1. Class 12
2. Important Question for exams Class 12
3. Chapter 6 Class 12 Application of Derivatives

Transcript

Example 35 Find the shortest distance of the point (0, c) from the parabola ð¦=ð¥2, where 0 â¤ c â¤ 5. Let ï·â ,ðï·¯ be any point on the parabola ð¦=ð¥2 Let D be the Required Distance between ï·â , ðï·¯ & ï·ð , ðï·¯ D = ï·ï·®ï·ï·0ââï·¯ï·®2ï·¯+ï·ï·ð âðï·¯ï·®2ï·¯ï·¯ D = ï·ï·®ï·ï·ââï·¯ï·®2ï·¯+ï·ï·ð âðï·¯ï·®2ï·¯ï·¯ D = ï·ï·®ï·âï·®2ï·¯+ï·ï·ð âðï·¯ï·®2ï·¯ï·¯ Since point ï·â , ðï·¯ is on the parabola ð¦=ð¥2 ï·â , ðï·¯ will satisfy the equation of parabola Putting ð¥=â ð¦=ð in equation â ð=ï·âï·®2ï·¯ Putting value of ð=ï·âï·®2ï·¯ in (1) D = ï·ï·®ð+ï·ï·ðâðï·¯ï·®2ï·¯ï·¯ We need to minimize distance D & find point ï·â , ðï·¯ Finding Dâï·ðï·¯ D = ï·ï·®ð+ï·ï·ðâðï·¯ï·®2ï·¯ï·¯ Diff w.r.t k Dâ ï·ðï·¯=ï·1ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Ã ï·ðï·ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·®ððï·¯ = ï·1ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Ã ï·1+2ï·ðâðï·¯.ï·ðï·ð â ðï·¯ï·®ððï·¯ï·¯ = ï·1ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Ã ï·1+2ï·ðâðï·¯.ï·0â1ï·¯ï·¯ = ï·1ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Ã ï·1+2ï·ðâðï·¯.ï·â1ï·¯ï·¯ = ï·1ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Ã ï·1â2ï·ðâðï·¯ï·¯ = ï·1 â 2ï·ð â ðï·¯ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯ Putting Dâ ï·ðï·¯=0 ï·1 â 2ï·ð â ðï·¯ï·®2ï·ï·®ð + ï·ï·ð â ðï·¯ï·®2ï·¯ï·¯ï·¯=0 1 â 2ï·ðâðï·¯=0 2ï·ðâðï·¯=1 c â k = ï·1ï·®2ï·¯ c â ï·1ï·®2ï·¯ = k ð=câï·1ï·®2ï·¯ Now, finding sign of Dâ(k) for values less than & greater than ð=câï·1ï·®2ï·¯ Thus, D is Minimum at ð=c âï·1 ï·®2ï·¯ Now, From (2) ð=ï·âï·®2ï·¯ ï·âï·®2ï·¯ = k ï·âï·®2ï·¯=c âï·1ï·®2ï·¯ Shortest distance is D = ï·ï·®ï·âï·®2ï·¯+ï·ï·ðâðï·¯ï·®2ï·¯ï·¯ Putting ï·âï·®2ï·¯=ðâï·1ï·®2ï·¯ & ð=ðâï·1ï·®2ï·¯ D = ï·ï·®ï·ð âï·1ï·®2ï·¯ï·¯+ï·ï·ðâï·ðâï·1ï·®2ï·¯ï·¯ï·¯ï·®2ï·¯ï·¯ = ï·ï·®ï·ðâï·1ï·®2ï·¯ï·¯+ï·ï·ï·1ï·®2ï·¯ï·¯ï·®2ï·¯ï·¯ = ï·ï·®ï·ð âï·1ï·®2ï·¯ï·¯+ï·1ï·®4ï·¯ï·¯ = ï·ï·®ð âï·1ï·®2ï·¯+ï·1ï·®4ï·¯ï·¯ = ï·ï·®ð+ï·ï·â 2 + 1ï·®4ï·¯ï·¯ï·¯ = ï·ï·®ðâï·1ï·®4ï·¯ï·¯ = ï·ï·®ï·4ð â 1ï·®4ï·¯ï·¯ Hence, shortest distance is ï·ï·®ï·ðð â ðï·®ðï·¯ ï·¯

Chapter 6 Class 12 Application of Derivatives

Class 12
Important Question for exams Class 12