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

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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 = 𝑥 & OB = 𝑦 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. 𝑑𝑥/𝑑𝑡 = 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 𝑑𝑦/𝑑𝑡 when 𝑥 = 4 cm. Wall OB is perpendicular to the ground OA Using Pythagoras theorem (OB)2 + (OA)2 = (AB)2 𝑦2 + 𝑥2 = (5)2 𝑦2 + 𝑥2 = 25 Differentiating w.r.t time (𝑑(𝑦2 + 𝑥2))/𝑑𝑡 = (𝑑(25))/𝑑𝑡 (𝑑 (𝑦2))/𝑑𝑡 + (𝑑(𝑥2))/𝑑𝑡 = 0 (𝑑(𝑦2))/𝑑𝑡 × 𝑑𝑦/𝑑𝑦 + (𝑑(𝑥2))/𝑑𝑡 × 𝑑𝑥/𝑑𝑥 = 0 2𝑦 × 𝑑𝑦/𝑑𝑡 + 2𝑥 × 𝑑𝑥/𝑑𝑡 = 0 2𝑦 × 𝑑𝑦/𝑑𝑥 + 2𝑥 × (2) = 0 2𝑦 𝑑𝑦/𝑑𝑡 + 4𝑥 = 0 2𝑦 𝑑𝑦/𝑑𝑡 = – 4 𝑥 𝑑𝑦/𝑑𝑡 = (− 4𝑥)/2𝑦 ("From (1): " 𝑑𝑦/𝑑𝑥=2𝑚/𝑠𝑒𝑐) We need to find 𝑑𝑦/𝑑𝑡 when 𝑥 = 4 cm ├ 𝑑𝑦/𝑑𝑡┤|_(𝑥 = 4) =(− 4 × 4)/2𝑦 ├ 𝑑𝑦/𝑑𝑡┤|_(𝑥 = 4) =(− 16)/2𝑦 Finding value of y From (2) 𝑥2 + 𝑦2 = 25 Putting 𝑥 =4 (4)2 + 𝑦2 = 25 𝑦2 = 25 – 16 𝑦2 = 9 𝑦 = 3 Now we need to find 𝑑𝑦/𝑑𝑡 at y = 3 𝑑𝑦/𝑑𝑡=(− 16)/2𝑦 𝑑𝑦/𝑑𝑡 = (− 16)/(2 × 3) 𝑑𝑦/𝑑𝑡 = (− 8)/3 Since length of OB is in cm & time is in sec So, 𝑑𝑦/𝑑𝑡 = (− 8 𝑐𝑚)/(3 𝑠𝑒𝑐) 𝑑𝑦/𝑑𝑡 = (−𝟖)/𝟑 cm/sec Hence, the height of the ladder on the wall is decreasing at the rate of 8/3 cm/sec