# Find the curvature K of the curve at the point r(t)=e^tcos tcdot i+e^tsin tcdot j+e^tk,P(1,0,1)

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

2020-12-29
$$r(t)=e^t\cos t\cdot i+e^t\sin t\cdot j+e^tk$$
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
$$k(t)=\frac{\sqrt2}{(3e^t)}$$

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