The unit tangent vector T → \overrightarrow{T} T N → \overrightarrow{N} N and the unit binormal vector B → \overrightarrow{B} B are three mutually perpendicular vectors used to describe a curve in two or three dimensions. This moving coordinate system is attached to the curve and describes the shape of the curve independent of any parameterization.
If the curve is given parametrically by
x = x ( t ) y = y ( t ) x=x(t)\quad\quad y=y(t) x = x ( t ) y = y ( t )
the position, unit tangent, unit normal, and unit binormal vectors are
r → ( t ) = x ( t ) ⋅ i → + y ( t ) ⋅ j → T → = r → ′ ( t ) ∣ r → ′ ( t ) ∣ N → = T → ′ ( t ) ∣ T → ′ ( t ) ∣ B → = T → ( t ) × N → \overrightarrow{r}(t)=x(t)\cdot \overrightarrow{i}+y(t)\cdot \overrightarrow{j}\\[0.3cm]
\overrightarrow{T}=\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|} \\[0.3cm]
\overrightarrow{N}=\frac{\overrightarrow{T}'(t)}{\left|\overrightarrow{T}'(t)\right|} \\[0.3cm]
\overrightarrow{B}=\overrightarrow{T}(t)\times\overrightarrow{N}\\[0.3cm] r ( t ) = x ( t ) ⋅ i + y ( t ) ⋅ j T = ∣ ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) N = ∣ ∣ T ′ ( t ) ∣ ∣ T ′ ( t ) B = T ( t ) × N 1 case :
r → ( t ) = t 3 3 ⋅ i → + t 2 2 ⋅ j → , t > 0 r → ′ ( t ) = d d t ( t 3 3 ⋅ i → + t 2 2 ⋅ j → ) = 3 t 2 3 ⋅ i → + 2 t 2 ⋅ j → r → ′ ( t ) = t 2 ⋅ i → + t ⋅ j → ∣ r → ′ ( t ) ∣ = ( t 2 ) 2 + t 2 = t 2 ⋅ ( t 2 + 1 ) = ∣ t ∣ ⋅ t 2 + 1 ∣ r → ′ ( t ) ∣ = [ t > 0 ] = t ⋅ t 2 + 1 T → ( t ) = r → ′ ( t ) ∣ r → ′ ( t ) ∣ = t 2 ⋅ i → + t ⋅ j → t ⋅ t 2 + 1 = t 2 ⋅ i → t ⋅ t 2 + 1 + t ⋅ j → t ⋅ t 2 + 1 T → ( t ) = t ⋅ i → t 2 + 1 + j → t 2 + 1 T → ′ ( t ) = d d t ( t ⋅ i → t 2 + 1 + j → t 2 + 1 ) = = ( 1 t 2 + 1 + t ⋅ ( − 1 2 ⋅ 2 t ( t 2 + 1 ) 3 / 2 ) ) ⋅ i → + ( − 1 2 ⋅ 2 t ( t 2 + 1 ) 3 / 2 ) ⋅ j → = = ( t 2 + 1 − t 2 ( t 2 + 1 ) 3 / 2 ) ⋅ i → + ( − t ( t 2 + 1 ) 3 / 2 ) ⋅ j → = = ( 1 ( t 2 + 1 ) 3 / 2 ) ⋅ i → + ( − t ( t 2 + 1 ) 3 / 2 ) ⋅ j → ∣ T → ( t ) ∣ = ( 1 ( t 2 + 1 ) 3 / 2 ) 2 + ( t ( t 2 + 1 ) 3 / 2 ) 2 = = t 2 + 1 ( t 2 + 1 ) 3 = 1 ( t 2 + 1 ) 2 = 1 t 2 + 1 ⟶ 1 ∣ T → ′ ( t ) ∣ = ( t 2 + 1 ) N → ( t ) = T → ′ ( t ) ∣ T → ′ ( t ) ∣ = ( t 2 + 1 ) ⋅ ( ( 1 ( t 2 + 1 ) 3 / 2 ) ⋅ i → + ( − t ( t 2 + 1 ) 3 / 2 ) ⋅ j → ) = ( 1 ( t 2 + 1 ) 1 / 2 ) ⋅ i → + ( − t ( t 2 + 1 ) 1 / 2 ) ⋅ j → N → ( t ) = i → t 2 + 1 − t ⋅ j → t 2 + 1 B → = T → × N → = ∣ i → j → k → T x T y T z = 0 N x N y N z = 0 ∣ = k → ⋅ ( T x N y − T y N x ) = = k → ⋅ ( ( t t 2 + 1 ) ⋅ ( − t t 2 + 1 ) − ( 1 t 2 + 1 ) ⋅ ( 1 t 2 + 1 ) ) = = k → ⋅ ( − ( t 2 + 1 ) t 2 + 1 ) = − 1 ⋅ k → B → = − 1 ⋅ k → \overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t>0\\[0.3cm]
\overrightarrow{r}'(t)=\frac{d}{dt}\left(\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j}\right)=\frac{3t^2}{3}\cdot\overrightarrow{i}+\frac{2t}{2}\cdot\overrightarrow{j}\\[0.3cm]
\boxed{\overrightarrow{r}'(t)=t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}\\[0.3cm]
\left|\overrightarrow{r}'(t)\right|=\sqrt{(t^2)^2+t^2}=\sqrt{t^2\cdot\left(t^2+1\right)}=|t|\cdot\sqrt{t^2+1}\\[0.3cm]
\left|\overrightarrow{r}'(t)\right|=[t>0]=t\cdot\sqrt{t^2+1}\\[0.3cm]
\overrightarrow{T}(t)=\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|}=\frac{t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}{t\cdot\sqrt{t^2+1}}=\frac{t^2\cdot\overrightarrow{i}}{t\cdot\sqrt{t^2+1}}+\frac{t\cdot\overrightarrow{j}}{t\cdot\sqrt{t^2+1}}\\[0.3cm]
\boxed{\overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}}\\[0.3cm]
\overrightarrow{T}'(t)=\frac{d}{dt}\left(\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}\right)=\\[0.3cm]
=\left(\frac{1}{\sqrt{t^2+1}}+t\cdot\left(-\frac{1}{2}\cdot\frac{2t}{\left(t^2+1\right)^{3/2}}\right)\right)\cdot\overrightarrow{i}+\left(-\frac{1}{2}\cdot\frac{2t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}=\\[0.3cm]
=\left(\frac{t^2+1-t^2}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}=\\[0.3cm]
=\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}\\[0.3cm]
\left|\overrightarrow{T}(t)\right|=\sqrt{\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)^2+\left(\frac{t}{\left(t^2+1\right)^{3/2}}\right)^2}=\\[0.3cm]
=\sqrt{\frac{t^2+1}{\left(t^2+1\right)^3}}=\sqrt{\frac{1}{\left(t^2+1\right)^2}}=\frac{1}{t^2+1}\longrightarrow\frac{1}{\left|\overrightarrow{T}'(t)\right|}=(t^2+1)\\[0.3cm]
\overrightarrow{N}(t)=\frac{\overrightarrow{T}'(t)}{\left|\overrightarrow{T}'(t)\right|}=(t^2+1)\cdot\left(\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}\right)\\[0.3cm]
=\left(\frac{1}{\left(t^2+1\right)^{1/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{1/2}}\right)\cdot\overrightarrow{j}\\[0.3cm]
\boxed{\overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}}}\\[0.3cm]
\overrightarrow{B}=\overrightarrow{T}\times\overrightarrow{N}=\left|\begin{array}{ccc}
\overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\
T_x&T_y&T_z=0\\
N_x&N_y&N_z=0
\end{array}\right|=\overrightarrow{k}\cdot(T_xN_y-T_yN_x)=\\[0.3cm]
=\overrightarrow{k}\cdot\left(\left(\frac{t}{\sqrt{t^2+1}}\right)\cdot\left(\frac{-t}{\sqrt{t^2+1}}\right)-\left(\frac{1}{\sqrt{t^2+1}}\right)\cdot\left(\frac{1}{\sqrt{t^2+1}}\right)\right)=\\[0.3cm]
=\overrightarrow{k}\cdot\left(\frac{-(t^2+1)}{t^2+1}\right)=-1\cdot\overrightarrow{k}\\[0.3cm]
\boxed{\overrightarrow{B}=-1\cdot\overrightarrow{k}} r ( t ) = 3 t 3 ⋅ i + 2 t 2 ⋅ j , t > 0 r ′ ( t ) = d t d ( 3 t 3 ⋅ i + 2 t 2 ⋅ j ) = 3 3 t 2 ⋅ i + 2 2 t ⋅ j r ′ ( t ) = t 2 ⋅ i + t ⋅ j ∣ ∣ r ′ ( t ) ∣ ∣ = ( t 2 ) 2 + t 2 = t 2 ⋅ ( t 2 + 1 ) = ∣ t ∣ ⋅ t 2 + 1 ∣ ∣ r ′ ( t ) ∣ ∣ = [ t > 0 ] = t ⋅ t 2 + 1 T ( t ) = ∣ ∣ r ′ ( t ) ∣ ∣ r ′ ( t ) = t ⋅ t 2 + 1 t 2 ⋅ i + t ⋅ j = t ⋅ t 2 + 1 t 2 ⋅ i + t ⋅ t 2 + 1 t ⋅ j T ( t ) = t 2 + 1 t ⋅ i + t 2 + 1 j T ′ ( t ) = d t d ( t 2 + 1 t ⋅ i + t 2 + 1 j ) = = ( t 2 + 1 1 + t ⋅ ( − 2 1 ⋅ ( t 2 + 1 ) 3/2 2 t ) ) ⋅ i + ( − 2 1 ⋅ ( t 2 + 1 ) 3/2 2 t ) ⋅ j = = ( ( t 2 + 1 ) 3/2 t 2 + 1 − t 2 ) ⋅ i + ( − ( t 2 + 1 ) 3/2 t ) ⋅ j = = ( ( t 2 + 1 ) 3/2 1 ) ⋅ i + ( − ( t 2 + 1 ) 3/2 t ) ⋅ j ∣ ∣ T ( t ) ∣ ∣ = ( ( t 2 + 1 ) 3/2 1 ) 2 + ( ( t 2 + 1 ) 3/2 t ) 2 = = ( t 2 + 1 ) 3 t 2 + 1 = ( t 2 + 1 ) 2 1 = t 2 + 1 1 ⟶ ∣ ∣ T ′ ( t ) ∣ ∣ 1 = ( t 2 + 1 ) N ( t ) = ∣ ∣ T ′ ( t ) ∣ ∣ T ′ ( t ) = ( t 2 + 1 ) ⋅ ( ( ( t 2 + 1 ) 3/2 1 ) ⋅ i + ( − ( t 2 + 1 ) 3/2 t ) ⋅ j ) = ( ( t 2 + 1 ) 1/2 1 ) ⋅ i + ( − ( t 2 + 1 ) 1/2 t ) ⋅ j N ( t ) = t 2 + 1 i − t 2 + 1 t ⋅ j B = T × N = ∣ ∣ i T x N x j T y N y k T z = 0 N z = 0 ∣ ∣ = k ⋅ ( T x N y − T y N x ) = = k ⋅ ( ( t 2 + 1 t ) ⋅ ( t 2 + 1 − t ) − ( t 2 + 1 1 ) ⋅ ( t 2 + 1 1 ) ) = = k ⋅ ( t 2 + 1 − ( t 2 + 1 ) ) = − 1 ⋅ k B = − 1 ⋅ k
2 case :
r → ( t ) = t 3 3 ⋅ i → + t 2 2 ⋅ j → , t ≥ 0 r → ′ ( t ) = d d t ( t 3 3 ⋅ i → + t 2 2 ⋅ j → ) = 3 t 2 3 ⋅ i → + 2 t 2 ⋅ j → r → ′ ( t ) = t 2 ⋅ i → + t ⋅ j → ∣ r → ′ ( t ) ∣ = ( t 2 ) 2 + t 2 = t 2 ⋅ ( t 2 + 1 ) = ∣ t ∣ ⋅ t 2 + 1 ∣ r → ′ ( t ) ∣ = [ t ≥ 0 ] = t ⋅ t 2 + 1 \overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t\ge0\\[0.3cm]
\overrightarrow{r}'(t)=\frac{d}{dt}\left(\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j}\right)=\frac{3t^2}{3}\cdot\overrightarrow{i}+\frac{2t}{2}\cdot\overrightarrow{j}\\[0.3cm]
\boxed{\overrightarrow{r}'(t)=t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}\\[0.3cm]
\left|\overrightarrow{r}'(t)\right|=\sqrt{(t^2)^2+t^2}=\sqrt{t^2\cdot\left(t^2+1\right)}=|t|\cdot\sqrt{t^2+1}\\[0.3cm]
\left|\overrightarrow{r}'(t)\right|=[t\ge0]=t\cdot\sqrt{t^2+1}\\[0.3cm] r ( t ) = 3 t 3 ⋅ i + 2 t 2 ⋅ j , t ≥ 0 r ′ ( t ) = d t d ( 3 t 3 ⋅ i + 2 t 2 ⋅ j ) = 3 3 t 2 ⋅ i + 2 2 t ⋅ j r ′ ( t ) = t 2 ⋅ i + t ⋅ j ∣ ∣ r ′ ( t ) ∣ ∣ = ( t 2 ) 2 + t 2 = t 2 ⋅ ( t 2 + 1 ) = ∣ t ∣ ⋅ t 2 + 1 ∣ ∣ r ′ ( t ) ∣ ∣ = [ t ≥ 0 ] = t ⋅ t 2 + 1 There is a problem with the parameter value t = 0 t=0 t = 0
r → ′ ( 0 ) = 0 2 ⋅ i → + 0 ⋅ j → ∣ r → ′ ( 0 ) ∣ = 0 ⋅ 0 2 + 1 = 0 T → ( 0 ) = 0 0 ⋅ i → + 0 0 ⋅ j → − formal entry \overrightarrow{r}'(0)=0^2\cdot\overrightarrow{i}+0\cdot\overrightarrow{j}\\[0.3cm]
\left|\overrightarrow{r}'(0)\right|=0\cdot\sqrt{0^2+1}=0\\[0.3cm]
\overrightarrow{T}(0)=\frac{0}{0}\cdot\overrightarrow{i}+\frac{0}{0}\cdot\overrightarrow{j}-\text{formal entry} r ′ ( 0 ) = 0 2 ⋅ i + 0 ⋅ j ∣ ∣ r ′ ( 0 ) ∣ ∣ = 0 ⋅ 0 2 + 1 = 0 T ( 0 ) = 0 0 ⋅ i + 0 0 ⋅ j − formal entry
It means that ∄ T → ( 0 ) \nexists\overrightarrow{T}(0) ∄ T ( 0 ) ∄ N → ( 0 ) \nexists\overrightarrow{N}(0) ∄ N ( 0 ) ∄ B → ( 0 ) \nexists\overrightarrow{B}(0) ∄ B ( 0 )
ANSWER
r → ( t ) = t 3 3 ⋅ i → + t 2 2 ⋅ j → , t > 0 T → ( t ) = t ⋅ i → t 2 + 1 + j → t 2 + 1 N → ( t ) = i → t 2 + 1 − t ⋅ j → t 2 + 1 B → = − 1 ⋅ k → r → ( t ) = t 3 3 ⋅ i → + t 2 2 ⋅ j → , t ≥ 0 T → ( t ) = t ⋅ i → t 2 + 1 + j → t 2 + 1 , t > 0 and ∄ T → ( 0 ) N → ( t ) = i → t 2 + 1 − t ⋅ j → t 2 + 1 , t > 0 and ∄ N → ( 0 ) B → = − 1 ⋅ k → , t > 0 and ∄ B → ( 0 ) \overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t>0\\[0.3cm]
\overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}\\[0.3cm]
\overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}}\\[0.3cm]
\overrightarrow{B}=-1\cdot\overrightarrow{k}\\[0.3cm]
\overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t\ge0\\[0.3cm]
\overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{T}(0)\\[0.3cm]
\overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{N}(0)\\[0.3cm]
\overrightarrow{B}=-1\cdot\overrightarrow{k},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{B}(0) r ( t ) = 3 t 3 ⋅ i + 2 t 2 ⋅ j , t > 0 T ( t ) = t 2 + 1 t ⋅ i + t 2 + 1 j N ( t ) = t 2 + 1 i − t 2 + 1 t ⋅ j B = − 1 ⋅ k r ( t ) = 3 t 3 ⋅ i + 2 t 2 ⋅ j , t ≥ 0 T ( t ) = t 2 + 1 t ⋅ i + t 2 + 1 j , t > 0 and ∄ T ( 0 ) N ( t ) = t 2 + 1 i − t 2 + 1 t ⋅ j , t > 0 and ∄ N ( 0 ) B = − 1 ⋅ k , t > 0 and ∄ B ( 0 )
#339914 on Dec 2023