Question

derivative (too big to handle)

Answer 1

\[<\frac{ 1 }{ \sqrt{5t^2+1} },\frac{ t }{ \sqrt{5t^2+1}}\, \,\frac{ 2t }{ \sqrt{5t^2+1} }>\]

Answer 2

i got \[<\frac{ -5t }{ (5t^2+1)^\frac{ 3 }{ 2 } }, \frac{ 1-5t^2 }{ (5t^2+1)^\frac{ 3 }{ 2 } }, \frac{ 8-5t^2 }{ (5t^2+1)^\frac{ 3 }{ 2 } }>\]

Answer 3

now find the unit vector for thAT

Answer 4

at the very first comment, is it not a unit vector of <1,t,t^2> ?
what is the goal of the stuff? please, post the original question

Answer 5

Find the unit tangent vector T(t) and unit normal vector N(t). Then use the formula to find the curvature.
r(t) = (t, 1/2 t^2, t^2).

Answer 6

ok, unit tangent vector
\[T(t)=\dfrac{<r'(t)>}{|r'(t)|}\\=\dfrac {<1,t,2t>}{\sqrt{1+5t^2}}\]
like what you have,

Answer 7

the curvature
\[K= \dfrac{|r'(t)~~cross ~r"(t)|}{|r'(t)|^3}\]
now, take second derivative of r(t) , just r(t) , (not T(t) ), it is <0,1,2> , then cross r'(t) and r"(t) to get the numerator of K,
quite easy, right?

Answer 8

but we also need to find N(t)

Answer 9

oh, yea, hehehe. forgot

Answer 10

Normal vector =\[\frac{ T'(t) }{ |T'(t)| }\]

Answer 11

plz see the 2nd one

Answer 12

that's what I got N(t)

Answer 13

give me time to take a look at my note ;) hopefully I can help

Answer 14

the second term is not right, it is 1/(1+5t^2)^3/2 http://www.wolframalpha.com/input/?i=derivative+of+%28%28t%2F%28sqrt%281%2B5t^2%29%29

Answer 15

so, the last term is 2/(1+5t^2)^3/2

Answer 16

so far, you have
\[T'(t)=<\frac{ -5t }{ (5t^2+1)^\frac{ 3 }{ 2 } }, \frac{ 1 }{ (5t^2+1)^\frac{ 3 }{ 2 } }, \frac{ 2 }{ (5t^2+1)^\frac{ 3 }{ 2 } }>\]

Answer 17

and magnitude of it is
the first term square = 25t^2/(5t^2+1)^3
the second term square = 1/(5t^2+1)^3
the last term square = 4/(5t^2 +1)^3
add them together = (25t^2 +5 )/(5t^2+1)^3)
factor 5 out from the numerator and cancel out with the denominator, to get 5/(5t^2+1)^2
now, put it under the square to get the magnitude of |T'(t)|= \(\dfrac{\sqrt{5}}{5t^2+1}\)
haaaaaaaaaaaaaaaaa!!!
put everything under the form or N(t)
N(t) = \(\dfrac{T'(t)}{|T'(t)|}\)