Answer on Question #72690 – Math – Differential Geometry | Topology
Question
Find equation of the osculating plane and osculating circle of the curve at the given point.
γ ( t ) = ( 2 sin 3 t , t , 2 cos 3 t ) , ( 0 , π , − 2 ) \gamma(t) = (2 \sin 3t, t, 2 \cos 3t), (0, \pi, -2) γ ( t ) = ( 2 sin 3 t , t , 2 cos 3 t ) , ( 0 , π , − 2 ) Solution
r ¨ ( t ) = 2 sin 3 t i ⃗ + t j ⃗ + 2 cos 3 t k ⃗ r ′ → ( t ) = 6 cos 3 t i ⃗ + j ⃗ − 6 sin 3 t k ⃗ ∣ r ′ → ( t ) ∣ = ( 6 cos 3 t ) 2 + ( 1 ) 2 + ( − 6 sin 3 t ) 2 = = 36 cos 2 3 t + 36 sin 2 3 t + 1 = 36 + 1 = 37 \begin{array}{l}
\ddot{r}(t) = 2 \sin 3t \vec{i} + t \vec{j} + 2 \cos 3t \vec{k} \\
\overrightarrow{r'}(t) = 6 \cos 3t \vec{i} + \vec{j} - 6 \sin 3t \vec{k} \\
\left| \overrightarrow{r'}(t) \right| = \sqrt{(6 \cos 3t)^2 + (1)^2 + (-6 \sin 3t)^2} = \\
= \sqrt{36 \cos^2 3t + 36 \sin^2 3t + 1} = \sqrt{36 + 1} = \sqrt{37}
\end{array} r ¨ ( t ) = 2 sin 3 t i + t j + 2 cos 3 t k r ′ ( t ) = 6 cos 3 t i + j − 6 sin 3 t k ∣ ∣ r ′ ( t ) ∣ ∣ = ( 6 cos 3 t ) 2 + ( 1 ) 2 + ( − 6 sin 3 t ) 2 = = 36 cos 2 3 t + 36 sin 2 3 t + 1 = 36 + 1 = 37
Unit tangent vector
T ⃗ ( t ) = r ′ → ( t ) ∣ r ′ → ( t ) ∣ = 6 37 cos 3 t i ⃗ + 1 37 j ⃗ − 6 37 sin 3 t k ⃗ T ′ ⃗ ( t ) = − 18 37 sin 3 t i ⃗ + ( 0 ) j ⃗ − 18 37 cos 3 t k ⃗ ∣ T ′ → ( t ) ∣ = ( − 18 37 sin 3 t ) 2 + ( − 18 37 cos 3 t ) 2 = = 324 37 sin 2 3 t + 324 37 cos 2 3 t = 18 37 \begin{array}{l}
\vec{T}(t) = \frac{\overrightarrow{r'}(t)}{\left| \overrightarrow{r'}(t) \right|} = \frac{6}{\sqrt{37}} \cos 3t \vec{i} + \frac{1}{\sqrt{37}} \vec{j} - \frac{6}{\sqrt{37}} \sin 3t \vec{k} \\
\vec{T'}(t) = -\frac{18}{\sqrt{37}} \sin 3t \vec{i} + (0) \vec{j} - \frac{18}{\sqrt{37}} \cos 3t \vec{k} \\
\left| \overrightarrow{T'}(t) \right| = \sqrt{\left(-\frac{18}{\sqrt{37}} \sin 3t\right)^2 + \left(-\frac{18}{\sqrt{37}} \cos 3t\right)^2} = \\
= \sqrt{\frac{324}{37} \sin^2 3t + \frac{324}{37} \cos^2 3t} = \frac{18}{\sqrt{37}}
\end{array} T ( t ) = ∣ ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) = 37 6 cos 3 t i + 37 1 j − 37 6 sin 3 t k T ′ ( t ) = − 37 18 sin 3 t i + ( 0 ) j − 37 18 cos 3 t k ∣ ∣ T ′ ( t ) ∣ ∣ = ( − 37 18 sin 3 t ) 2 + ( − 37 18 cos 3 t ) 2 = = 37 324 sin 2 3 t + 37 324 cos 2 3 t = 37 18
Unit normal vector
N ⃗ ( t ) = T ′ → ( t ) ∣ T ′ → ( t ) ∣ = − sin 3 t i ⃗ − cos 3 t k ⃗ \vec{N}(t) = \frac{\overrightarrow{T'}(t)}{\left| \overrightarrow{T'}(t) \right|} = -\sin 3t \vec{i} - \cos 3t \vec{k} N ( t ) = ∣ ∣ T ′ ( t ) ∣ ∣ T ′ ( t ) = − sin 3 t i − cos 3 t k
"Binormal vector": unit normal vector to osculating plane
B ⃗ ( t ) = T ⃗ ( t ) × N ⃗ ( t ) B ⃗ ( t ) = ∣ i ⃗ j ⃗ k ⃗ 6 37 cos 3 t 1 37 − 6 37 sin 3 t − sin 3 t 0 − cos 3 t ∣ = = i ⃗ ( − cos 3 t 37 + 0 ) + j ⃗ ( 6 37 sin 2 3 t + 6 37 cos 2 3 t ) + k ⃗ ( 0 + sin 3 t 37 ) = = − cos 3 t 37 i ⃗ + 6 37 j ⃗ + sin 3 t 37 k ⃗ \begin{array}{l}
\vec{B}(t) = \vec{T}(t) \times \vec{N}(t) \\
\vec{B}(t) = \left| \begin{array}{ccc}
\vec{i} & \vec{j} & \vec{k} \\
\frac{6}{\sqrt{37}} \cos 3t & \frac{1}{\sqrt{37}} & -\frac{6}{\sqrt{37}} \sin 3t \\
-\sin 3t & 0 & -\cos 3t
\end{array} \right| = \\
= \vec{i} \left(-\frac{\cos 3t}{\sqrt{37}} + 0\right) + \vec{j} \left(\frac{6}{\sqrt{37}} \sin^2 3t + \frac{6}{\sqrt{37}} \cos^2 3t\right) + \vec{k} \left(0 + \frac{\sin 3t}{\sqrt{37}}\right) = \\
= -\frac{\cos 3t}{\sqrt{37}} \vec{i} + \frac{6}{\sqrt{37}} \vec{j} + \frac{\sin 3t}{\sqrt{37}} \vec{k}
\end{array} B ( t ) = T ( t ) × N ( t ) B ( t ) = ∣ ∣ i 37 6 cos 3 t − sin 3 t j 37 1 0 k − 37 6 sin 3 t − cos 3 t ∣ ∣ = = i ( − 37 c o s 3 t + 0 ) + j ( 37 6 sin 2 3 t + 37 6 cos 2 3 t ) + k ( 0 + 37 s i n 3 t ) = = − 37 c o s 3 t i + 37 6 j + 37 s i n 3 t k
Point P ( 0 , π , − 2 ) P(0, \pi, -2) P ( 0 , π , − 2 ) t = π t = \pi t = π
r ( π ) = { 0 , π , − 2 } r(\pi) = \{0, \pi, -2\} r ( π ) = { 0 , π , − 2 } r ′ → ( π ) = 6 cos ( 3 π ) i ⃗ + j ⃗ − 6 sin ( 3 π ) k ⃗ \overrightarrow{r'}(\pi) = 6 \cos(3\pi) \vec{i} + \vec{j} - 6 \sin(3\pi) \vec{k} r ′ ( π ) = 6 cos ( 3 π ) i + j − 6 sin ( 3 π ) k r ⃗ ′ ( π ) = ⟨ − 6 , 1 , 0 ⟩ B ⃗ ( t ) = − cos ( 3 π ) 37 i ⃗ + 6 37 j ⃗ + sin ( 3 π ) 37 k ⃗ B ⃗ ( t ) = ⟨ 1 37 , 6 37 , 0 ⟩ \begin{array}{l}
\vec{r}'(\pi) = \langle -6, 1, 0 \rangle \\
\vec{B}(t) = -\frac{\cos(3\pi)}{\sqrt{37}} \vec{i} + \frac{6}{\sqrt{37}} \vec{j} + \frac{\sin(3\pi)}{\sqrt{37}} \vec{k} \\
\vec{B}(t) = \langle \frac{1}{\sqrt{37}}, \frac{6}{\sqrt{37}}, 0 \rangle \\
\end{array} r ′ ( π ) = ⟨ − 6 , 1 , 0 ⟩ B ( t ) = − 37 c o s ( 3 π ) i + 37 6 j + 37 s i n ( 3 π ) k B ( t ) = ⟨ 37 1 , 37 6 , 0 ⟩
Osculating plane
a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 1 37 ( x − 0 ) + 6 37 ( y − π ) + ( 0 ) ( z − ( − 2 ) ) = 0 x + 6 y − 6 π = 0 or x + 6 y = 6 π \begin{array}{l}
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \\
\frac{1}{\sqrt{37}}(x - 0) + \frac{6}{\sqrt{37}}(y - \pi) + (0)(z - (-2)) = 0 \\
x + 6y - 6\pi = 0 \quad \text{or} \quad x + 6y = 6\pi \\
\end{array} a ( x − x 0 ) + b ( y − y 0 ) + c ( z − z 0 ) = 0 37 1 ( x − 0 ) + 37 6 ( y − π ) + ( 0 ) ( z − ( − 2 )) = 0 x + 6 y − 6 π = 0 or x + 6 y = 6 π
Osculating normal circle to the curve at the given point P P P P P P
with the same unit tangent vector as curve,
the same unit normal vector as curve,
and the same curvature as curve
Properties:
Circle has radius R R R
R = 1 curvature = 1 k ( P ) R = \frac{1}{\text{curvature}} = \frac{1}{k(P)} R = curvature 1 = k ( P ) 1
Center is distance 1 / k ( P ) 1/k(P) 1/ k ( P ) P P P N ⃗ ( P ) \vec{N}(P) N ( P )
r C ⃗ ( t ) = r ⃗ ( P ) + 1 k ( P ) N ⃗ ( P ) \vec{r_C}(t) = \vec{r}(P) + \frac{1}{k(P)} \vec{N}(P) r C ( t ) = r ( P ) + k ( P ) 1 N ( P ) k ( t ) = ∣ T ′ → ( t ) ∣ ∣ r ′ → ( t ) ∣ k ( t ) = 18 37 37 = 18 37 k ( π ) = 18 37 N ⃗ ( π ) = ⟨ 0 , 0 , 1 ⟩ T ⃗ ( π ) = ⟨ − 6 37 , 1 37 , 0 ⟩ \begin{array}{l}
k(t) = \frac{|\overrightarrow{T'}(t)|}{|\overrightarrow{r'}(t)|} \\
k(t) = \frac{\frac{18}{\sqrt{37}}}{\sqrt{37}} = \frac{18}{37} \\
k(\pi) = \frac{18}{37} \\
\vec{N}(\pi) = \langle 0, 0, 1 \rangle \\
\vec{T}(\pi) = \langle -\frac{6}{\sqrt{37}}, \frac{1}{\sqrt{37}}, 0 \rangle \\
\end{array} k ( t ) = ∣ r ′ ( t ) ∣ ∣ T ′ ( t ) ∣ k ( t ) = 37 37 18 = 37 18 k ( π ) = 37 18 N ( π ) = ⟨ 0 , 0 , 1 ⟩ T ( π ) = ⟨ − 37 6 , 37 1 , 0 ⟩
Osculating circle has radius
R = 1 18 / 37 = 37 18 R = \frac{1}{18/37} = \frac{37}{18} R = 18/37 1 = 18 37
Osculating circle has center
⟨ 0 , π , − 2 ⟩ + 37 18 ⟨ 0 , 0 , 1 ⟩ = ⟨ 0 , π , 1 18 ⟩ \langle 0, \pi, -2 \rangle + \frac{37}{18} \langle 0, 0, 1 \rangle = \langle 0, \pi, \frac{1}{18} \rangle ⟨ 0 , π , − 2 ⟩ + 18 37 ⟨ 0 , 0 , 1 ⟩ = ⟨ 0 , π , 18 1 ⟩
Osculating circle parametrized by
q ( θ ) = ⟨ 0 , π , 1 18 ⟩ + 37 18 ( cos θ ⟨ 0 , 0 , 1 ⟩ + sin θ ⟨ − 6 37 , 1 37 , 0 ⟩ ) q(\theta) = \langle 0, \pi, \frac{1}{18} \rangle + \frac{37}{18} \left( \cos \theta \langle 0, 0, 1 \rangle + \sin \theta \langle -\frac{6}{\sqrt{37}}, \frac{1}{\sqrt{37}}, 0 \rangle \right) q ( θ ) = ⟨ 0 , π , 18 1 ⟩ + 18 37 ( cos θ ⟨ 0 , 0 , 1 ⟩ + sin θ ⟨ − 37 6 , 37 1 , 0 ⟩ )
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#340153 on Dec 2023