91Ó°ÊÓ

The givenr(t)is$\left(\cos t,\sin t,\frac{1}{t}\right),t>0.$

To show that $\underset{t→∞}{\lim }k=1\mathrm{and}\underset{t→0^{+}}{\lim }k=0,$we will use the formula for the Curvature of a Space Curve $\mathrm{κ}=\frac{||r\text{'}\left(t\right)×r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3}.$

So,

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

$$\begin{gathered} \mathrm{κ}=\frac{{\displaystyle \frac{\sqrt{t^6+t^2+4}}{t^3}}}{{\left({\displaystyle \frac{\sqrt{t^4+1}}{t^2}}\right)}^3} \\ \mathrm{κ}=\frac{\sqrt{t^6+t^2+4}}{t^3}{\left(\frac{t^2}{\sqrt{t^4+1}}\right)}^3 \\ \mathrm{κ}=t^3\sqrt{\frac{t^6+t^2+4}{{\left(t^4+1\right)}^3}} \end{gathered}$$

Let's find the limits,

$$\begin{gathered} \underset{t→∞}{\lim }\mathrm{κ} \\ =\underset{t→∞}{\lim }{t}^3\sqrt{\frac{t^6+t^2+4}{{\left(t^4+1\right)}^3}} \\ =\underset{t→∞}{\lim }{t}^3\sqrt{\frac{t^6\left(1+{\displaystyle \frac{1}{t^4}}+{\displaystyle \frac{4}{t^6}}\right)}{{\left(t^4\left(1+{\displaystyle \frac{1}{t^4}}\right)\right)}^3}} \\ =\underset{t→∞}{\lim }\frac{t^3}{t^3}\sqrt{\frac{\left(1+{\displaystyle \frac{1}{t^4}}+{\displaystyle \frac{4}{t^6}}\right)}{{\left(1+{\displaystyle \frac{1}{t^4}}\right)}^3}} \\ =\sqrt{\frac{1+0+0}{{\left(1+0\right)}^3}} \\ =1 \end{gathered}$$

Hence proved.

Now, let's find another limit,

$$\begin{gathered} \underset{t→0^{+}}{\lim }k \\ =\underset{t→0^{+}}{\lim }{t}^3\sqrt{\frac{t^6+t^2+4}{{\left(t^4+1\right)}^3}} \\ =0\sqrt{\frac{0+0+4}{1}} \\ =0 \end{gathered}$$

Hence proved.