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December 19, 2014

December 19, 2014

Posted by **Elizabeth** on Wednesday, July 18, 2012 at 5:51pm.

f(x)=x^2 (2,1/2)

- Calculus -
**MathMate**, Wednesday, July 18, 2012 at 6:13pmFrom the given point P0(x0,y0)=(2,1/2), we draw a line to join a point on the curve

f(x)=x^2 at P(x,y).

The distance P-P0 will be shortest when P-P0 is perpendicular to the tangent of f(x) at P.

The slope of P-P0 is given by:

m1=(y0-y)/(x0-x)

=(y0-x^2)/(x0-x) since P is on curve f(x)=x^2.

The tangent of the curve f(x) at P is given by

m2=f'(x)=2x

Perpendicularity requires that m1*m2=-1, or

[(y0-x^2)/(x0-x)]*(2x)=-1

which simplifies to a cubic equation

x³-1=0 which factorizes easily to

(x-1)(x²+x+1)=0

The first factor gives x=1, and the second factor results in two complex roots which do not concern us.

Therefore the required point is P(1,1), or P(1,1) is the point on f(x) that is closest to the given point (1, 1/2).

- Calculus -
**Elizabeth**, Wednesday, July 18, 2012 at 9:19pmThanks!

- Calculus :) -
**MathMate**, Wednesday, July 18, 2012 at 9:34pmYou're welcome!

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