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 /s                    (B) 3 cm 2 /s

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

 

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

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Question 1 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) 10√𝟑 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 𝑥^2 Finding rate of change of area Differentiating A w.r.t.x 𝑑𝐴/𝑑𝑡 = √3/4 (𝑥^2 )′ 𝑑𝐴/𝑑𝑡 = √3/4 × (𝑑〖(𝑥〗^2))/𝑑𝑥 × 𝑑𝑥/𝑑𝑡 𝑑𝐴/𝑑𝑡 = √3/4 (2𝑥) 𝑑𝑥/𝑑𝑡 𝒅𝑨/𝒅𝒕 = (√𝟑 𝒙)/𝟐 𝒅𝒙/𝒅𝒕 Putting 𝒅𝒙/𝒅𝒕 = 2, from equation (1) 𝑑𝐴/𝑑𝑡 = (√3 𝑥)/2 × 2 𝑑𝐴/𝑑𝑡 = √𝟑 𝒙 Since, we have to find rate of change of area when side is 10 cm ∴ Putting 𝒙 = 10 cm in 𝑑𝐴/𝑑𝑡 𝒅𝑨/𝒅𝒕 = 10 √𝟑 cm2/sec Hence, area increases at the rate of 10 √𝟑 cm2/sec So, the correct answer is (C) 𝒅𝑨/𝒅𝒕 = (√𝟑 𝒙)/𝟐 𝒅𝒙/𝒅𝒕 Putting 𝒅𝒙/𝒅𝒕 = 2, from equation (1) 𝑑𝐴/𝑑𝑡 = (√3 𝑥)/2 × 2 𝑑𝐴/𝑑𝑡 = √𝟑 𝒙 Since, we have to find rate of change of area when side is 10 cm ∴ Putting 𝒙 = 10 cm in 𝑑𝐴/𝑑𝑡 𝒅𝑨/𝒅𝒕 = 10 √𝟑 cm2/sec Hence, area increases at the rate of 10 √𝟑 cm2/sec So, the correct answer is (C)

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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.