|
[VIEWED 7802
TIMES]
|
SAVE! for ease of future access.
|
|
|
|
comingsoon
Please log in to subscribe to comingsoon's postings.
Posted on 02-28-11 12:21
PM
Reply
[Subscribe]
|
Login in to Rate this Post:
0 
? 
|
| |
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 4/3
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
|
| |
|
|
|
|
Grace_S
Please log in to subscribe to Grace_S's postings.
Posted on 02-28-11 1:32
PM [Snapshot: 83]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
|
| |
|
|
comingsoon
Please log in to subscribe to comingsoon's postings.
Posted on 02-28-11 1:44
PM [Snapshot: 117]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
ok thanks actually I have not that error in my paper (typing error)
so that is right equation ?
|
| |
|
|
terobaaje
Please log in to subscribe to terobaaje's postings.
Posted on 02-28-11 1:48
PM [Snapshot: 130]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
Please log in to subscribe to Grace_S's postings.
Posted on 02-28-11 1:56
PM [Snapshot: 144]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
Please log in to subscribe to comingsoon's postings.
Posted on 02-28-11 2:02
PM [Snapshot: 163]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
yas , thats i did. But the ans seems like a jokker. 
|
| |
|
|
comingsoon
Please log in to subscribe to comingsoon's postings.
Posted on 02-28-11 2:08
PM [Snapshot: 192]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
@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
Please log in to subscribe to Countryboy's postings.
Posted on 03-01-11 8:27
AM [Snapshot: 499]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
@terobaaje :: that is soo gay daju, made my morning though
|
| |
|
|
Grace_S
Please log in to subscribe to Grace_S's postings.
Posted on 03-02-11 12:35
PM [Snapshot: 712]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
Please log in to subscribe to ANS's postings.
Posted on 03-02-11 2:17
PM [Snapshot: 784]
Reply
[Subscribe]
|
| Login in to Rate this Post:
0
?
|
| |
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
|
| |
.png)
