Question

what is the curvature of r(t)=(sint-tcos t )i+ (cost+tsin t)j+ (t^2)k

Answer 1

can you get the unit tangent vector?

Answer 2

i have the derivative of r(t) as tsint i + t cost j + 2t k

Answer 3

\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]is probably the easiest formula to use here
what you have written so far is r'(t)

Answer 4

i am having a togh time figuring out the unit tangent

Answer 5

and I have not checked that r'(t) yet
once you have it though, divide it by the magnitude |r'(t)|
that is the unit tangent T(t)

Answer 6

magnitude of r'(t)=t(sqrt5)

Answer 7

yeah I agree with your r'(t)
so now just divide by the magnitude

Answer 8

divide, that is your unit tangent

Answer 9

\[\vec T(t)=\frac{\vec r'(t)}{\|\vec r'(t)\|}=?\]

Answer 10

T=[(sint - tcost)i +(cost + tsint)j + 2t]/(tsqrt5)
this is where i get confused.
I dont know if i need to use the second derivative of r(t)???

Answer 11

\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]

Answer 12

no you divided r(t) by the magnitude of r'(t)

Answer 13

ok.let me try to solve that!

Answer 14

k=1/5t

Answer 15

I got k=t, but I could have made a mistake

Answer 16

oh wait\[\kappa=\frac1{5t}\]?yeah that's what I get

Answer 17

is that what you meant with t on the bottom?

Answer 18

yes

Answer 19

sweet, then we're good :)

Answer 20

thank you...first time and it may not be the last one

Answer 21

hope not, see ya!