example of curvature (space curve)Example space curves and calculating their curvatures using the formulaκ(t)=∥𝐫′(t)×𝐫′′(t)∥∥𝐫′(t)∥31. 𝐫(t)=3ti^+t2j^-4t2k^the first derivative𝐫′(t)=3i^+2tj^-8tk^vector magnitude of the derivative∥𝐫′(t)∥=32+(2t)2+(-8t)2∥𝐫′(t)∥=9+4t2+16t2=9+20t2the second derivative𝐫′′(t)=2j^-8k^the cross product𝐫′(t)×𝐫′′(t)=|i^j^k^32t-8t02-8|=(-16t+16t)i^-(-24)j^+6k^𝐫′(t)×𝐫′′(t)=24j^+6k^∥𝐫′(t)×𝐫′′(t)∥=576+36=612=2153∥𝐫′(t)∥3=(9+20t2)3/2κ(t)=2153(9+20t2)3/22. Calculate the curvature of the right circular helix as given in the plot below and defined as𝐫(t)=costi^+sintj^+tk^𝐫′(t)=-sinti^+costj^+k^∥𝐫′(t)∥=sin2t+cos2t+12=2𝐫′′(t)=-costi^-sintj^𝐫′(t)×𝐫′′(t)=|i^j^k^-sintcost1-cost-sint0|=sinti^-costj^+(sin2t+cos2t)k^𝐫′(t)×𝐫′′(t)=sinti^-costj^+k^∥𝐫′(t)×𝐫′′(t)∥=sin2t+cos2t+12=2∥𝐫′(t)∥3=23/2κ(t)=223/2=12