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3. [-/10.25 Points] (a) Show that $dB/ds$ is perpendicular to B. $|B|=1 \Rightarrow B \cdot B = \boxed{1}$ $\frac{d}{ds}(B \cdot B) = \boxed{0}$ $\frac{dB}{ds} \cdot B + B \cdot \frac{dB}{ds} = 0$ $2 \frac{dB}{ds} \cdot B = 0 \Rightarrow \frac{dB}{ds} \cdot B = 0$ (b) Show that $dB/ds$ is perpendicular to T. $B = T \times N$ $\frac{dB}{ds} = \frac{d}{ds}(T \times N) = \frac{d}{dt}(T \times N) \frac{dt}{ds}$ $= ([T' \times N] + [T \times N']) \frac{1}{|r'(t)|}$ $= ([T' \times N] + [T \times N']) \frac{1}{|r'(t)|}$ $= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$ $= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$ $= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$ (c) Deduce from parts (a) and (b) that $dB/ds = -\tau(s)N$ for some number $\tau(s)$ called the torsion of the curve. (The torsion measures the degree to which the curve twists out of the osculating plane.) $B = T \times N \Rightarrow T \perp N, B \perp T$ and $B \perp N$. So $B, T$ and $N$ form $\boxed{an orthonormal}$ set of vectors in the three dimensional space $\mathbb{R}^3$. From parts (a) and (b), $dB/ds$ is perpendicular to both T and B. Therefore, $dB/ds = \boxed{a scalar multiple of N}$. So $dB/ds = -\tau(s)N$, where $\tau(s)$ is scalar. (d) Show that for a plane curve the torsion is $\tau(s) = 0$. Since $B = T \times N$, T and N are unit vectors, B is a unit vector mutually $\boxed{perpendicular}$ to both T and N. For a plane curve, the unit vector always $\boxed{lies in the plane}$. Thus $dB/ds = 0$, but $dB/ds = -\tau(s)N$, and $N \neq 0$, so $\tau(s) = 0$.

Name: 3 1025 points a show that dbds is perpendicular to b b1 rightarrow b cdot b boxed1 fracddsb cdot b boxed0 fracdbds cdot b b cdot fracdbds 0 2 fracdbds cdot b 0 rightarrow fracdbds cdot b 0 b 89663
Uploaded: 2022-01-11T02:00:49-08:00
Duration: 1 min 4 s
Channel: Carson Merrill
Description: 3 1025 points a show that dbds is perpendicular to b b1 rightarrow b cdot b boxed1 fracddsb cdot b boxed0 fracdbds cdot b b cdot fracdbds 0 2 fracdbds cdot b 0 rightarrow fracdbds cdot b 0 b 89663

          3. [-/10.25 Points]
(a) Show that $dB/ds$ is perpendicular to B.
$|B|=1 \Rightarrow B \cdot B = \boxed{1}$
$\frac{d}{ds}(B \cdot B) = \boxed{0}$
$\frac{dB}{ds} \cdot B + B \cdot \frac{dB}{ds} = 0$
$2 \frac{dB}{ds} \cdot B = 0 \Rightarrow \frac{dB}{ds} \cdot B = 0$
(b) Show that $dB/ds$ is perpendicular to T.
$B = T \times N$
$\frac{dB}{ds} = \frac{d}{ds}(T \times N) = \frac{d}{dt}(T \times N) \frac{dt}{ds}$
$= ([T' \times N] + [T \times N']) \frac{1}{|r'(t)|}$
$= ([T' \times N] + [T \times N']) \frac{1}{|r'(t)|}$
$= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$
$= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$
$= ([\frac{T'}{|T'|} \times N] + [T \times N']) \frac{1}{|r'(t)|}$
(c) Deduce from parts (a) and (b) that $dB/ds = -\tau(s)N$ for some number $\tau(s)$ called the torsion of the curve. (The torsion measures the degree to which the curve twists out of the osculating plane.)
$B = T \times N \Rightarrow T \perp N, B \perp T$ and $B \perp N$. So $B, T$ and $N$ form $\boxed{an orthonormal}$ set of vectors in the three dimensional space $\mathbb{R}^3$. From parts (a) and (b), $dB/ds$ is perpendicular to both T and B. Therefore, $dB/ds = \boxed{a scalar multiple of N}$. So $dB/ds = -\tau(s)N$, where $\tau(s)$ is scalar.
(d) Show that for a plane curve the torsion is $\tau(s) = 0$.
Since $B = T \times N$, T and N are unit vectors, B is a unit vector mutually $\boxed{perpendicular}$ to both T and N. For a plane curve, the unit vector always $\boxed{lies in the plane}$. Thus $dB/ds = 0$, but $dB/ds = -\tau(s)N$, and $N \neq 0$, so $\tau(s) = 0$.

$3 1025 points a show that dbds is perpendicular to b b1 rightarrow b cdot b boxed1 fracddsb cdot b boxed0 fracdbds cdot b b cdot fracdbds 0 2 fracdbds cdot b 0 rightarrow fracdbds cdot b 0 b 89663$

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