Last updated at May 6, 2021 by Teachoo

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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 of isosceles triangle i.e. AB = AC = π And, Base = BC = b Given that equal side of Triangle decreasing at 3 cm per second i.e. ππ₯/ππ‘= β 3 cm/sec. We need to find how fast is the area decreasing when the two equal sides are equal to the base i.e. π π¨/π π when π = b Finding Area Letβs draw perpendicular AD to BC i.e. AD β₯ BC In Isosceles triangle, perpendicular from vertex to the side bisects the side i.e. D is the mid point of BC Thus, we can write BD = DC = π/π In β ADB Using Pythagoras theorem (π΄π΅)^2=(π΄π·)^2+(π΅π·)^2 (π₯)^2=(π΄π·)^2+ (π/2)^2 π₯2 β (π/2)^2=(π΄π·)^2 (π΄π·)^2 = π₯2 β (π/2)^2 π¨π«=β(ππβ(π/π)^π ) We know that Area of isosceles triangle = 1/2 Γ Base Γ Height A = 1/2 Γ b Γ β(π₯2β(π/2)^2 ) A = π/π Γ b Γ β(ππβπ^π/π) Finding π π¨/π π Differentiating w.r.t. t ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ‘ ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ‘ Γππ₯/ππ₯ ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ₯ Γπ π/π π ππ΄/ππ‘= 1/2 π . π(β(π₯^2 β π^2/4))/ππ₯ Γ π ππ΄/ππ‘= 1/2 π [1/(2β(π₯2 β π^2/4)) Γ π(π₯^2 β π^2/4)/ππ₯]Γ 3" " ππ΄/ππ‘= 1/2 π [1/(2β(π₯2 β π^2/4)) Γ(2π₯β0)]Γ 3" " π π¨/π π= πππ/(πβ(ππ β π^π/π)) Finding π π¨/π π at π = b β ππ΄/ππ‘β€|_(π₯ = π)=(3π^2)/(2β(π^2 β π^2/4)) = (6π^2)/(4β((4π^2 β π^2)/4))= (6π^2)/(4β((3π^2)/4))= (6π^2)/((4β3 π)/2)= (6π^2)/(2β3 π)= 3π/β3 =πβ3 Since dimension of area is cm2 and time is seconds β΄ ππ΄/ππ‘ = πβπ cm2/s

Miscellaneous

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.