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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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question 7 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 = π‘₯ cm, OB = 𝑦 cm & βˆ π΄π΅π‘‚= πœƒ Given that Top of the ladder slides downwards at the rate of 10 cm/sec i.e. 𝒅𝒙/𝒅𝒕 = βˆ’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 π’…πœƒ" " /𝒅𝒕 when π’š = 2 m or π’š=𝟐𝟎𝟎 π’„π’Ž. (Negative sign shows decreasing) Now, We need to find π’…πœ½/𝒅𝒕 , so for that we will differentiate π‘₯ w.r.t. 𝑑, since value of 𝒅𝒙/𝒅𝒕 is given. sin⁑〖 πœƒ=(π‘Άπ’‘π’‘π’π’”π’Šπ’•π’”π’† π‘Ίπ’Šπ’…π’†)/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† " " γ€— sin⁑〖 πœƒ=𝑂𝐴/𝐴𝐡 " " γ€— sin⁑〖 πœƒ=π‘₯/500 " " γ€— 𝒙=πŸ“πŸŽπŸŽ 𝐬𝐒𝐧⁑𝜽 cos⁑〖 πœƒ=(𝑨𝒅𝒋𝒂𝒄𝒆𝒏𝒕 π‘Ίπ’Šπ’…π’†)/π‘―π’šπ’‘π’π’•π’†π’π’–π’”π’† " " γ€— cos⁑〖 πœƒ=𝑂𝐡/𝐴𝐡 " " γ€— cos⁑〖 πœƒ=𝑦/500 " " γ€— π’š=πŸ“πŸŽπŸŽ π’„π’π’”β‘πœ½ Differentiating 𝒙 w.r.t. 𝒕 π‘₯=500 sinβ‘πœƒ 𝒅𝒙/𝒅𝒕 = 500 Γ— (𝑑(sinβ‘πœƒ))/𝑑𝑑 𝑑π‘₯/𝑑𝑑 = 500 Γ— (𝒅(π’”π’Šπ’β‘πœ½))/π’…πœ½ Γ— π‘‘πœƒ/𝑑𝑑 𝒅𝒙/𝒅𝒕 = 500 Γ— π‘π‘œπ‘ β‘πœƒ Γ— π‘‘πœƒ/𝑑𝑑 Putting 𝒅𝒙/𝒅𝒕 = -10 cm/sec βˆ’10 = 500 cosβ‘πœƒ Γ— π‘‘πœƒ/𝑑𝑑 (βˆ’10)/(500 π‘π‘œπ‘ β‘πœƒ )=π‘‘πœƒ/𝑑𝑑 π‘‘πœƒ/𝑑𝑑= (βˆ’10)/(500 cosβ‘πœƒ ) π‘‘πœƒ/𝑑𝑑= (βˆ’1)/(50 cosβ‘πœƒ ) Putting 𝐜𝐨𝐬⁑𝜽=π’š/πŸ“πŸŽπŸŽ from equation (2) π’…πœ½/𝒅𝒕= (βˆ’1)/(50 Γ— π’š/πŸ“πŸŽπŸŽ) π’…πœ½/𝒅𝒕 = (βˆ’1)/( 𝑦/10) π’…πœ½/𝒅𝒕= (βˆ’πŸπŸŽ)/( π’š) Putting π’š=𝟐𝟎𝟎 β”œ π’…πœ½/𝒅𝒕─|_(π’š =𝟐𝟎𝟎) " "=(βˆ’10)/200 β”œ π‘‘πœƒ/𝑑𝑑─|_(𝑦 =200) " "=(βˆ’πŸ)/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
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.