Check sibling questions

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.