what is the curvature of r(t)=(sint-tcos t )i+ (cost+tsin t)j+ (t^2)k
can you get the unit tangent vector?
i have the derivative of r(t) as tsint i + t cost j + 2t k
\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]is probably the easiest formula to use here what you have written so far is r'(t)
i am having a togh time figuring out the unit tangent
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)
magnitude of r'(t)=t(sqrt5)
yeah I agree with your r'(t) so now just divide by the magnitude
divide, that is your unit tangent
\[\vec T(t)=\frac{\vec r'(t)}{\|\vec r'(t)\|}=?\]
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)???
\[\kappa=\frac{\|\vec T'(t)\|}{\|\vec r'(t)\|}\]
no you divided r(t) by the magnitude of r'(t)
ok.let me try to solve that!
k=1/5t
I got k=t, but I could have made a mistake
oh wait\[\kappa=\frac1{5t}\]?yeah that's what I get
is that what you meant with t on the bottom?
yes
sweet, then we're good :)
thank you...first time and it may not be the last one
hope not, see ya!
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