# Ex 6.1,10 - Chapter 6 Class 12 Application of Derivatives

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 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 = & OB = 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. = 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 when = 4 cm. Wall OB is perpendicular to the ground OA using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 2 + 2 = (5)2 2 + 2 = 25 Different w. r. t time ( 2+ 2) = (25) 2 + ( 2) = 0 ( 2) + ( 2) = 0 2 + 2 = 0 2 + 2 (2) = 0 2 + 4 = 0 2 = 4 = 4 2 We need to find when = 4 cm = 4 4 4 2 = 4 16 2 2 + 2 = 25 Putting = (4)2 + 2 = 25 2 = 25 16 2 = 9 = 3 Now we need to find at y = 3 = 16 2 3 = 8 3 Since length of OB is in cm & time is in sec So, = 8 3 = 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.