Question

Find the curvature of each of the functions defined by the parametric equations in Exercises 36–38.

$x=\cos t+t\sin t,y=\sin t-t\cos t$

Short answer

The curvature is$\frac{1}{t}.$

Step-by-step solution

Step 1. Given information.

The given function is,

$x=\cos t+t\sin t,y=\sin t-t\cos t$

Step 2. Curvature.

Now,

$$\begin{gathered} r\left(t\right)=⟨\cos t+t\sin t,\sin t-t\cos t⟩ \\ {r}^{\text{'}}\left(t\right)=⟨t\cos t,t\sin t⟩ \\ ||r^{\text{'}}\left(t\right)||=\sqrt{(t\cos t{)}^2+(t\sin t{)}^2} \\ =\sqrt{t^2\left({\cos }^{2}t+{\sin }^{2}t\right)}=t \\ T\left(t\right)=\frac{r^{\text{'}}\left(t\right)}{||r^{\text{'}}\left(t\right)||} \\ =\frac{⟨t\cos t,t\sin t⟩}{t} \\ =⟨\cos t,\sin t⟩ \\ ||T^{\text{'}}\left(t\right)||=\sqrt{(-\sin t{)}^2+{\cos }^{2}t}=1 \\ Curvature=k=\frac{||T^{\text{'}}\left(t\right)||}{||r^{\text{'}}\left(t\right)||} \\ =\frac{1}{t} \end{gathered}$$