Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12     1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Examples

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Example 41 An Apache helicopter of enemy is flying along the curve given by 𝑦= 𝑥^2 + 7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance. The curve is given as y = x2 + 7 Let helicopter be at point (x, y) = (x, x2 + 7) Let d be the distance between helicopter soldier at (3, 7) d = √((𝑥2−𝑥1)2+(𝑦2−𝑦1)2) d = √((𝑥−3)2+(〖(𝑥〗^2+7)−7)2) d = √((𝑥−3)2+(𝑥)4) We need to find nearest distance i.e. minimum value of d Let f(x) = d2 f(x) = (x − 3)2 + x4 When f(x) is minimum, d is minimum Finding f’(x) f’(x) = 2(x − 3) + 4x3 = 2x − 6 + 4x3 = 4x3 − 2x − 6 (To make calculation easy) Factorizing f’(x) f’(1) = 4(1)3 − 2(1) − 6 = 4 + 2 − 6 = 0 Hence, (x – 1) is a factor of 4x3 − 2x − 6 Thus, f’(x) = (x − 1) (4x2 + 4x + 6) Hence f’ (x) = 0 gives x − 1 = 0 x = 1 2x2 + 2x + 3 = 0 x = (−2 ± √(4 − 4(2)(3)))/4 x = (−2 ± √(4 − 24))/4 x = (−2 ± √(−20))/4 This is not possible as there are no real roots. Hence there is only one point x = 1 This is either the maxima or minima. Hence we find f”(x) f’’ (x) = (4x3 + 2x − 6)’ f’’(x) = 12x2 + 2 Finding value at x = 1,. f’’(1) = 12(1)2 + 2 = 12 + 2 = 14 Since f’’(x) > 0 ∴ x = 1 is the minima. The value of f(1) is f(1) = (1 − 3)2 + 14 = (−2)2 + 1 = 4 + 1 = 5 Hence, minimum distance between soldier & Helicopter d = √(𝑓(1)) d = √𝟓

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