Answered

Consider the vector function given below.

r(t)=(3t^2, sin(t)-tcos(t), cos(t)+tsin(t)), t>0


Do the following

AFind the unit tangent and unit normal vectors T(t) and N(t)

T(t) = < , , >

N(t) = < , , >

BFind the curvature

k(t) =

Verified Answer :

The tangent vector is by definition the derivative of r(t) with respect to t: 

T' = dr/dt = <6t, tsin(t), tcos(t)> 

The unit vector T = T'/|T'| = <6t, tsin(t), tcos(t)>/sqrt(36t^2 + t^tsin(t)^2 +t^2cos(t^2)) 

T = <6t, tsin(t), tcos(t)>/(t*sqrt(37)) = <6, sin(t), cos(t)>/sqrt(37) 

Now the normal unit vector N is perpendicular to r/|r| and T. It is the second derivative of r/|r| with repsect to time 

N' = d^2r/dt^2 = <6, sin(t) + tcos(t), cos(t) - tsin(t)> 


N= N'/|N'| = <6, sin(t) + tcos(t), cos(t) - tsin(t)>/sqrt(36 + sin^2t +2tsin(t)cos(t)+t^2cos^2t + cos^2(t) -2tcos(t) sin(t) +t^2sin^2t) 

N = <6, sin(t) + tcos(t), cos(t) - tsin(t)>/sqrt(37 +t^2)

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