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)

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.

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