Ex 6.1

Chapter 6 Class 12 Application of Derivatives
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 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/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 = 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๐ + ๐๐ = 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/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 = (โ๐๐)/๐ Putting ๐ฆ = 4 cm โ ๐๐ฅ/๐๐กโค|_(๐ฆ = 4) =(โ2 ร 4)/๐ฅ โ ๐๐/๐๐โค|_(๐ = ๐) =(โ๐)/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=(โ8)/๐ฅ ๐๐ฅ/๐๐ก = (โ8)/3 Since x is in cm & time is in sec โด ๐๐ฅ/๐๐ก = (โ๐)/๐ cm/sec Hence, height of ladder on the wall is decreasing at rate of ๐/๐ cm/sec