# Misc 3 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 3 The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm per second. How fast is the area decreasing when the two equal sides are equal to the base ? Let x be the equal sides isosceles triangle with fixed base b. i.e. AB = AC = 𝑥 & BC = b Given that side of triangle decreasing when 𝑥 = b i.e. 𝑑𝑥𝑑𝑡= 3cm/sec. We need to find how fast area is decreasing when 𝑥 = b i.e. 𝑑𝐴𝑑𝑡 when 𝑥 = b Finding Area Draw a perpendicular AD to BC ⇒ i.e. AD ⊥ BC In isosceles triangle, perpendicular from vertex to the side bisects the side i.e. D is the mid point of BC BD = DC ∴ BD = DC = 𝑏2 In ∆ 𝐴𝐷𝐵 Using Pythagoras theorem 𝐴𝐵2= 𝐴𝐷2+ 𝐵𝐷2 𝑥2= 𝐴𝐷2+ 𝑏22 𝑥2 – 𝑏22= 𝐴𝐷2 𝐴𝐷2 = 𝑥2 – 𝑏22 𝐴𝐷= 𝑥2− 𝑏22 We know that Area of isosceles triangle = 12 × 𝑏𝑎𝑠𝑒× ℎ𝑒𝑖𝑔ℎ𝑡 A = 12 × BC × AD A = 12 × b × 𝑥2− 𝑏22 A = 12 × b × 𝑥2− 𝑏24 We need 𝑑𝐴𝑑𝑡 Diff w.r.t t 𝑑𝐴𝑑𝑡= 12𝑏 . 𝑑 𝑥2 − 𝑏24𝑑𝑡 𝑑𝐴𝑑𝑡= 12𝑏 12 𝑥2 − 𝑏24 × 𝑑 𝑥2 − 𝑏24𝑑𝑡 𝑑𝐴𝑑𝑡= 12𝑏 12 𝑥2 − 𝑏24 × 𝑑 𝑥2𝑑𝑡−0 𝑑𝐴𝑑𝑡= 12𝑏 12 𝑥2 − 𝑏24 × 𝑑 𝑥2𝑑𝑥 × 𝑑𝑥𝑑𝑡 𝑑𝐴𝑑𝑡= 12𝑏 12 𝑥2 − 𝑏24 ×2𝑥 × 𝑑𝑥𝑑𝑡 𝑑𝐴𝑑𝑡= 12𝑏 12 𝑥2 − 𝑏24 ×2𝑥 ×3 𝑑𝐴𝑑𝑡= 𝑏4 𝑥2 − 𝑏24 × 6𝑥 Finding 𝑑𝐴𝑑𝑡 At 𝑥 = b 𝑑𝐴𝑑𝑡𝑥 = 𝑏= 6 𝑏24 𝑏2 − 𝑏24= 6 𝑏24 4 𝑏2 − 𝑏24= 6 𝑏24 3 𝑏24= 6 𝑏2 4 3 𝑏2= 6 𝑏22 3 𝑏 = 3𝑏 3=𝑏 3 Since dimension of area is cm2 s ⇒ 𝑑𝐴𝑑𝑡 at 𝑥 = b is 𝒃 𝟑 cm2/s

Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.