Question

Let C be the graph of a vector function $r\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$ in the xy-plane, where x(t) and y(t) are twice-differentiable functions oft. Show that the curvature κ of C at a point on the curve is given by$\kappa =\frac{|x\text{'}(t)y\text{'}\text{'}(t)-x\text{'}\text{'}(t)y\text{'}(t\left)\right|}{\left(\right(x\text{'}{\left(t\right))}^2+(y\text{'}{\left(t\right))}^2{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}.$

This is part (b) of Theorem 11.24.

Short answer

The given statement is proved.

Step-by-step solution

Step 1. Given Information. 

It is given thatC is the graph of a vector function $r\left(t\right)=\left(x\left(t\right),y\left(t\right)\right)$in the xy-plane, wherex(t) andy(t) are twice-differentiable functions oft.

Step 2. Prove. 

We have to show that the curvature κ of C at a point on the curve is given by

$\kappa =\frac{|x\text{'}(t)y\text{'}\text{'}(t)-x\text{'}\text{'}(t)y\text{'}(t\left)\right|}{\left(\right(x\text{'}{\left(t\right))}^2+(y\text{'}{\left(t\right))}^2{)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.$

To show the curvature of the graph of κ of C at a point on the curve,we will use the formula for the curvature of a space curve.

The curvature of a space curve is $\kappa =\frac{||r\text{'}\left(t\right)\times r\text{'}\text{'}\left(t\right)||}{{||r\text{'}\left(t\right)||}^3}.$

So,

Step 3. Calculate. 

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

$$\begin{gathered} \kappa =\frac{\left|x\text{'}\left(t\right)y\text{'}\text{'}\left(t\right)-x\text{'}\text{'}\left(t\right)y\text{'}\left(t\right)\right|}{{\left(\sqrt{{\left(x\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\left(t\right)\right)}^2}\right)}^3} \\ \kappa =\frac{\left|x\text{'}\left(t\right)y\text{'}\text{'}\left(t\right)-x\text{'}\text{'}\left(t\right)y\text{'}\left(t\right)\right|}{{\left({\left(x\text{'}\left(t\right)\right)}^2+{\left(y\text{'}\left(t\right)\right)}^2\right)}^{{\displaystyle \frac{3}{2}}}} \end{gathered}$$

Hence proved.