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The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:

(A)10 cm 2 /sΒ  Β  Β  Β  Β  Β  Β  Β  Β  Β  (B) 3 cm 2 /s

(C) 10 √ 3 cm 2 /s               (D) 10/3 cm 2 /s

Β 

This question is similar to Ex 6.1, 1 - Chapter 6 Class 12 - Application of Derivatives

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Transcript

Question 6 The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is: 10 cm2/s (B) 3 cm2/s (C) 10βˆšπŸ‘ cm2/s (D) 10/3 cm2/s Let Area of equilateral triangle = A cm2 & let Side = 𝒙 cm Given that Sides of equilateral triangle are increasing at the rate of 2 cm/sec ∴ 𝒅𝒙/𝒅𝒕 = 2 We need to find rate of change of area w.r.t. side i.e., we need to find 𝒅𝑨/𝒅𝒕 We know that Area of equilateral triangle = A = √3/4 π‘₯^2 Finding rate of change of area Differentiating A w.r.t.x 𝑑𝐴/𝑑𝑑 = √3/4 (π‘₯^2 )β€² 𝑑𝐴/𝑑𝑑 = √3/4 Γ— (𝑑〖(π‘₯γ€—^2))/𝑑π‘₯ Γ— 𝑑π‘₯/𝑑𝑑 𝑑𝐴/𝑑𝑑 = √3/4 (2π‘₯) 𝑑π‘₯/𝑑𝑑 𝒅𝑨/𝒅𝒕 = (βˆšπŸ‘ 𝒙)/𝟐 𝒅𝒙/𝒅𝒕 Putting 𝒅𝒙/𝒅𝒕 = 2, from equation (1) 𝑑𝐴/𝑑𝑑 = (√3 π‘₯)/2 Γ— 2 𝑑𝐴/𝑑𝑑 = βˆšπŸ‘ 𝒙 Since, we have to find rate of change of area when side is 10 cm ∴ Putting 𝒙 = 10 cm in 𝑑𝐴/𝑑𝑑 𝒅𝑨/𝒅𝒕 = 10 βˆšπŸ‘ cm2/sec Hence, area increases at the rate of 10 βˆšπŸ‘ cm2/sec So, the correct answer is (C) 𝒅𝑨/𝒅𝒕 = (βˆšπŸ‘ 𝒙)/𝟐 𝒅𝒙/𝒅𝒕 Putting 𝒅𝒙/𝒅𝒕 = 2, from equation (1) 𝑑𝐴/𝑑𝑑 = (√3 π‘₯)/2 Γ— 2 𝑑𝐴/𝑑𝑑 = βˆšπŸ‘ 𝒙 Since, we have to find rate of change of area when side is 10 cm ∴ Putting 𝒙 = 10 cm in 𝑑𝐴/𝑑𝑑 𝒅𝑨/𝒅𝒕 = 10 βˆšπŸ‘ cm2/sec Hence, area increases at the rate of 10 βˆšπŸ‘ cm2/sec So, the correct answer is (C)

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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.