# Example 41 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 41 An Apache helicopter of enemy is flying along the curve given by = 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 = 2 1 2+ 2 1 2 d = 3 2+ ( 2 +7) 7 2 d = 3 2+ 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 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 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 = (1) d =

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Chapter 6 Class 12 Application of Derivatives

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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 9 years. He provides courses for Maths and Science at Teachoo.