Answer to Question #271940 in Differential Geometry | Topology for Angel Nodado

Question

Find the curvature, the radius and the center of curvature at a point.

x = t ^ 2 , y = t ^ 3 ; t = 1/2

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:

$$R=\dfrac{1}{K}=62.5$$ $$y'_x=\dfrac{y'_t}{x'_t}=\dfrac{3t^2-3}{6t}=\dfrac{t^2-1}{2t}$$ $$(y'_x)'_t=\dfrac{2t^2-t^2+1}{2t^2}=\dfrac{t^2+1}{2t^2}$$ $$y''_{xx}=\dfrac{t^2+1}{2t^2(6t)}=\dfrac{t^2+1}{12t^3}$$ $$x_C=x-\dfrac{y'(1+(y')^2)}{y''}$$ $$y_C=y+\dfrac{1+(y')^2}{y''}$$

$t=2$

$$x=3(2)^2=12, y=(2)^3-3(2)=2$$ $$y'=\dfrac{(2)^2-1}{2(2)}=\dfrac{1}{4}$$ $$y''=\dfrac{(2)^2+1}{12(2)^3}=\dfrac{5}{96}$$ $$x_C=12-\dfrac{\dfrac{1}{4}(1+(\dfrac{1}{4})^2)}{\dfrac{5}{96}}=6.9$$ $$y_C=2+\dfrac{1+(\dfrac{1}{4})^2}{\dfrac{5}{96}}=32.4$$

$C(6.9, 32.4)$