# Ex 6.1,2

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 ð¥ be the edge of a cube. Let V be the volume of cube. & S be the surface are of cube. Given that Volume of a cube is increasing at the rate of 8 cm3/ sec i.e. ðð£ï·®ðð¡ï·¯ = 8 cm3/ sec We need to calculate how fast the surface Area is increasing when length of an edge is 12 cm i.e. We need to calculate ðð ï·®ðð¡ï·¯ when ð¥ = 12 cm. Now, Volume of cube = (edge)3 ð(ð¥3)ï·®ðð¡ï·¯ = 8 ð(ð¥3)ï·®ðð¡ï·¯ Ã ðð¥ï·®ðð¥ï·¯= 8 ð(ð¥3)ï·®ðð¥ï·¯ Ã ðð¥ï·®ðð¡ï·¯= 8 3x2 . ðð¥ï·®ðð¡ï·¯= 8 ðð¥ï·®ðð¡ï·¯ = 8ï·®3ð¥2ï·¯ We know that Surface Area of cube = 6 (edge)2 S =6ð¥2 Different w.r.t ð¡ ðð ï·®ðð¡ï·¯ = ð 6ð¥2ï·¯ï·®ðð¡ï·¯ ðð ï·®ðð¡ï·¯ = 6ð ð¥2ï·¯ï·®ðð¡ï·¯ ðð ï·®ðð¡ï·¯ = 6 . ð ð¥2ï·¯ï·®ðð¡ï·¯ Ã ðð¥ï·®ðð¥ï·¯ ðð ï·®ðð¡ï·¯ = 6 . ð ð¥2ï·¯ï·®ðð¥ï·¯ Ã ðð¥ï·®ðð¡ï·¯ ðð ï·®ðð¡ï·¯ = 6 . 2ð¥ . ðð¥ï·®ðð¡ï·¯ ðð ï·®ðð¡ï·¯ = 12ð¥ . ð ðï·®ð ðï·¯ ðð ï·®ðð¡ï·¯ = 12 ð¥. ðï·®ðððï·¯ï·¯ ðð ï·®ðð¡ï·¯ = 32ï·®ð¥ï·¯ When ð¥ = 12 ðð ï·®ðð¡ï·¯ï·¯ï·®ð¥ =12ï·¯ = 32ï·®12ï·¯ ðð ï·®ðð¡ï·¯ï·¯ï·®ð¥ =12ï·¯ = 8ï·®3ï·¯ Since surface area is in cm2 & time is in sec ðð ï·®ðð¡ï·¯ = 8ï·®3ï·¯ ðð2ï·®ð ððï·¯ ðð ï·®ðð¡ï·¯ = ðï·®ðï·¯ cm2/sec Hence, surface area is increasing at the rate of 8ï·®3ï·¯ cm2/sec when edge is 12 cm

Chapter 6 Class 12 Application of Derivatives

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .