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Question 26 A kite is flying at a height of 3 metres and 5 metres of string is out. If the kite is moving away horizontally at the rate of 200" " cm/s, find the rate at which the string is being released.Let AB be the string of kite & OA be the height of kite & OA be the ground. Given Height of kite is 5 m OA = 3 cm Since kite is moving horizontally, Height is fixed, And string length and horizontal distance is variable Let OB = 𝑥 cm & OB = 𝑦 cm Let OB = 𝑥 cm & AB = 𝑦 cm Given that kite is moving away horizontally at the rate of 200" " cm/s, i.e. 𝒅𝒙/𝒅𝒕 = 200 cm/sec We need to calculate rate at which the string is being released when 5 metres of string is out i.e. We need to calculate 𝒅𝒚/𝒅𝒕 when 𝒚 = 5 cm. Since ∆AOB is a right angled triangle Using Pythagoras theorem (AB)2 = (OB)2 + (OA)2 𝑦2= 𝑥2+3^2 𝒚𝟐= 𝒙𝟐+𝟗 Now, Differentiating w.r.t time (𝑑(𝑦2))/𝑑𝑡 =(𝑑(𝑥2))/𝑑𝑡+(𝑑(9))/𝑑𝑡 (𝑑 (𝑦2))/𝑑𝑡 = (𝑑(𝑥2))/𝑑𝑡+0 (𝑑(𝑦2))/𝑑𝑡 × 𝑑𝑦/𝑑𝑦 = (𝑑(𝑥2))/𝑑𝑡 × 𝑑𝑥/𝑑𝑥 (𝑑(𝑦2))/𝑑𝑦 × 𝑑𝑦/𝑑𝑡 = (𝑑(𝑥2))/𝑑𝑥 × 𝑑𝑥/𝑑𝑡 2𝑦 × 𝒅𝒚/𝒅𝒕 = 2𝑥 × 𝑑𝑥/𝑑𝑡 𝒚 × 𝒅𝒚/𝒅𝒕 = 𝒙 × 𝒅𝒙/𝒅𝒕 Now, 𝒅𝒙/𝒅𝒕 = 200, y = 5, but we do not know x Finding x when y = 5 From (1) 𝑦2= 𝑥2+9 52= 𝑥2+9 25 = 𝑥2+9 25 – 9 = 𝑥2 16 = 𝑥2 𝑥2=16 𝑥2=4^2 𝒙=𝟒 Putting x = 4, 𝒅𝒙/𝒅𝒕 = 200, y = 5 in equation (2) 𝑦 × 𝑑𝑦/𝑑𝑡 = 𝑥 × 𝑑𝑥/𝑑𝑡 5 × 𝒅𝒚/𝒅𝒕 = 4 × 200 𝑑𝑦/𝑑𝑡 = 4 × 200 × 1/5 𝑑𝑦/𝑑𝑡 = 4 × 40 𝒅𝒚/𝒅𝒕 = 160 Thus, rate at which the string is being released is 160 cm/s

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.