Posted by: ANS March 2, 2011

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**0****?**
Is this correct ??

the point is (3, 3 ^{2/3}).................. [³√3²=3 ^{2/3}]

Now for slope m,

dy/dx=d(x^{2/3})/dx= 2/3 x^{-1/3}

at poin x=3

2* 3 ^{-1/3}/3= 2*(3 ^{-1/3}* 3 ^{-1}) = 2*3 ^{-4}^{/3}

The line is perpendicular is m=-2^{-}^{1}*3 ^{4/3}

finally the equation of the normal line is

y-3^{2/3}=-2^{-}^{1}*3 ^{4/3}(x-3)

y = -2^{-}^{1}*3 ^{4/3} x + 2^{-}^{1}*3^{2/3 }**+**2^{-}^{1}*******3 ^{7/}^{3}**

OR

y=(-3

OR

y=(-3

^{4/3}/2 )x +**3**

^{2/3}**+**

**3**

^{7/}^{3}/**2**