Find the radius of curvature and the center of curvature for f (t) = <t cos t, t sin t, t> at t =pi/2?

Question

Find the radius of curvature and the center of curvature for f (t) = at t =pi/2

Answer 1

x= t cos t
y=tsin t
dx= (cos t-tsin t)dt
dy =(sint+tcost) dt so dy/dx = (sint+tcost)/(cost-tsint)=1/(-pi/2)

r= (1+y´^2)^3/2 /y´´
dy´/dx = =dy´/dt *dt/dx = 1/(cos t-t sin t)^2 =*[(cost-tsint)(cost+cos t-tsint )-(sin t+tcost)(-sint-sint-tcost)] *
1/(cost-tsint)

=1/(pi/2)^2 *[(-pi/2)^2-(pi/2(-2)]*(-2/pi) = 2/(pi/2)^2-1/(pi/2)

r=(1+1/(pi/2)^2)3/2 /[2/(pi/2)^2-1/(pi/2)
Maybe it is shorter passing to polar coordinates
x^2+y^2=t^2 so r=t and y/x =tan@=tan t so @=t
and
r=@ is the polar equation
radius= =(@^2+1)^3/2/[@^2+2]=((pi/2)^2+1))^3/2/((pi/2)^2+2))

and the center has coordinates
xc=-pi/2 yc=pi/2

Answer 2

you have to show f(x)-sinx=0 for all pi/2

Answer 3

they are both there
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