Ex 6.1

Ex 6.1, 1
Deleted for CBSE Board 2022 Exams

Ex 6.1,2 Deleted for CBSE Board 2022 Exams You are here

Ex 6.1,3 Deleted for CBSE Board 2022 Exams

Ex 6.1,4 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,5 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,6 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 6.1,10 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,11 Important 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

Ex 6.1,14 Deleted for CBSE Board 2022 Exams

Ex 6.1,15 Important Deleted for CBSE Board 2022 Exams

Ex 6.1,16 Deleted for CBSE Board 2022 Exams

Ex 6.1,17 (MCQ) Deleted for CBSE Board 2022 Exams

Ex 6.1, 18 (MCQ) Important Deleted for CBSE Board 2022 Exams

Last updated at April 19, 2021 by Teachoo

Ex 6.1, 2 The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?Let 𝒙 be length of side V be Volume t be time per second We know that Volume of cube = (Side)3 V = 𝒙𝟑 Given that Volume of cube is increasing at rate of 8 cm3/sec. Therefore 𝒅𝑽/𝒅𝒕 = 8 Putting V = 𝒙𝟑 (〖𝑑(𝑥〗^3))/𝑑𝑡 = 8 〖𝑑𝑥〗^3/𝑑𝑡 . 𝑑𝑥/𝑑𝑥 = 8 〖𝑑𝑥〗^3/𝑑𝑥 . 𝑑𝑥/𝑑𝑡 = 8 3𝒙𝟐 . 𝑑𝑥/𝑑𝑡 = 8 𝒅𝒙/𝒅𝒕 = 𝟖/〖𝟑𝒙〗^𝟐 Now, We need to find fast is the surface area increasing when the length of an edge is 12 centimeters i.e. 𝒅𝑺/𝒅𝒕 for x = 12 We know that Surface area of cube = 6 × Side2 S = 6𝑥2 Finding 𝒅𝑺/𝒅𝒕 𝑑𝑆/𝑑𝑡 = (𝑑(6𝑥^2))/𝑑𝑡 = (𝑑(6𝑥2))/𝑑𝑡 . 𝑑𝑥/𝑑𝑥 = 6. (𝑑(𝑥2))/𝑑𝑥 . 𝑑𝑥/𝑑𝑡 = 6 . (2x) . 𝑑𝑥/𝑑𝑡 = 12𝑥 . 𝒅𝒙/𝒅𝒕 = 12𝑥 . 𝟖/𝟑𝒙𝟐 = 𝟑𝟐/𝒙 For 𝑥= 12 cm 𝑑𝑆/𝑑𝑡 = 32/12 (From (1): 𝒅𝒙/𝒅𝒕 = 𝟖/(𝟑𝒙^𝟐 )) 𝑑𝑆/𝑑𝑡 = 8/3 Since surface area is in cm2 & time is in seconds, 𝒅𝑺/𝒅𝒕 = 𝟖/𝟑 cm2 /s