Question

Find the unit tangent vector at the indicated point of the vector function
r(t)=e^(12t)cost i+e^(12t)sint j+e^(12t) k

T(π/2)=<_,_,_>

Answer 1

We have-

$\mathbf{r(t)=e^{12t}cos(t)\;i+e^{12t}sin(t)\;j+e^{12t}\;k}$

$\mathbf{ \therefore \;v(t)=r'(t)=(12e^{12t}cos(t)+e^{12t}(-sin(t)))\;i}$

$\mathbf{+(12e^{12t}sin(t)+e^{12t}cos(t))\;j+12e^{12t}\; k}$

$\implies \mathbf{r'(t)=e^{12t}[(12cos(t)-sin(t))\;i+(12sin(t)+cos(t))\;j}$

$\mathbf{+12\;k]}$

and $\mathbf{\mid\mid v(t) \mid\mid=\sqrt{e^{24t}[(144.cos^2(t)+sin^2(t)-24.cos(t)sin(t))+}}$

$\mathbf{\sqrt{\\ \\+(144.sin^2(t)+cos^2(t)+24.cos(t)sin(t))+144]}}$

$\mathbf{\implies \mid\mid v(t) \mid\mid=e^{12t}\sqrt{144(cos^2(t)+sin^2(t))+(cos^2(t)+sin^2(t))+144}}$

$\mathbf{\implies \mid\mid v(t) \mid\mid=e^{12t}\sqrt{144+1+144}=e^{12t}\sqrt{289}=17e^{12t}}$

To find the unit tangent vector, we just divide

$\mathbf{T(t)=\dfrac{v(t)}{||v(t)||}=\dfrac{e^{12t}[(12cos(t)-sin(t))\;i+(12sin(t)+cos(t))\;j+12\;k]}{17e^{12t}}}$

$\mathbf{=\dfrac{[(12cos(t)-sin(t))\;i+(12sin(t)+cos(t))\;j+12\;k]}{17}}$

To find T(pi/2), plug in (pi/2) to get-

$\mathbf{T(\pi/2)=\dfrac{[(12cos(\pi/2)-sin(\pi/2))\;i+(12sin(\pi/2)+cos(\pi/2))\;j+12\;k]}{17}}$

$\mathbf{=\dfrac{(0-1)\;i+(12+0)\;j+12\;k}{17}=\dfrac{-\;i+12\;j+12\;k}{17}}$

$\mathbf{\implies T(\pi/2)=\bigg(\dfrac{-1}{17}\bigg)i+\bigg(\dfrac{12}{17}\bigg)j+\bigg(\dfrac{12}{17}\bigg)k}$ ....................Ans.