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Differential Calculus

Cool goIITian

Joined:	14 Apr 2008
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21 May 2008 10:48:12 IST

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Calculus

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The point (0,3) is nearest to the curve x^2=2y at

ans=(2,2)

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ashish banga

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21 May 2008 10:53:17 IST

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LET A POINT ON THE CURVE ( 2T, 2T^2)
NOW DISTANCE OF THE POINT FROM ( 0,3 )

D^2 = 4T^2 + ( 2T^2 - 3)^2
NOW IT IS THE NEAREST POINT , SO DIFFERENTIATE IT WITH RESPECT TO T AND PUT IT 0 , TO GET T

U WILL GET THE DESIRED POINT

Nikhil Krishna

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21 May 2008 14:09:35 IST

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let the point be P $({x}_{1}, {y}_{1})$

distance of the point from (0,3) is

${({{x}_{1}}^{2} + {{y}_{1}}^{2} - 6{y}_{1} + 9)}^{\frac{1}{2}}$

now this value should be minimum

thus

$f(y) = {({{y}_{1}}^{2} - 4{y}_{1} + 9)}^{\frac{1}{2}}$

because

${{x}_{1}}^{2} = 2{y}_{1}$

now differentiate f(y) w.r.t ' y '

u will get

${f}^{'}(y) = \frac{1}{2}{({{y}_{1}}^{2} - 4{y}_{1} + 9)}^{\frac{-1}{2}}(2{y}_{1} - 4)$

Equate f^'(y) = 0

y₁ = 2 but the other quadratic eqn has imaginary roots

hence

P is (2,2).

Manasi

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Joined: 1 Dec 2006

Posts: 2108

21 May 2008 14:29:09 IST

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well, above method is absolutely right... but i am telling u onemore method..

write the parametric equation of the normal of the given curve....here its a parabola, so write the equation of the parabola at the point (t, t²/2) ..

now we knw, shortest distance between two points is the perpndicular distance between thm..that is the given point should lie on the normal to the curve at the required point...

thus the point should satisfy, the equation of the normal... frm here, u'll get an equation in t, which upon solving will give u the value of t and hence the point !!!

any doubt in this method, do knock back :)

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