# Find the Curvature

• Mar 18th 2011, 10:13 AM
sderosa518
Find the Curvature
Find the curvature of the space curve.
r(t) = -4i + (4+ 2t)j + (t^2 +2)k

* I was just wondering if my solution looks good so far????

r ' (t)= 0i +2j + 2tk

ll r ' (t)ll = sqrt(4 +4t^2) = 2 + 2t^2

= T(t) = r ' (t) / ll r ' (t) ll = 2j + 2tk / 2j + 2t^2K

* After this I use the quationat rule??? 99% sure???
• Mar 18th 2011, 11:21 AM
alexmahone
Quote:

Originally Posted by sderosa518
Find the curvature of the space curve.
r(t) = -4i + (4+ 2t)j + (t^2 +2)k

* I was just wondering if my solution looks good so far????

r ' (t)= 0i +2j + 2tk

ll r ' (t)ll = sqrt(4 +4t^2) = 2 + 2t^2

= T(t) = r ' (t) / ll r ' (t) ll = 2j + 2tk / 2j + 2t^2K

* After this I use the quationat rule??? 99% sure???

$\displaystyle \sqrt{4+4t^2}=\sqrt{4(1+t^2)}=2\sqrt{1+t^2}$
• Mar 18th 2011, 11:38 AM
HallsofIvy
Also, you do NOT divide a vector by a vector. The denominator is the number $\displaystyle 2\sqrt{1+ t^2}$, not a vector.
• Mar 18th 2011, 08:52 PM
sderosa518
here is my solution to T ' (t).

T(t) = r '(t)/ llr '(t)ll = 2j + 2tk / 2(1+t^2)^ (1/2)

T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2)
• Mar 19th 2011, 07:42 AM
alexmahone
Quote:

Originally Posted by sderosa518
here is my solution to T ' (t).

T(t) = r '(t)/ llr '(t)ll = 2j + 2tk / 2(1+t^2)^ (1/2)

T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2)

Please learn to use $\displaystyle LaTeX$.

In any case, your computation of T'(t) seems to be incorrect. Remember that the formula for curvature is $\displaystyle \displaystyle \kappa=\frac{1}{v}|\frac{dT}{dt}|$.
• Mar 19th 2011, 11:38 AM
sderosa518
[Math]
T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2) [Math]