Example 41 - An Apache helicopter of enemy is flying along

Example 41 - Chapter 6 Class 12 Application of Derivatives - Part 2
Example 41 - Chapter 6 Class 12 Application of Derivatives - Part 3
Example 41 - Chapter 6 Class 12 Application of Derivatives - Part 4
Example 41 - Chapter 6 Class 12 Application of Derivatives - Part 5
Example 41 - Chapter 6 Class 12 Application of Derivatives - Part 6

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

Transcript

Example 41 An Apache helicopter of enemy is flying along the curve given by 𝑦= π‘₯^2 + 7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance.Given curve y = x2 + 7 Let (π‘₯,𝑦) be any point on parabola 𝑦=π‘₯2+7 Let D be required Distance between (π‘₯,𝑦) & (3 , 7) D = √((πŸ‘βˆ’π’™)^𝟐+(πŸ• βˆ’π’š)^𝟐 ) = √(9+π‘₯^2βˆ’6π‘₯+49+𝑦^2βˆ’14𝑦) = √(π‘₯^2+𝑦^2βˆ’6π‘₯βˆ’14𝑦+58) Since point (π‘₯ , 𝑦) is on the parabola 𝑦=π‘₯2+7 (𝒙 , π’š) will satisfy the equation of parabola Putting π‘₯ and 𝑦 in equation π’š=𝒙^𝟐+πŸ• Putting value of 𝑦=π‘₯^2+7 D = √(π‘₯^2+𝑦^2βˆ’6π‘₯βˆ’14𝑦+58) D = √(π‘₯^2+γ€–(π‘₯^2+7)γ€—^2 βˆ’ 6π‘₯βˆ’14(π‘₯^2+7)+58) D = √(π‘₯^2+π‘₯^4+49+14π‘₯^2 βˆ’ 6π‘₯βˆ’14π‘₯^2βˆ’98+58) D = √(𝒙^πŸ’+𝒙^𝟐 βˆ’πŸ”π’™+πŸ—) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = 𝒙^πŸ’+𝒙^𝟐 βˆ’πŸ”π’™+πŸ— Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =π‘₯^4+π‘₯^2 βˆ’6π‘₯+9 Differentiating w.r.t. h Z’ = 𝑑(π‘₯^4 + π‘₯^2 βˆ’ 6π‘₯ + 9)/π‘‘β„Ž Z’ = 4π‘₯^3+2π‘₯ βˆ’6 Putting Z’ = 0 4π‘₯^3+2π‘₯ βˆ’6=0 Factorizing Z’ Z’(1) = 4(1)3 + 2(1) βˆ’ 6 = 4 + 2 βˆ’ 6 = 0 Hence, (x – 1) is a factor of 4x3 βˆ’ 2x βˆ’ 6 Thus, 4π‘₯^3+2π‘₯ βˆ’6=0 (π‘₯βˆ’1)(4π‘₯^2+4π‘₯+6)=0 2x2 + 2x + 3 = 0 x = (βˆ’2 Β± √(4 βˆ’ 4(2)(3)))/4 = (βˆ’2 Β± √(βˆ’πŸπŸŽ))/4 This is not possible as there are no real roots. Checking sign of 𝒁^β€²β€² " " 𝑑𝑍/𝑑π‘₯=4π‘₯^3+2π‘₯ βˆ’6 Differentiating again w.r.t x (𝑑^2 𝑍)/(𝑑π‘₯^2 )=4 Γ— 3π‘₯^2+2 (𝑑^2 𝑍)/(𝑑π‘₯^2 )=12π‘₯^2+2 Since 𝐙^β€²β€² > 0 for x = 1 ∴ Z is minimum when x = 1 Thus, D is Minimum at x = 1 Finding Minimum value of D D = √(𝒙^πŸ’+𝒙^𝟐 βˆ’πŸ”π’™+πŸ—) Putting x = 1 D = √(1^4+1^2βˆ’6(1)+9) D = βˆšπŸ“ Hence, shortest distance isβˆšπŸ“

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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.