Question

Show that the curvature is constant at every point on the circular helix defined by $\mathit{r}\left(t\right)=\left(a\cos t,a\sin t,bt\right),$ where a and b are positive constants.

Short answer

The curvature is constant at every point on the circular helix defined by$r\left(t\right).$

Step-by-step solution

Step 1. Given Information. 

The given circular helix is$\mathit{r}\left(t\right)=\left(a\cos t,a\sin t,bt\right).$

Step 2. Showing.

To show that the curvature is constant at every point on the circular helix defined by $\mathit{r}\left(t\right)=\left(a\cos t,a\sin t,bt\right),$we will use the formula for the Curvature of a Space Curve $\kappa =\frac{||r\text{'}\left(t\right)\times r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3}.$

So,

Step 3. Calculate.

Put the values of (a) and (b) in the formula of Curvature of a space curve,

$$\begin{gathered} \kappa =\frac{\sqrt{a^2\left(a^2+b^2\right)}}{{\left(\sqrt{a^2+b^2}\right)}^3} \\ \kappa =\frac{a\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}\left(a^2+b^2\right)} \\ \kappa =\frac{a}{\left(a^2+b^2\right)} \end{gathered}$$

Thus, kis constant.