Misc 3 - Two equal sides of isosceles triangle, fixed base b

Misc 3 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 3 - Chapter 6 Class 12 Application of Derivatives - Part 3
Misc 3 - Chapter 6 Class 12 Application of Derivatives - Part 4
Misc 3 - Chapter 6 Class 12 Application of Derivatives - Part 5
Misc 3 - Chapter 6 Class 12 Application of Derivatives - Part 6

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

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 of isosceles triangle i.e. AB = AC = 𝒙 And, Base = BC = b Given that equal side of Triangle decreasing at 3 cm per second i.e. 𝑑π‘₯/𝑑𝑑= βˆ’ 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. 𝒅𝑨/𝒅𝒕 when 𝒙 = b Finding Area Let’s draw perpendicular AD to BC i.e. AD βŠ₯ 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 = 𝒃/𝟐 In βˆ† ADB Using Pythagoras theorem (𝐴𝐡)^2=(𝐴𝐷)^2+(𝐡𝐷)^2 (π‘₯)^2=(𝐴𝐷)^2+ (𝑏/2)^2 π‘₯2 – (𝑏/2)^2=(𝐴𝐷)^2 (𝐴𝐷)^2 = π‘₯2 – (𝑏/2)^2 𝑨𝑫=√(π’™πŸβˆ’(𝒃/𝟐)^𝟐 ) We know that Area of isosceles triangle = 1/2 Γ— Base Γ— Height A = 1/2 Γ— b Γ— √(π‘₯2βˆ’(𝑏/2)^2 ) A = 𝟏/𝟐 Γ— b Γ— √(π’™πŸβˆ’π’ƒ^𝟐/πŸ’) Finding 𝒅𝑨/𝒅𝒕 Differentiating w.r.t. t 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑𝑑 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑𝑑 ×𝑑π‘₯/𝑑π‘₯ 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑π‘₯ ×𝒅𝒙/𝒅𝒕 𝑑𝐴/𝑑𝑑= 1/2 𝑏 . 𝑑(√(π‘₯^2 βˆ’ 𝑏^2/4))/𝑑π‘₯ Γ— πŸ‘ 𝑑𝐴/𝑑𝑑= 1/2 𝑏 [1/(2√(π‘₯2 βˆ’ 𝑏^2/4)) Γ— 𝑑(π‘₯^2 βˆ’ 𝑏^2/4)/𝑑π‘₯]Γ— 3" " 𝑑𝐴/𝑑𝑑= 1/2 𝑏 [1/(2√(π‘₯2 βˆ’ 𝑏^2/4)) Γ—(2π‘₯βˆ’0)]Γ— 3" " 𝒅𝑨/𝒅𝒕= πŸ‘π’ƒπ’™/(𝟐√(π’™πŸ βˆ’ 𝒃^𝟐/πŸ’)) Finding 𝒅𝑨/𝒅𝒕 at 𝒙 = b β”œ 𝑑𝐴/𝑑𝑑─|_(π‘₯ = 𝑏)=(3𝑏^2)/(2√(𝑏^2 βˆ’ 𝑏^2/4)) = (6𝑏^2)/(4√((4𝑏^2 βˆ’ 𝑏^2)/4))= (6𝑏^2)/(4√((3𝑏^2)/4))= (6𝑏^2)/((4√3 𝑏)/2)= (6𝑏^2)/(2√3 𝑏)= 3𝑏/√3 =π‘βˆš3 Since dimension of area is cm2 and time is seconds ∴ 𝑑𝐴/𝑑𝑑 = π’ƒβˆšπŸ‘ cm2/s

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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.