- Solution 1: Note that (cost+sint)2+(sint−cost)2=2 for all t. So, the points (x,y) of our curve lie on x2+y2=2, which is a circle of radius 2. Indeed
x(t)y(t)=cost+sint=2[costcos4π+sintsin4π]=2cos(t−4π)=sint−cost=2[sintcos4π−costsin4π]=2sin(t−4π)
So, r(t) circumnavigates a circle of radius 2 and consequently has curvature κ=21.
- Solution 2: We use the formula κ=(dtds)3∣v(t)×a(t)∣, remembering that v(t)=r′(t), a(t)=r′′(t), and dtds=∣r′(t)∣.
v(t)a(t)v(t)×a(t)dtdsκ=r′(t)=(−sint+cost,cost+sint)=r′′(t)=(−cost−sint,−sint+cost)=[(−sint+cost)2+(cost+sint)2]k^=2k^=dtdv=(−sint+cost)2+(cost+sint)2=2=(dtds)3v(t)×a(t)=232k^=21