The graph of r(t)=<t- 3sin(t)/2,1- 3cos(t)/2,t> For which values of t is the curvature largest?

Question

I don’t know why my answer is not matching the answer key. Follow up for Oobleck
R’(t)= 1-3cos(t)/2, 3sin(t)/2, 1
R’’(t)=3sin(t)/2,3cos(t)/2,0
This is what I did
Curvature formula K(t)=lf’’(t)l/[1+(f’(t))^2]^(3/2)
r’(t)xr’’(t) ( 1-3cos(t)/2, 3sin(t)/2, 1)x(3sin(t)/2,3cos(t)/2,0)
[(3sin(t)/2*0)- (1* 3cos(t)/2)]i= -3cos(t)/2i
[(1-3cos(t)/2 *0)- ( 3sin(t)/2*1)]j= -3sin(t)/2j
[(1-3cos(t)/2*3cos(t)/2)- (-3sin(t)/2*3sin(t)/2)]k->3cos(t)/2-9cos^2(t)/4- 9sin^2(t)/4k= 3cos(t)/2 -9/4k
lr’(t)xr’’(t)l= √[(-3cos(t)/2)^2+(-3sin(t)/2)^2+(3cos(t)/2 -9/4)^2]
=√(9cos^2(t)/4+9sin^2(t)/4+ 9cos^2(t)/4 -27cos(t)/8 -27cos(t)/8+ 81/16)
=√(9/4cos^2(t)+sin^2(t)+ 9cos^2(t)/4 -54cos(t)/8+ 81/16)
=√(9cos^2(t)/4 -54cos(t)/8+ 81/16+ 9/4)
=√(9cos^2(t)/4 -54cos(t)/8+ 81/16+ 36/16)
lr’(t)xr’’(t)l= √(9cos^2(t)/4 -54cos(t)/8+ 117/16)
[1+(f’(t))^2]^(3/2)= [1+()= [(1-3cos(t)/2)^2+ (3sin(t)/2)^2+( 1)^2]^(3/2)
=[1 -3cos(t)/2 -3cos(t)/2+ 9cos^2(t)/4+ 9sin^2(t)/4) + 1)]^(3/2)
=[-6cos(t)/2+ 9/4cos^2(t)+sin^2(t))+ 1+1)]^(3/2)
=[-6cos(t)/2+ 9/4 + 2)]^(3/2)
=[-6cos(t)/2+ 17/4]^(3/2)
My answer
[√(9cos^2(t)/4 -54cos(t)/8+ 117/16)]/ =[-6cos(t)/2+ 17/4]^(3/2)
Answer key
[√(4cos^2(t)-12cos(t)+13)]/[17-12cos(t)]^(3/2)

Answer 1

typo at these parts
[1+(f’(t))^2]^(3/2)= [1+ (1-3cos(t)/2)^2+ (3sin(t)/2)^2+( 1)^2]^(3/2)
=[1+1 -3cos(t)/2 -3cos(t)/2+ 9cos^2(t)/4+ 9sin^2(t)/4) + 1)]^(3/2)
=[-6cos(t)/2+ 9/4cos^2(t)+sin^2(t))+ 1+1+1)]^(3/2)
=[-6cos(t)/2+ 9/4 + 3)]^(3/2)
=[-6cos(t)/2+ 21/4]^(3/2)
My answer
[√(9cos^2(t)/4 -54cos(t)/8+ 117/16)]/ =[-6cos(t)/2+ 21/4]^(3/2)
Answer key
[√(4cos^2(t)-12cos(t)+13)]/[17-12cos(t)]^(3/2)

Answer 2

The curvature formula you used is for y = f(x).
For general parametric equations, which this is,

k = (x'y" - y'x")/(x'^2 + y'^2)^(3/2) = 6(3-2cost)/(13-12cost)^(3/2)
so, dk/dt = 24sint(3cost-7) / (13-12cost)^(5/2)
dk/dt=0 at all multiples of π
k is a max of 6 at even multiples of π
k is a min of 6/25 at odd multiples of π