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

Transcript

Example 36 Let AP and BQ be two vertical poles at points A and B, respectively. If AP = 16 m, BQ = 22 m and AB = 20 m, then find the distance of a point R on AB from the point A such that RP2 + RQ2 is minimumGiven AP & BQ be two poles where AP = 16m, BQ = 22m & AB = 20m Let R be a point on AB Let AR = 𝒙 m. RB = AB – AR RB = 20 – 𝒙 Here, ∠ A = 90Β° & ∠ B = 90Β° as they are vertical poles In βˆ†ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βˆ†BRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 – x)2 + 222 We need to find distance of R on AB from point A such that RP2 + RQ2 is minimum i.e. we need to find 𝒙 such that RP2 + RQ2 is minimum In βˆ†ARP , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RP2 = AR2 + AP2 RP2 = x2 + 162 In βˆ†BRQ , Using Pythagoras theorem (Hypotenuse)2 = (Base)2 + (Height)2 RQ2 = RB2 + BQ2 RQ2 = (20 – x)2 + 222 We need to find distance of R on AB from point A such that RP2 + RQ2 is maximum i.e. we need to find π‘₯ such that RP2 + RQ2 is maximum Let S(x) = RP2 + RQ2 From (1) & (2) S(x) = π‘₯^2+(16)^2+(20βˆ’π‘₯)^2+(22)^2 = π‘₯^2+(16)^2+(20βˆ’π‘₯)^2+(22)^2 = π‘₯^2+256+((20)^2+(π‘₯)^2βˆ’2(20)(π‘₯))+484 = π‘₯^2+256+(400+π‘₯^2βˆ’40π‘₯ )+484 = γ€–πŸπ’™γ€—^πŸβˆ’πŸ’πŸŽπ’™+πŸπŸπŸ’πŸŽ Finding S’(𝒙) S’(π‘₯)=𝑑(2π‘₯^2 βˆ’ 40π‘₯ + 1140)/𝑑π‘₯ S’(π‘₯)=4π‘₯βˆ’40+0 S’(π‘₯)=4(π‘₯βˆ’10) Putting S’(𝒙)=𝟎 4(π‘₯βˆ’10)=0 π‘₯βˆ’10=0 𝒙=𝟏𝟎 Finding S’’(π‘₯) S’’(π‘₯)=4 𝑑(π‘₯ βˆ’ 10)/𝑑π‘₯ S’’(π‘₯)=4(1βˆ’0) S’’(π‘₯)=4 Since S’’(π‘₯)>0 at π‘₯=10 ∴ 𝒙=𝟏𝟎 is point of minima Hence, S(π‘₯) is minimum when π‘₯=10 Thus, the distance of R from A on AB is AR = x = 10 π’Ž

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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.