Curvature of curve at the given parametric value

Question

The given is: $$r(t)=t \pmb i +\frac{1}{36}t^3 \pmb j \space,t=3$$ Using the following formula where $K$ is the curvature: $$K=\frac{||T^{'}(t)||}{||v(t)||}$$ $$T(t)= \frac{v(t)}{||v(t)||} =\frac{\sqrt {145}}{12}\pmb i+\frac{\sqrt {145}}{144}t^2\pmb j$$ $$T^{'}(t)=\frac{\sqrt{145}}{72}t\pmb j$$ So now we have $K$ to be: $$K=\frac{\sqrt{145}}{72}*\frac{12}{\sqrt{145}}$$ $$K=\frac{1}{6}$$ But this isn't correct, I've went through this three times and only now got $\frac{1}{6}$. I'm not sure where I went wrong.

Answer 1

$r(t)=t \pmb i +\frac{1}{36}t^3 \pmb j$

$v(t) = r'(t) = (1, \frac{t^2}{12})$

$||v(t)|| = \frac{\sqrt {t^4 + 144}} {12}$ ...(1)

So, unit tangent vector of $r(t),$ $\, T(t) = (\frac {12} {\sqrt {t^4 + 144}}, \frac {t^2} {\sqrt {t^4 + 144}})$

Normal vector $\, T'(t) = \displaystyle (- \frac {24t^3} {(t^4 + 144)^{3/2}}, \frac {2t} {\sqrt {t^4 + 144}} - \frac {2t^5} {(t^4 + 144)^{3/2}})$ ...(ii)

At $\, t = 3$,

From (i), $||v(t)|| = \frac{5} {4}$

From (ii), $T'(t) = \displaystyle (- \frac {24} {125}, \frac {32} {125})$

$||T'(t)|| = \displaystyle \frac {8} {25}$