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

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^{t} k, P (1, 0, 1)

Answer 1

r (t) = e^{t} \cos t \cdot i + e^{t} \sin t \cdot j + e^{t} k

Differentiate

r^{'} (t) = e^{t} (\cos t - \sin t) i + e^{t} (\sin t + \cos t) j + e^{t} k

Find Magnitude

| r^{'} (t) | = e^{t} \sqrt{(\cos t - \sin t)^{2} + (\sin t + \cos t)^{2} + 1}

| r^{'} (t) | = e^{t} \sqrt{(\cos^{2} t - \sin^{2} t - 2 \cos t \sin t)^{2} + (\sin^{2} t + \cos^{2} t + 2 \cos t \sin t)^{2} + 1}

| r^{'} (t) | = e^{t} \sqrt{3}

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^{t} k)}{(e^{t} \sqrt{3})}

T (t) = \frac{1}{\sqrt{3} [(\cos t - \sin t) i + (\sin t + \cos t) j + k]}

Differentiate

T^{'} (t) = \frac{1}{\sqrt{3} [(- \sin t - \cos t) i + (\cos t - \sin t) j + 0 k]}

Find Magnitude

| T^{'} (t) | = \frac{1}{\sqrt{3} \sqrt{(- \sin t - \cos t)^{2} + (\cos t - \sin t)^{2} + 0)}}

| T^{'} (t) | = \frac{1}{\sqrt{3} \sqrt{(\sin^{2} t + \cos^{2} t + 2 \sin t \cos t) + (\cos^{2} t + \sin^{2} t - 2 \cos t \sin t) + 0)}}

| T^{'} (t) | = \frac{\sqrt{2}}{\sqrt{3}}

Recall that

k (t) = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{(e^{t} \sqrt{3})} = \frac{\sqrt{2}}{(3 e^{t})}

Result

k (t) = \frac{\sqrt{2}}{(3 e^{t})}

Expert Answer