A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is:

(A) 1/10 radian/sec            (B) 1/20 radian/sec

(C) 20 radian/sec               (D) 10 radian/sec

 

This question is similar to Ex 6.1, 10 - Chapter 6 Class 12 - Application of Derivatives

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Transcript

Question 2 A ladder, 5 meter long, standing on a horizontal floor, leans against a vertical wall. If the top of the ladder slides downwards at the rate of 10 cm/sec, then the rate at which the angle between the floor and the ladder is decreasing when lower end of ladder is 2 metres from the wall is: (A) 1/10 radian/sec (B) 1/20 radian/sec (C) 20 radian/sec (D) 10 radian/sec Let AB be the ladder & OA be the wall & OB be the ground. Given Length of ladder is 5 m AB = 5 m AB = 500 cm Let OA = 𝑥 cm, OB = 𝑦 cm & ∠𝐴𝐵𝑂= 𝜃 Given that Top of the ladder slides downwards at the rate of 10 cm/sec i.e. 𝒅𝒙/𝒅𝒕 = −10 cm/sec We need to calculate at which the rate the angle between the floor and the ladder decreases when lower end of ladder is 2 metres from the wall. i.e. We need to calculate 𝒅𝜃" " /𝒅𝒕 when 𝒚 = 2 m or 𝒚=𝟐𝟎𝟎 𝒄𝒎. Now, We need to find 𝒅𝜽/𝒅𝒕 , so for that we will differentiate 𝑥 w.r.t. 𝑡, since value of 𝒅𝒙/𝒅𝒕 is given. sin⁡〖 𝜃=(𝑶𝒑𝒑𝒐𝒔𝒊𝒕𝒔𝒆 𝑺𝒊𝒅𝒆)/𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 " " 〗 sin⁡〖 𝜃=𝑂𝐴/𝐴𝐵 " " 〗 sin⁡〖 𝜃=𝑥/500 " " 〗 𝒙=𝟓𝟎𝟎 𝐬𝐢𝐧⁡𝜽 cos⁡〖 𝜃=(𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 𝑺𝒊𝒅𝒆)/𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆 " " 〗 cos⁡〖 𝜃=𝑂𝐵/𝐴𝐵 " " 〗 cos⁡〖 𝜃=𝑦/500 " " 〗 𝒚=𝟓𝟎𝟎 𝒄𝒐𝒔⁡𝜽 Differentiating 𝒙 w.r.t. 𝒕 𝑥=500 sin⁡𝜃 𝒅𝒙/𝒅𝒕 = 500 × (𝑑(sin⁡𝜃))/𝑑𝑡 𝑑𝑥/𝑑𝑡 = 500 × (𝒅(𝒔𝒊𝒏⁡𝜽))/𝒅𝜽 × 𝑑𝜃/𝑑𝑡 𝒅𝒙/𝒅𝒕 = 500 × 𝑐𝑜𝑠⁡𝜃 × 𝑑𝜃/𝑑𝑡 Putting 𝒅𝒙/𝒅𝒕 = -10 cm/sec −10 = 500 cos⁡𝜃 × 𝑑𝜃/𝑑𝑡 (−10)/(500 𝑐𝑜𝑠⁡𝜃 )=𝑑𝜃/𝑑𝑡 𝑑𝜃/𝑑𝑡= (−10)/(500 cos⁡𝜃 ) 𝑑𝜃/𝑑𝑡= (−1)/(50 cos⁡𝜃 ) Putting 𝐜𝐨𝐬⁡𝜽=𝒚/𝟓𝟎𝟎 from equation (2) 𝒅𝜽/𝒅𝒕= (−1)/(50 × 𝒚/𝟓𝟎𝟎) 𝒅𝜽/𝒅𝒕 = (−1)/( 𝑦/10) 𝒅𝜽/𝒅𝒕= (−𝟏𝟎)/( 𝒚) Putting 𝒚=𝟐𝟎𝟎 ├ 𝒅𝜽/𝒅𝒕┤|_(𝒚 =𝟐𝟎𝟎) " "=(−10)/200 ├ 𝑑𝜃/𝑑𝑡┤|_(𝑦 =200) " "=(−𝟏)/𝟐𝟎 Hence, angle between ladder and the ground is decreasing at rate of 𝟏/𝟐𝟎 radian/sec. So, the correct answer is (B)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.