Answer
(a) $\kappa =\dfrac{6 \pi}{5} \sqrt {10}$
(b) $\kappa =\pi$
Work Step by Step
(a) When $r(t)$ is a smooth curve, then the curvature can be defined as: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|...(1)$
where $T=\dfrac{v}{|v|} ;$ a unit tangent vector.
We have: $v=\dfrac{dr}{dt}=-3 \sin t \ i+3\cos t j+k$ and $|v|=\sqrt {(-3 \sin t)^2+(3\cos t)^2+(1)^2}=\sqrt {10}$
So, $T=\dfrac{v}{|v|}=\dfrac{-3 \sin t \ i+3\cos t j+k}{\sqrt {10}}$ and $|\dfrac{dT}{dt}|=\dfrac{3}{\sqrt {10}}$
Equation (1) becomes: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|=\dfrac{1}{\sqrt {10}} \times \dfrac{3}{\sqrt {10}}=\dfrac{3}{10}$
Now, the total curvature is: $\kappa =\int_{t_0}^{t_1} \kappa |v| dt=\int_{0}^{4 \pi} \dfrac{3}{10} \sqrt {10} \ dt=\dfrac{6 \pi}{5} \sqrt {10}$
(b) We will use the parametric equations $x=t; y =t^2$, so $r(t)=t \ i +t^2 \ j$; then the curvature can be defined as: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|...(2)$
where $T=\dfrac{v}{|v|} ;$ a unit tangent vector.
We have: $v=\dfrac{dr}{dt}=i+2t \ j$ and $|v|=\sqrt {(1)^2+( 2t)^2}=\sqrt {1+4t^2}$
So, $T=\dfrac{v}{|v|}=\dfrac{i+2t \ j}{\sqrt {1+4t^2}}$ and $\dfrac{dT}{dt}=\dfrac{-4t}{(1+4t^2)^{3/2}}i+\dfrac{2}{(1+4t^2) \sqrt {1+4t^2}}j$
and $|\dfrac{dT}{dt}|=\sqrt {\dfrac{-4t}{(1+4t^2)^{3/2}})^2+(\dfrac{2}{(1+4t^2) \sqrt {1+4t^2}})^2}=\dfrac{2}{1+4t^2}$
Equation (2) becomes: $\kappa =\dfrac{1}{|v|}|\dfrac{dT}{dt}|=\dfrac{1}{\sqrt {1+4t^2}} \times \dfrac{2}{\sqrt {1+4t^2}}=\dfrac{2 \sqrt {4t^2+1}}{(1+4t^2)^2}$
Now, the total curvature is: $\kappa =\int_{t_0}^{t_1} \kappa |v| dt=\int_{-\infty}^{ \infty} \dfrac{2 \sqrt {4t^2+1}}{(1+4t^2)^2} \times \sqrt {1+4t^2} \ dt=[\tan^{1} (2t)]_{-\infty}^{ \infty}=2 \tan^{-1} (\infty)=(2) (\dfrac{\pi}{2})=\pi$
