Misc 3 - Chapter 6 Class 12 Application of Derivatives

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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 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﷯= 3cm/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﷯ × BC × AD A = 1﷮2﷯ × b × ﷮𝑥2− 𝑏﷮2﷯﷯﷮2﷯﷯ A = 1﷮2﷯ × b × ﷮𝑥2− 𝑏﷮2﷯﷮4﷯﷯ We need 𝑑𝐴﷮𝑑𝑡﷯ Diff w.r.t t 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﷯= 6 𝑏﷮2﷯﷮4 ﷮ 𝑏﷮2﷯ − 𝑏﷮2﷯﷮4﷯﷯﷯= 6 𝑏﷮2﷯﷮4 ﷮ 4 𝑏﷮2﷯ − 𝑏﷮2﷯﷮4﷯﷯﷯= 6 𝑏﷮2﷯﷮4 ﷮ 3 𝑏﷮2﷯﷮4﷯﷯﷯= 6 𝑏﷮2﷯﷮ 4 ﷮3﷯ 𝑏﷮2﷯﷯= 6 𝑏﷮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