Question

proof

Answer 1

show that the osculating plane r(t)=<t+2,1-t,.5t^2> at every point on the curve is the same plane

Answer 2

@phi

Answer 3

just to clarify, is r(t) the curve ?

Answer 4

yes

Answer 5

\[ r(t)= <t+2, 1-t, \frac{t^2}{2}>? \]

Answer 6

yes

Answer 7

\[r(t)=,2+t,1-t,\frac{t^2}{2}\\r'(t)=<1,-1,t>\]
we need to find the vector equation of the tangent line to the curve at a general point,and show that it depends on only two variables

Answer 8

any point?

Answer 9

they said at every point ,we can chooose \(P(x_0,y_0,z_0)\)

Answer 10

so,T= (x0,y0,z0)+t<1,-1,t>

Answer 11

i think theres a better way to do this,this is not the best way to do it...i was just making an observation

Answer 12

but why u took tangent ?
i mean osculating plane is not the same as tangent

Answer 13

u need to find a plane s.t all point of r(t) lies on it

Answer 14

if there exist such a plane then its called osculating plane...

Answer 15

try to sketch r(t)

Answer 16

the problem is its bretty easy but i gtg sry xD

Answer 17

osculating plane :r'(t)×r''(t)

Answer 18

see http://www.encyclopediaofmath.org/index.php/Osculating_plane

plug into their formula and you get an equation of a plane independent of t

Answer 19

we need to work with unit vectors right?

Answer 20

\[\left| \begin{matrix}x - (t+2) & y-(1-t) & z - \frac{ t^2 }{ 2 } \\ 1 & -1 & t \\ 0 & 0 & 1\end{matrix} \right|=0\]

Answer 21

Is that determinant ?

Answer 22

yes. But I have not studied this, so I am no help in explaining its derivation.

Answer 23

so the idea is finding a perpendiculer vector to r(t)

Answer 24

@ikram002p how do u do that

Answer 25

so, I need to find unit tangent and unit normal and cross them for point (x0,y0,z0)

Answer 26

unit tangent
r'(t) / || r'(t)

Answer 27

I got unit tangnt= \[<\frac{ 1 }{ \sqrt{t^2+2}},\frac{ -1 }{ \sqrt{t^2+2} },\frac{ t }{ \sqrt{t^2+2} }\]

Answer 28

\(\Huge \frac {r'(t)}{r'(t)}=\frac{<1,-1,10t>}{\sqrt {2+100t^2}} \)

Answer 29

that is .5 not 5

Answer 30

sry dint see
so that was T(t)

Answer 31

for normal vector
\(\Huge N(t)= \frac {T'(t)}{T'(t)} \)

Answer 32

okay so N(t) = \[<-t,t,2>\frac{ 1 }{ (t^2+2)^{3/2}}\]

Answer 33

now need to find length of ^

Answer 34

it oonly need work !

Answer 35

is there a shortcut to find length

Answer 36

lets try somthing like this
\(\large ge T(t)\times N(t)=\)
\(\Huge \frac {r'(t)}{||r'(t)||}\times \frac {T'(t)}{||T'(t)||} = \)

Answer 37

to try that i need to get unit normal vector

Answer 38

no wait

Answer 39

im trying to find shortcuts

Answer 40

I got \[<-t,t,2>\frac{ 1 }{ 2t+2}\]

Answer 41

not sure

Answer 42

@ikram002p or can u help me to evaluate the (phi's) determinant

Answer 43

I am getting -x-y+3=0

Answer 44

never mind got it

Answer 45

so what its is ?