example of curvature (space curve)


Example space curves and calculating their curvaturesMathworldPlanetmathPlanetmath using the formula

κ⁢(t)=∥𝐫′⁢(t)×𝐫′′⁢(t)∥∥𝐫′⁢(t)∥3

1. 𝐫⁢(t)=3⁢t⁢i^+t2⁢j^-4⁢t2⁢k^

𝐫′⁢(t)=3⁢i^+2⁢t⁢j^-8⁢t⁢k^

∥𝐫′⁢(t)∥=32+(2⁢t)2+(-8⁢t)2

∥𝐫′⁢(t)∥=9+4⁢t2+16⁢t2=9+20⁢t2

the second derivative

𝐫′′⁢(t)=2⁢j^-8⁢k^

𝐫′⁢(t)×𝐫′′⁢(t)=|i^j^k^32⁢t-8⁢t02-8|=(-16⁢t+16⁢t)⁢i^-(-24)⁢j^+6⁢k^

𝐫′⁢(t)×𝐫′′⁢(t)=24⁢j^+6⁢k^

∥𝐫′⁢(t)×𝐫′′⁢(t)∥=576+36=612=2⁢153

∥𝐫′⁢(t)∥3=(9+20⁢t2)3/2

κ⁢(t)=2⁢153(9+20⁢t2)3/2

2. Calculate the curvature of the right circular helix as given in the plot below and defined as

𝐫⁢(t)=cos⁡t⁢i^+sin⁡t⁢j^+t⁢k^

𝐫′⁢(t)=-sin⁡t⁢i^+cos⁡t⁢j^+k^

∥𝐫′⁢(t)∥=sin2⁡t+cos2⁡t+12=2

𝐫′′⁢(t)=-cos⁡t⁢i^-sin⁡t⁢j^

𝐫′⁢(t)×𝐫′′⁢(t)=|i^j^k^-sin⁡tcos⁡t1-cos⁡t-sin⁡t0|=sin⁡t⁢i^-cos⁡t⁢j^+(sin2⁡t+cos2⁡t)⁢k^

𝐫′⁢(t)×𝐫′′⁢(t)=sin⁡t⁢i^-cos⁡t⁢j^+k^

∥𝐫′⁢(t)×𝐫′′⁢(t)∥=sin2⁡t+cos2⁡t+12=2

∥𝐫′⁢(t)∥3=23/2

κ⁢(t)=223/2=12

Title example of curvature (space curve)
Canonical name ExampleOfCurvaturespaceCurve
Date of creation 2013-03-22 15:40:58
Last modified on 2013-03-22 15:40:58
Owner bloftin (6104)
Last modified by bloftin (6104)
Numerical id 8
Author bloftin (6104)
Entry type Example
Classification msc 53A04
Related topic PositionVector