where is my mistake ?

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1) A normal line to the graph of a function f at the point (x,f(x)) is defined to be the line perpendicular to the tangent line at that point . Find the equation of the normal line to the curve y= ³√x² at the point where x=3
Solution:
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* 3 ^1/3) = 2/3 ⁴^/3The line is perpendicular is m=-3 ^4/3/2
finally the equation of the normal line is
y-3^2/3/2=3 ^4/3/2(x-3)
y=(-3 ^4/3/2 )x +3 ^7/2 +3 ^2/3

Grace_S

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A simple mathematical error is what I see (happens all the time). See the last step of your calculation. You missed the denominator 2 in the second term.

y-3^2/3/2=3 ^4/3/2(x-3)
y=(-3 ^4/3/2 )x +3 ^7/2/2 +3 ^2/3

comingsoon

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ok thanks actually I have not that error in my paper (typing error)
so that is right equation ?

terobaaje

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math to me is like "kalo ackcheer bhaisi barabar" lol, wow...das ma das jode bis hunchha timro ra mero oth jde kiss hunchh...thats my math right there lol

Grace_S

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1) Find the slope of the tangent (lets say m)
2) The slope of the line perpendicular to the tangent would be -1/m
3) Use the slope in the equation: y-y1 = m(x-x1) at the given points.

Good luck!

comingsoon

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yas , thats i did. But the ans seems like a jokker.

comingsoon

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@Grace_S ,
can you give me rough idea for this question,
when f is defined by f(x)=√x , find a so that f'(a) is three times value of f'(2)
I find f'(2)=1/(2√2)
so , f'(a)=3/(2√2)
but I don't know how to find a ? do you have any idea on it?

Countryboy

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@terobaaje :: that is soo gay daju, made my morning though

Grace_S

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comingsoon, did you solve for an 'a'? These kinds of problems are like solving mysteries; the more you do, the better is the fun!

ANS

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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 ^4/3/2 )x +3 ^2/3 +3 ^7/³/2