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 v\:* {behavior:url(#default#VML);} o\:* {behavior:url(#default#VML);} w\:* {behavior:url(#default#VML);} .shape {behavior:url(#default#VML);} Normal 0 false false false EN-US X-NONE X-NONE MicrosoftInternetExplorer4 /* Style Definitions */ table.MsoNormalTable {mso-style-name:"Table Normal"; mso-tstyle-rowband-size:0; mso-tstyle-colband-size:0; mso-style-noshow:yes; mso-style-priority:99; mso-style-qformat:yes; mso-style-parent:""; mso-padding-alt:0in 5.4pt 0in 5.4pt; mso-para-margin-top:0in; mso-para-margin-right:0in; mso-para-margin-bottom:10.0pt; mso-para-margin-left:0in; line-height:115%; mso-pagination:widow-orphan; font-size:11.0pt; font-family:"Calibri","sans-serif"; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:"Times New Roman"; mso-fareast-theme-font:minor-fareast; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin;}

Grace_S · Feb 28, 2011 1:32 PM

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

comingsoon · Feb 28, 2011 1:44 PM

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

terobaaje · Feb 28, 2011 1:48 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!

Grace_S · Feb 28, 2011 1:56 PM

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

comingsoon · Feb 28, 2011 2:02 PM

@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?

comingsoon · Feb 28, 2011 2:08 PM

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

Countryboy · Mar 1, 2011 8:27 AM

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

Grace_s · Mar 2, 2011 12:35 PM

Is this correct ?? 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 -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

ANS · Mar 2, 2011 2:17 PM