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1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Miscellaneous

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. 𝑑𝑥/𝑑𝑡= 3 cm/sec. We need to find how fast area is decreasing when 𝑥 = b i.e. 𝑑𝐴/𝑑𝑡 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 = 𝑏/2 In ∆ 𝐴𝐷𝐵 Using Pythagoras theorem (𝐴𝐵)^2=(𝐴𝐷)^2+(𝐵𝐷)^2 (𝑥)^2=(𝐴𝐷)^2+ (𝑏/2)^2 𝑥2 – (𝑏/2)^2=(𝐴𝐷)^2 (𝐴𝐷)^2 = 𝑥2 – (𝑏/2)^2 𝐴𝐷=√(𝑥2−(𝑏/2)^2 ) We know that Area of isosceles triangle = 1/2 × Base × Height A = 1/2 × b × √(𝑥2−(𝑏/2)^2 ) A = 1/2 × b × /2 √(𝑥2−𝑏^2/4) We need 𝑑𝐴/𝑑𝑡 Diff w.r.t.x 𝑑𝐴/𝑑𝑡= 1/2 𝑏 . 𝑑(√(𝑥^2 − 𝑏^2/4))/𝑑𝑡 𝑑𝐴/𝑑𝑡= 1/2 𝑏 [1/(2√(𝑥2 − 𝑏^2/4)) × 𝑑(𝑥^2 − 𝑏^2/4)/𝑑𝑡] 𝑑𝐴/𝑑𝑡= 1/2 𝑏 [1/(2√(𝑥2 − 𝑏^2/4)) ×(𝑑(𝑥2)/𝑑𝑡−0)] 𝑑𝐴/𝑑𝑡= 1/2 𝑏 [1/(2√(𝑥2 − 𝑏^2/4)) ×(𝑑(𝑥2)/𝑑𝑥 × 𝑑𝑥/𝑑𝑡)] 𝑑𝐴/𝑑𝑡= 1/2 𝑏 [1/(2√(𝑥2 − 𝑏^2/4)) × 2𝑥 × 𝑑𝑥/𝑑𝑡] 𝑑𝐴/𝑑𝑡= 1/2 𝑏 [1/(2√(𝑥2 − 𝑏^2/4)) × 2𝑥 × 3] 𝑑𝐴/𝑑𝑡= 𝑏/(4√(𝑥2 − 𝑏^2/4)) × 6𝑥 Finding 𝑑𝐴/𝑑𝑡 At 𝑥 = b ├ 𝑑𝐴/𝑑𝑡┤|_(𝑥 = 𝑏)=(6𝑏^2)/(4√(𝑏^2 − 𝑏^2/4))= (6𝑏^2)/(4√((4𝑏^2 − 𝑏^2)/4))= (6𝑏^2)/(4√((3𝑏^2)/4))= (6𝑏^2)/((4√3 𝑏)/2)= (6𝑏^2)/(2√3 𝑏) = 3𝑏/√3=𝑏√3 Since dimension of area is cm2 s ⇒ 𝑑𝐴/𝑑𝑡 at 𝑥 = b is 𝒃√𝟑 cm2/s

Miscellaneous 