# Ex 6.1,10

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

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 = 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๏ทฎ๐๐ก๏ทฏ = 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๏ทฎ๐๐ก๏ทฏ 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 Different w. r. t time ๐(๐ฆ2+๐ฅ2)๏ทฎ๐๐ก๏ทฏ = ๐(25)๏ทฎ๐๐ก๏ทฏ ๐ ๐ฆ2๏ทฏ๏ทฎ๐๐ก๏ทฏ + ๐(๐ฅ2)๏ทฎ๐๐๐ก๏ทฏ = 0 ๐(๐ฆ2)๏ทฎ๐๐ก๏ทฏ ร ๐๐ฆ๏ทฎ๐๐ฆ๏ทฏ + ๐(๐ฅ2)๏ทฎ๐๐ก๏ทฏ ร ๐๐ฅ๏ทฎ๐๐ฅ๏ทฏ = 0 2๐ฆ ร ๐๐ฆ๏ทฎ๐๐ก๏ทฏ + 2๐ฅ ร ๐๐ฅ๏ทฎ๐๐ก๏ทฏ = 0 2๐ฆ ร ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ + 2๐ฅ ร (2) = 0 2๐ฆ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ + 4๐ฅ = 0 2๐ฆ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ 4 ๐ฅ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ 4๐ฅ๏ทฎ2๐ฆ๏ทฏ We need to find ๐๐ฆ๏ทฎ๐๐ก๏ทฏ when ๐ฅ = 4 cm ๐๐ฆ๏ทฎ๐๐ก๏ทฏ๏ทฏ๏ทฎ๐ฅ = 4๏ทฏ โ 4 ร 4๏ทฎ2๐ฆ๏ทฏ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ๏ทฏ๏ทฎ๐ฅ = 4๏ทฏ โ 16๏ทฎ2๐ฆ๏ทฏ ๐ฅ2 + ๐ฆ2 = 25 Putting ๐ฅ = ๐ฆ (4)2 + ๐ฆ2 = 25 ๐ฆ2 = 25 โ 16 ๐ฆ2 = 9 ๐ฆ = 3 Now we need to find ๐๐ฆ๏ทฎ๐๐ก๏ทฏ at y = 3 ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ 16๏ทฎ2 ร 3๏ทฏ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ 8๏ทฎ3๏ทฏ Since length of OB is in cm & time is in sec So, ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ 8 ๐๐๏ทฎ3 ๐ ๐๐๏ทฏ ๐๐ฆ๏ทฎ๐๐ก๏ทฏ = โ ๐๏ทฎ๐๏ทฏ cm/sec Hence, the height of the ladder on the wall is decreasing at the rate of 8๏ทฎ3๏ทฏ cm/sec

Chapter 6 Class 12 Application of Derivatives

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.