Examples

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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Example 23 Find the shortest distance of the point (0, c) from the parabola š¦=š„2, where 0 ā¤ c ā¤ 5. Let (ā ,š) be any point on parabola š¦=š„2 Let D be required Distance between (ā , š) & (0 , š) D = ā((0āā)^2+(š āš)^2 ) D = ā((āā)^2+(š āš)^2 ) D = ā(š^š+(š āš)^š ) Also, Since point (ā , š) is on the parabola š¦=š„2 (š , š) will satisfy the equation of parabola Putting š„=ā , š¦=š in equation š=š^š Putting value of š=ā^2 D = ā(ā^2+(š āš)^2 ) D = ā(š+(šāš)^š ) We need to minimize D, but D has a square root Which will be difficult to differentiate Let Z = D2 Z = š+(šāš)^2 Since D is positive, D is minimum if D2 is minimum So, we minimize Z = D2 Differentiating Z Z =š+(šāš)^2 Differentiating w.r.t. k Zā = š(š + (š ā š)^2 )/šš Zā = 1 + 2 (c ā k) Ć (c ā k)ā Zā = 1 + 2 (c ā k) Ć (0 ā 1) Zā = 1 ā 2 (c ā k) Zā = 1 ā 2c ā 2k Putting Zā = 0 1 ā 2c ā 2k = 0 2k = 2c ā 1 k = (šš ā š)/š Now, checking sign of š^ā²ā² " " šš/šš=4šā2š Differentiating again w.r.t k (š^2 š)/(šā^2 ) = 4 ā0 (š^š š)/(šš^š ) = š Since š^ā²ā² > 0 for k = (2š ā 1)/2 ā“ Z is minimum when k = (2š ā 1)/2 Thus, D is Minimum at š=(šš ā š)/š Finding Minimum value of D D = ā(š+(šāš)^2 ) Putting š=(2š ā 1)/2 D = ā(((2š ā 1)/2)+(šā((2š ā 1)/2))^2 ) D = ā(((2š ā 1)/2)+((2š ā 2š ā 1)/2)^2 ) D = ā(((2š ā 1)/2)+((ā1)/2)^2 ) D = ā(((2š ā 1)/2)+1/4) D = ā(šā1/2+1/4) D = ā(šā1/4) D = ā(4š ā 1)/2 Hence, shortest distance is ā(šš ā š)/š