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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


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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.