Question

Find the curvature, the radius of curvature and the center of curvature having a parametric equations at the given point:

x = 3t ^ 2 , y = t ^ 3 - 3t at t = 2

Answer 1

$$K=\dfrac{|x'y''-x''y'|}{[(x')^2+(y')^2]^{3/2}}$$

$$x'_t=6t, x''_{tt}=6$$

$$y'_t=3t^2-3,y''_{tt}=6t-3$$

$$K=\dfrac{|6t(6t-3)-6(3t^2-3)|}{[(6t)^2+(3t^2-3)^2]^{3/2}}$$

$t=2$

$$K=\dfrac{|6(2)(6(2)-3)-6(3(2)^2-3)|}{[(6(2))^2+(3(2)^2-3)^2]^{3/2}}=\dfrac{2}{125}$$

$$=0.016$$

The radius of curvature of a curve at a point is called the inverse of the curvature $K$ of the curve at this point:

$$R=\dfrac{1}{K}=62.5$$

$$y'_x=\dfrac{y'_t}{x'_t}=\dfrac{3t^2-3}{6t}=\dfrac{t^2-1}{2t}$$

$$(y'_x)'_t=\dfrac{2t^2-t^2+1}{2t^2}=\dfrac{t^2+1}{2t^2}$$

$$y''_{xx}=\dfrac{t^2+1}{2t^2(6t)}=\dfrac{t^2+1}{12t^3}$$

$$x_C=x-\dfrac{y'(1+(y')^2)}{y''}$$

$$y_C=y+\dfrac{1+(y')^2}{y''}$$

$t=2$

$$x=3(2)^2=12, y=(2)^3-3(2)=2$$

$$y'=\dfrac{(2)^2-1}{2(2)}=\dfrac{1}{4}$$

$$y''=\dfrac{(2)^2+1}{12(2)^3}=\dfrac{5}{96}$$

$$x_C=12-\dfrac{\dfrac{1}{4}(1+(\dfrac{1}{4})^2)}{\dfrac{5}{96}}=6.9$$

$$y_C=2+\dfrac{1+(\dfrac{1}{4})^2}{\dfrac{5}{96}}=32.4$$

$C(6.9, 32.4)$