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Ex 6.1, 3 - Radius of a circle is increasing uniformly at 3 cm/s

Ex 6.1,3 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,3 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.1,3 - Chapter 6 Class 12 Application of Derivatives - Part 4


Transcript

Ex 6.1, 3 The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.Let r be the radius of circle . & A be the Area of circle. Given that Radius of a circle is increasing at the rate of 3 cm/s Thus, ๐’…๐’“/๐’…๐’• = 3 cm /sec We need to find rate of change of area of circle w. r. t time when r = 10 cm i.e. we need to find ๐’…๐‘จ/๐’…๐’• when r = 10 cm We know that Area of circle = ฯ€r2 A = ฯ€r2 Differentiating w.r.t time ๐’…๐‘จ/๐’…๐’• = ๐’…(๐…๐’“๐Ÿ)/๐’…๐’• ๐‘‘๐ด/๐‘‘๐‘ก = ฯ€ ๐‘‘(๐‘Ÿ2)/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก = ฯ€ ๐‘‘(๐‘Ÿ2)/๐‘‘๐‘ก ร— ๐’…๐’“/๐’…๐’“ ๐‘‘๐ด/๐‘‘๐‘ก = ฯ€ ๐’…(๐’“๐Ÿ)/๐’…๐’“ ร— ๐‘‘๐‘Ÿ/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก = ฯ€. 2r . ๐‘‘๐‘Ÿ/๐‘‘๐‘ก ๐‘‘๐ด/๐‘‘๐‘ก = 2ฯ€r . ๐’…๐’“/๐’…๐’• ๐‘‘๐ด/๐‘‘๐‘ก = 2ฯ€r . 3 ๐‘‘๐ด/๐‘‘๐‘ก = 6ฯ€r When ๐’“ = 10 cm โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘Ÿ =10) = 6 ร— ฯ€ ร— 10 โ”œ ๐‘‘๐ด/๐‘‘๐‘กโ”ค|_(๐‘Ÿ =10) = 60 ฯ€ (From (1): ๐‘‘๐‘Ÿ/๐‘‘๐‘ก = 3) Since area is in cm2 & time is in sec ๐‘‘๐ด/๐‘‘๐‘ก = 60ฯ€ cm2/sec Hence, Area is increasing at the rate of 60ฯ€ cm2/sec when r = 10 cm

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.