Answer to Question #271915 in Differential Geometry | Topology for Angel Nodado

Question #271915

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


1
Expert's answer
2021-11-29T19:30:01-0500

"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)"



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