Answer to Question #117435 in Calculus for Olivia

Question

Find parametric equations for the tangent line to the curve r(t) =e^(−t)〈cos(t),sin(t),1〉at t= 0.

Answer 1

The parametric equation of the tangent, for $t=t_0$ , to the line given parametrically $r(t)$ has the form

$$r_{tan}(t)=r\left(t_0\right)+t\cdot r'\left(t_0\right)$$

or in coordinate form

$$r_{tan}(t)=\left\{\begin{array}{l} x_{tan}(t)=x\left(t_0\right)+t\cdot x'\left(t_0\right)\\[0.3cm] y_{tan}(t)=y\left(t_0\right)+t\cdot y'\left(t_0\right)\\[0.3cm] z_{tan}(t)=z\left(t_0\right)+t\cdot z'\left(t_0\right) \end{array}\right.$$

In our case,

$$r(t)=e^{-t}\cdot\langle\cos t,\sin t,1\rangle\longrightarrow\\[0.3cm] r(0)=e^0\cdot\langle\cos0,\sin0,1\rangle\longrightarrow\boxed{r(0)=\langle1,0,1\rangle}\\[0.3cm] r'(t)=-e^{-t}\cdot\langle\cos t,\sin t,1\rangle+e^{-t}\cdot\langle-\sin t,\cos t,0\rangle\\[0.3cm] r'(0)=-e^{-0}\cdot\langle\cos0,\sin0,1\rangle+e^{-0}\cdot\langle-\sin 0,\cos 0,0\rangle=\\[0.3cm] =-1\cdot\langle1,0,1\rangle+1\cdot\langle0,1,0\rangle=\langle-1,1,-1\rangle\\[0.3cm] \boxed{r'(0)=\langle-1,1,-1\rangle}$$

Then, The parametric equation of the tangent is

$$r_{tan}(t)=\langle1,0,1\rangle+t\cdot\langle-1,1,-1\rangle=\langle1-t,t,1-t\rangle\\[0.3cm] \boxed{r_{tan}(t)=\langle1-t,t,1-t\rangle}$$

or in coordinate form

$$r_{tan}(t)=\left\{\begin{array}{l} x_{tan}(t)=1-t\\[0.3cm] y_{tan}(t)=t\\[0.3cm] z_{tan}(t)=1-t \end{array}\right.$$

ANSWER

Vecotr form

$$r_{tan}(t)=\langle1-t,t,1-t\rangle$$

or in coordinate form

$$r_{tan}(t)=\left\{\begin{array}{l} x_{tan}(t)=1-t\\[0.3cm] y_{tan}(t)=t\\[0.3cm] z_{tan}(t)=1-t \end{array}\right.$$