The parametric equation of the tangent, for t = t 0 t=t_0 t = t 0 r ( t ) r(t) r ( t )
r t a n ( t ) = r ( t 0 ) + t ⋅ r ′ ( t 0 ) r_{tan}(t)=r\left(t_0\right)+t\cdot r'\left(t_0\right) r t an ( t ) = r ( t 0 ) + t ⋅ r ′ ( t 0 )
or in coordinate form
r t a n ( t ) = { x t a n ( t ) = x ( t 0 ) + t ⋅ x ′ ( t 0 ) y t a n ( t ) = y ( t 0 ) + t ⋅ y ′ ( t 0 ) z t a n ( t ) = z ( t 0 ) + t ⋅ z ′ ( t 0 ) r_{tan}(t)=\left\{\begin{array}{l}
x_{tan}(t)=x\left(t_0\right)+t\cdot x'\left(t_0\right)\\[0.3cm]
y_{tan}(t)=y\left(t_0\right)+t\cdot y'\left(t_0\right)\\[0.3cm]
z_{tan}(t)=z\left(t_0\right)+t\cdot z'\left(t_0\right)
\end{array}\right. r t an ( t ) = ⎩ ⎨ ⎧ x t an ( t ) = x ( t 0 ) + t ⋅ x ′ ( t 0 ) y t an ( t ) = y ( t 0 ) + t ⋅ y ′ ( t 0 ) z t an ( t ) = z ( t 0 ) + t ⋅ z ′ ( t 0 )
In our case,
r ( t ) = e − t ⋅ ⟨ cos t , sin t , 1 ⟩ ⟶ r ( 0 ) = e 0 ⋅ ⟨ cos 0 , sin 0 , 1 ⟩ ⟶ r ( 0 ) = ⟨ 1 , 0 , 1 ⟩ r ′ ( t ) = − e − t ⋅ ⟨ cos t , sin t , 1 ⟩ + e − t ⋅ ⟨ − sin t , cos t , 0 ⟩ r ′ ( 0 ) = − e − 0 ⋅ ⟨ cos 0 , sin 0 , 1 ⟩ + e − 0 ⋅ ⟨ − sin 0 , cos 0 , 0 ⟩ = = − 1 ⋅ ⟨ 1 , 0 , 1 ⟩ + 1 ⋅ ⟨ 0 , 1 , 0 ⟩ = ⟨ − 1 , 1 , − 1 ⟩ r ′ ( 0 ) = ⟨ − 1 , 1 , − 1 ⟩ r(t)=e^{-t}\cdot\langle\cos t,\sin t,1\rangle\longrightarrow\\[0.3cm]
r(0)=e^0\cdot\langle\cos0,\sin0,1\rangle\longrightarrow\boxed{r(0)=\langle1,0,1\rangle}\\[0.3cm]
r'(t)=-e^{-t}\cdot\langle\cos t,\sin t,1\rangle+e^{-t}\cdot\langle-\sin t,\cos t,0\rangle\\[0.3cm]
r'(0)=-e^{-0}\cdot\langle\cos0,\sin0,1\rangle+e^{-0}\cdot\langle-\sin 0,\cos 0,0\rangle=\\[0.3cm]
=-1\cdot\langle1,0,1\rangle+1\cdot\langle0,1,0\rangle=\langle-1,1,-1\rangle\\[0.3cm]
\boxed{r'(0)=\langle-1,1,-1\rangle} r ( t ) = e − t ⋅ ⟨ cos t , sin t , 1 ⟩ ⟶ r ( 0 ) = e 0 ⋅ ⟨ cos 0 , sin 0 , 1 ⟩ ⟶ r ( 0 ) = ⟨ 1 , 0 , 1 ⟩ r ′ ( t ) = − e − t ⋅ ⟨ cos t , sin t , 1 ⟩ + e − t ⋅ ⟨ − sin t , cos t , 0 ⟩ r ′ ( 0 ) = − e − 0 ⋅ ⟨ cos 0 , sin 0 , 1 ⟩ + e − 0 ⋅ ⟨ − sin 0 , cos 0 , 0 ⟩ = = − 1 ⋅ ⟨ 1 , 0 , 1 ⟩ + 1 ⋅ ⟨ 0 , 1 , 0 ⟩ = ⟨ − 1 , 1 , − 1 ⟩ r ′ ( 0 ) = ⟨ − 1 , 1 , − 1 ⟩
Then, The parametric equation of the tangent is
r t a n ( t ) = ⟨ 1 , 0 , 1 ⟩ + t ⋅ ⟨ − 1 , 1 , − 1 ⟩ = ⟨ 1 − t , t , 1 − t ⟩ r t a n ( t ) = ⟨ 1 − t , t , 1 − t ⟩ r_{tan}(t)=\langle1,0,1\rangle+t\cdot\langle-1,1,-1\rangle=\langle1-t,t,1-t\rangle\\[0.3cm]
\boxed{r_{tan}(t)=\langle1-t,t,1-t\rangle} r t an ( t ) = ⟨ 1 , 0 , 1 ⟩ + t ⋅ ⟨ − 1 , 1 , − 1 ⟩ = ⟨ 1 − t , t , 1 − t ⟩ r t an ( t ) = ⟨ 1 − t , t , 1 − t ⟩
or in coordinate form
r t a n ( t ) = { x t a n ( t ) = 1 − t y t a n ( t ) = t z t a n ( t ) = 1 − t r_{tan}(t)=\left\{\begin{array}{l}
x_{tan}(t)=1-t\\[0.3cm]
y_{tan}(t)=t\\[0.3cm]
z_{tan}(t)=1-t
\end{array}\right. r t an ( t ) = ⎩ ⎨ ⎧ x t an ( t ) = 1 − t y t an ( t ) = t z t an ( t ) = 1 − t
ANSWER
Vecotr form
r t a n ( t ) = ⟨ 1 − t , t , 1 − t ⟩ r_{tan}(t)=\langle1-t,t,1-t\rangle r t an ( t ) = ⟨ 1 − t , t , 1 − t ⟩
or in coordinate form
r t a n ( t ) = { x t a n ( t ) = 1 − t y t a n ( t ) = t z t a n ( t ) = 1 − t r_{tan}(t)=\left\{\begin{array}{l}
x_{tan}(t)=1-t\\[0.3cm]
y_{tan}(t)=t\\[0.3cm]
z_{tan}(t)=1-t
\end{array}\right. r t an ( t ) = ⎩ ⎨ ⎧ x t an ( t ) = 1 − t y t an ( t ) = t z t an ( t ) = 1 − t
#339914 on Dec 2023
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