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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

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 2cm/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 & length of ladder is 5 m i.e. AB = 5 & OB be the wall & OA be the ground. Let OA = ๐‘ฅ & OB = ๐‘ฆ Given that The bottom of the ladder is pulled along the ground, away from the wall at the rate of 2 cm/ s i.e. ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 2cm/sec We need to calculate at which rate height of ladder on the wall. Decreasing when foot of the ladder is 4 m away from the wall i.e. we need to calculate ๐‘‘๐‘ฆ/๐‘‘๐‘ก when ๐‘ฅ = 4 cm. Wall OB is perpendicular to the ground OA Using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 ๐‘ฆ2 + ๐‘ฅ2 = (5)2 ๐‘ฆ2 + ๐‘ฅ2 = 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 2๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ก + 2๐‘ฅ ร— ๐‘‘๐‘ฅ/๐‘‘๐‘ก = 0 2๐‘ฆ ร— ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ + 2๐‘ฅ ร— (2) = 0 2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ก + 4๐‘ฅ = 0 2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ก = โ€“ 4 ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (โˆ’ 4๐‘ฅ)/2๐‘ฆ ("From (1): " ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=2๐‘š/๐‘ ๐‘’๐‘) We need to find ๐‘‘๐‘ฆ/๐‘‘๐‘ก when ๐‘ฅ = 4 cm โ”œ ๐‘‘๐‘ฆ/๐‘‘๐‘กโ”ค|_(๐‘ฅ = 4) =(โˆ’ 4 ร— 4)/2๐‘ฆ โ”œ ๐‘‘๐‘ฆ/๐‘‘๐‘กโ”ค|_(๐‘ฅ = 4) =(โˆ’ 16)/2๐‘ฆ Finding value of y From (2) ๐‘ฅ2 + ๐‘ฆ2 = 25 Putting ๐‘ฅ =4 (4)2 + ๐‘ฆ2 = 25 ๐‘ฆ2 = 25 โ€“ 16 ๐‘ฆ2 = 9 ๐‘ฆ = 3 Now we need to find ๐‘‘๐‘ฆ/๐‘‘๐‘ก at y = 3 ๐‘‘๐‘ฆ/๐‘‘๐‘ก=(โˆ’ 16)/2๐‘ฆ ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (โˆ’ 16)/(2 ร— 3) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (โˆ’ 8)/3 Since length of OB is in cm & time is in sec So, ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (โˆ’ 8 ๐‘๐‘š)/(3 ๐‘ ๐‘’๐‘) ๐‘‘๐‘ฆ/๐‘‘๐‘ก = (โˆ’๐Ÿ–)/๐Ÿ‘ cm/sec Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 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 9 years. He provides courses for Maths and Science at Teachoo.