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Differential Calculus

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13 Mar 2010 14:29:51 IST

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two equations are xy=k, x=sqr(y) ,these curves intersecting each other at right angle ,then pr

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two equations are xy=k, x=sqr(y) ,these curves intersecting each other at right angle ,then prove that 1=8*sqr(k)

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ankitsingh

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13 Mar 2010 14:41:47 IST

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xy=k, x=sqry

=> y=k power1/3 , x=k power2/3

2ydy/dx= 1 =>dy/dx =1/2y

y+xdy/dx=0 => dy/dx= -y/x

since tangent are perpendicular, -y/x X 1/2y = -1

=>x= 1/2

k power 2/3 = 1/2

cubing both side

k power 2 =1/8

=> 8 sqr (k) =1, proved...!

charanjit ghai

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Joined: 23 Nov 2009

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13 Mar 2010 17:20:41 IST

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xy=k n y^2=x. solving them we get that they intersect at x=k raised to 2/3 n y = k raised to 1/3.

since the curves intersect at right angles this implies that the tangent drawn to them at these points to the curves are perpendicular to each other.supposing the equations to b y=mx+cand applying the condition of tangency differently for both the curves we get the equations of tangents respectively as x+k^1/3 y=2k^2/3 n x-2 k^1/3 y+k^2/3=0

applying the condition of tangency m1 x m2 =o, we get the desired result ie 8k^2=1 did u get it?

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