ππ/ππ‘β€|_(π₯ =10) = 900 Since value is in cm3 & time is in sec π π½/π π = 900 cm3/sec Hence, volume of a cube is increasing at the rate of 900 cm3/sec