Question

Find the curvature K of the curve at the point \(r(t)=e^t\cos t\cdot i+e^t\sin t\cdot j+e^tk,P(1,0,1)\)

Answer 1

\(r(t)=e^t\cos t\cdot i+e^t\sin t\cdot j+e^tk\)
Differentiate
\(r'(t)=e^t(\cos t-\sin t)i+e^t(\sin t+\cos t)j+e^tk\)
Find Magnitude
\(|r'(t)|=e^t\sqrt{(\cos t-\sin t)^2+(\sin t+\cos t)^2+1}\)
\(|r'(t)|=e^t\sqrt{(\cos^2t-\sin^2t-2\cos t\sin t)^2+(\sin^2t+\cos^2t+2\cos t\sin t)^2+1}\)
\(|r'(t)|=e^t\sqrt3\)
Recall that: \(T(t)=\frac{r'(t)}{|r'(t)|}\)
Therefore \(T(t)=\frac{(e^t(\cos t-\sin t)i+e^t(\sin t+\cos t)j+e^tk)}{(e^t\sqrt3)}\)
\(T(t)=\frac{1}{\sqrt3[(\cos t-\sin t)i+(\sin t+\cos t)j+k]}\)
Differentiate \(T'(t)=\frac{1}{\sqrt3[(-\sin t-\cos t)i+(\cos t-\sin t)j+0k]}\)
Find Magnitude \(|T'(t)|=\frac{1}{\sqrt3\sqrt{(-\sin t-\cos t)^2+(\cos t-\sin t)^2+0)}}\)
\(|T'(t)|=\frac{1}{\sqrt3\sqrt{(\sin^2t+\cos^2t+2\sin t\cos t)+(\cos^2t+\sin^2t-2\cos t\sin t)+0)}}\)
\(|T'(t)|=\frac{\sqrt2}{\sqrt3}\)
Recall that \(k(t)=\frac{\frac{\sqrt2}{\sqrt3}}{(e^t\sqrt3)}=\frac{\sqrt2}{(3e^t)}\)
Result
\(k(t)=\frac{\sqrt2}{(3e^t)}\)