Question

EXAMPLE 3 Show that the curvature of a circle of radius k is 1/k. SOLUTION We can take the circle to have center the origin, and then a parametrization is r(t) = k cos t i + k sin t j. Therefore r'(t) = and |r'(t)| = k so T(t) = r'(t) |r'(t)| = and T'(t) = . This gives |T'(t)| = 1, so using this equation, we have κ(t) = |T'(t)| |r'(t)| = __ .

Answer 1

Answer:

Therefore r'(t) =-k sin t i + k cos t j and |r'(t)| = k so T(t) = r'(t)/|r'(t)| = -sin t i + cos t j  and T'(t) = -cos t i- sin t j . This gives |T'(t)| = 1, so using this equation, we have κ(t) = |T'(t)|/|r'(t)| = 1/k.

Step-by-step explanation:

We are already given the definition of curvature and the parametrization needed to find the curvature of the circle. In genecral the curvature κ is equal to κ(t)=|T'(t)|/|r'(t)| where r(t) is a parametrization of the curve and T(t) is the normalized tangent vector respect to the parametrization, that is, T(t)=r'(t)/|r'(t)|.

Now, using the derivatives of sines and cosines, and the definition of norm,  we obtain that:

r(t) = k cos t i + k sin t j ⇒ r'(t)=-k sin t i + k cos t j ⇒|r'(t)|²=sin²t+cos²t=1

T(t) = r'(t)/|r'(t)|=-sin t i +cos t j ⇒ T'(t)= -cos t i - sin t j ⇒|T'(t)|²=cos²t+sin²t=1