where is my mistake ?

Original Post

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(x2/3)/dx= 2/3 x-1/3 at poin x=3 2* 3 -1/3/3= 2/(3* 3 1/3) = 2/3 4/3 The line is perpendicular is m=-3 4/3/2 finally the equation of the normal line is y-32/3/2=3 4/3/2(x-3) y=(-3 4/3/2 )x +3 7/2 +3 2/3

comingsoon

Feb 28, 2011 12:21 PM · 17,808 views

9 Replies

Grace_S Feb 28, 2011 1:32 PM

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-32/3/2=3 4/3/2(x-3)

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

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comingsoon Feb 28, 2011 1:44 PM

ok thanks actually I have not that error in my paper (typing error)

so that is right equation ?

terobaaje Feb 28, 2011 1:48 PM

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Grace_S Feb 28, 2011 1:56 PM

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 Feb 28, 2011 2:02 PM

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

comingsoon Feb 28, 2011 2:08 PM

@Grace_S ,

can you give me rough idea for this question,

when f is defined by f(x)=&radic;x , find a so that f'(a) is three times value of f'(2)

I find f'(2)=1/(2&radic;2)

so , f'(a)=3/(2&radic;2)

but I don't know how to find a ? do you have any idea on it?

Countryboy Mar 1, 2011 8:27 AM

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

Grace_s Mar 2, 2011 12:35 PM

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 Mar 2, 2011 2:17 PM

Is this correct ??

the point is (3, 3 2/3).................. [&sup3;&radic;3&sup2;=3 2/3]

Now for slope m,

dy/dx=d(x2/3)/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-32/3=-2-1*3 4/3(x-3)

y = -2-1*3 4/3 x + 2-1*32/3 +2-1*3 7/3

OR

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