Ex 6.1, 10 - A ladder 5 m long is leaning against a wall

Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives - Part 5

  1. Chapter 6 Class 12 Application of Derivatives
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Transcript

Ex 6.1, 10 A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2 cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall ?Let AB be the ladder & OB be the wall & OA be the ground. Given Length of ladder is 5 m AB = 5 cm Let OA = ๐‘ฅ cm & OB = ๐‘ฆ cm Given that Bottom of ladder is pulled away from the wall at the rate of 2 cm/ s i.e. ๐’…๐’š/๐’…๐’• = 2 cm/sec We need to calculate at which rate height of ladder on the wall is decreasing when foot of the ladder is 4 m away from the wall i.e. We need to calculate ๐’…๐’™/๐’…๐’• when ๐’š = 4 cm. Since Wall OB is perpendicular to the ground OA Using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 ๐‘ฆ2 + ๐‘ฅ2 = (5)2 ๐’š๐Ÿ + ๐’™๐Ÿ = 25 Differentiating w.r.t time (๐‘‘(๐‘ฆ2 + ๐‘ฅ2))/๐‘‘๐‘ก = (๐‘‘(25))/๐‘‘๐‘ก (๐‘‘ (๐‘ฆ2))/๐‘‘๐‘ก + (๐‘‘(๐‘ฅ2))/๐‘‘๐‘ก = 0 (๐‘‘(๐‘ฆ2))/๐‘‘๐‘ก ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ + (๐‘‘(๐‘ฅ2))/๐‘‘๐‘ก ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ = 0 (๐‘‘(๐‘ฆ2))/๐‘‘๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ก + (๐‘‘(๐‘ฅ2))/๐‘‘๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 0 2๐‘ฆ ร— ๐’…๐’š/๐’…๐’• + 2๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 0 2๐‘ฆ ร— 2 + 2๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 0 ("From (1): " ๐‘‘๐‘ฆ/๐‘‘๐‘ก= 2 m/s) 2๐‘ฅ ๐‘‘๐‘ฅ/๐‘‘๐‘ก = โ€“4๐‘ฆ ๐’…๐’™/๐’…๐’• = (โˆ’๐Ÿ๐’š)/๐’™ Putting ๐‘ฆ = 4 cm โ”œ ๐‘‘๐‘ฅ/๐‘‘๐‘กโ”ค|_(๐‘ฆ = 4) =(โˆ’2 ร— 4)/๐‘ฅ โ”œ ๐’…๐’™/๐’…๐’•โ”ค|_(๐’š = ๐Ÿ’) =(โˆ’๐Ÿ–)/๐’™ To find ๐‘‘๐‘ฅ/๐‘‘๐‘ก , we need to find value of x for y = 4 Finding value of x We know that ๐‘ฅ2 + ๐‘ฆ2 = 25 Putting ๐‘ฆ =4 ๐‘ฅ2 + 42 = 25 ๐‘ฅ2 = 25 โ€“ 16 ๐‘ฅ2 = 9 ๐‘ฅ = 3 Putting x = 3 in (2) ๐‘‘๐‘ฅ/๐‘‘๐‘ก=(โˆ’8)/๐‘ฅ ๐‘‘๐‘ฅ/๐‘‘๐‘ก = (โˆ’8)/3 Since x is in cm & time is in sec โˆด ๐‘‘๐‘ฅ/๐‘‘๐‘ก = (โˆ’๐Ÿ–)/๐Ÿ‘ cm/sec Hence, height of ladder on the wall is decreasing at rate of ๐Ÿ–/๐Ÿ‘ cm/sec

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.