Question

Find the curvature of each of the vector-valued functions defined in Exercises 39–44.

$\mathit{r}\left(t\right)=\left(t\sin t,t\cos t,\alpha t\right),\mathrm{where}\alpha \mathrm{is}\mathrm{a}\mathrm{constant}$

Short answer

The curvature of the given vector-valued function is$\kappa =\sqrt{\frac{t^4+\left(4+{\alpha }^2\right)t^2+4+4{\alpha }^2}{{\left(t^2+1+{\alpha }^2\right)}^3}}.$

Step-by-step solution

Step 1. Given Information. 

The given vector-valued function is$\mathit{r}\left(t\right)=\left(t\sin t,t\cos t,\alpha t\right),\mathrm{where}\alpha \mathrm{is}\mathrm{a}\mathrm{constant}.$

Step 2. Find the curvature. 

To find the curvature of the given vector-valued function, 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,

$$\begin{gathered} \kappa =\frac{\sqrt{t^4+\left(4+{\alpha }^2\right)t^2+4+4{\alpha }^2}}{{\left(\sqrt{t^2+1+{\alpha }^2}\right)}^3} \\ \kappa =\sqrt{\frac{t^4+\left(4+{\alpha }^2\right)t^2+4+4{\alpha }^2}{{\left(t^2+1+{\alpha }^2\right)}^3}} \end{gathered}$$