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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.