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Last updated at Jan. 7, 2020 by Teachoo

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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 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 = ๐ฅ & OB = ๐ฆ 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. ๐๐ฅ/๐๐ก = 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 ๐๐ฆ/๐๐ก when ๐ฅ = 4 cm. Wall OB is perpendicular to the ground OA Using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 ๐ฆ2 + ๐ฅ2 = (5)2 ๐ฆ2 + ๐ฅ2 = 25 Differentiating w.r.t time (๐(๐ฆ2 + ๐ฅ2))/๐๐ก = (๐(25))/๐๐ก (๐ (๐ฆ2))/๐๐ก + (๐(๐ฅ2))/๐๐ก = 0 (๐(๐ฆ2))/๐๐ก ร ๐๐ฆ/๐๐ฆ + (๐(๐ฅ2))/๐๐ก ร ๐๐ฅ/๐๐ฅ = 0 2๐ฆ ร ๐๐ฆ/๐๐ก + 2๐ฅ ร ๐๐ฅ/๐๐ก = 0 2๐ฆ ร ๐๐ฆ/๐๐ฅ + 2๐ฅ ร (2) = 0 2๐ฆ ๐๐ฆ/๐๐ก + 4๐ฅ = 0 2๐ฆ ๐๐ฆ/๐๐ก = โ 4 ๐ฅ ๐๐ฆ/๐๐ก = (โ 4๐ฅ)/2๐ฆ ("From (1): " ๐๐ฆ/๐๐ฅ=2๐/๐ ๐๐) We need to find ๐๐ฆ/๐๐ก when ๐ฅ = 4 cm โ ๐๐ฆ/๐๐กโค|_(๐ฅ = 4) =(โ 4 ร 4)/2๐ฆ โ ๐๐ฆ/๐๐กโค|_(๐ฅ = 4) =(โ 16)/2๐ฆ Finding value of y From (2) ๐ฅ2 + ๐ฆ2 = 25 Putting ๐ฅ =4 (4)2 + ๐ฆ2 = 25 ๐ฆ2 = 25 โ 16 ๐ฆ2 = 9 ๐ฆ = 3 Now we need to find ๐๐ฆ/๐๐ก at y = 3 ๐๐ฆ/๐๐ก=(โ 16)/2๐ฆ ๐๐ฆ/๐๐ก = (โ 16)/(2 ร 3) ๐๐ฆ/๐๐ก = (โ 8)/3 Since length of OB is in cm & time is in sec So, ๐๐ฆ/๐๐ก = (โ 8 ๐๐)/(3 ๐ ๐๐) ๐๐ฆ/๐๐ก = (โ๐)/๐ cm/sec Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/sec

Chapter 6 Class 12 Application of Derivatives

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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.