Questions: Determine the value of t that corresponds to the point (1,0,1). t=0 Write a vector equation r(t) for the given curve. r(t)=(e^(-7 t) cos(7 t), e^(-7 t) sin(7 t), e^(-7 t)) Find r'(t). r'(t)=sqrt[(-7 e^(-7 t) sin(7 t)-7 e^(-7 t) cos(7 t),-7 e^(-7 t) sin(7 t)+7 e^(-7 t) cos(7 t),-7 e^(-7 t))](7) Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x=e^(-7 t) cos(7 t), y=e^(-7 t) sin(7 t), z=e^(-7 t), (1,0,1) (x(t), y(t), z(t))=(square)

Determine the value of t that corresponds to the point (1,0,1).
t=0

Write a vector equation r(t) for the given curve.
r(t)=(e^(-7 t) cos(7 t), e^(-7 t) sin(7 t), e^(-7 t))

Find r'(t).
r'(t)=sqrt[(-7 e^(-7 t) sin(7 t)-7 e^(-7 t) cos(7 t),-7 e^(-7 t) sin(7 t)+7 e^(-7 t) cos(7 t),-7 e^(-7 t))](7)

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x=e^(-7 t) cos(7 t), y=e^(-7 t) sin(7 t), z=e^(-7 t), (1,0,1)
(x(t), y(t), z(t))=(square)

Transcript text: Determine the value of $t$ that corresponds to the point $(1,0,1)$. \[ t=0 \] Write a vector equation $r(t)$ for the given curve. \[ r(t)=\left(e^{-7 t} \cos (7 t), e^{-7 t} \sin (7 t), e^{-7 t}\right) \] Find $\mathbf{r}^{\prime}(t)$. \[ \left.\mathbf{r}^{\prime}(t)=\sqrt\left[\left(-7 e^{-7 t} \sin (7 t)-7 e^{-7 t} \cos (7 t),-7 t^{-7 t} \sin (7 t)+7 e^{-7 t} \cos (7 t),-7 e^{-7 t}\right), ~\right) ~\right]{(7)} \] Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. \[ \begin{array}{c} x=e^{-7 t} \cos (7 t), \quad y=e^{-7 t} \sin (7 t), \quad z=e^{-7 t}, \quad(1,0,1) \\ (x(t), y(t), z(t))=(\square) \end{array} \]

Solution

Solution Steps

Step 1: Determine the value of $ t $ that corresponds to the point $ (1,0,1) $

Given the point $ (1,0,1) $ and the parametric equations: \[ x = e^{-7t} \cos(7t), \quad y = e^{-7t} \sin(7t), \quad z = e^{-7t}, \] we substitute $ (1,0,1) $ into the equations: \[ 1 = e^{-7t} \cos(7t), \quad 0 = e^{-7t} \sin(7t), \quad 1 = e^{-7t}. \] From $ z = e^{-7t} = 1 $, we solve for $ t $: \[ e^{-7t} = 1 \implies -7t = 0 \implies t = 0. \] Thus, $ t = 0 $ corresponds to the point $ (1,0,1) $.

Step 2: Write a vector equation $ r(t) $ for the given curve

The vector equation $ r(t) $ for the curve is given by: \[ r(t) = \left(e^{-7t} \cos(7t), e^{-7t} \sin(7t), e^{-7t}\right). \]

Step 3: Find $ \mathbf{r}^{\prime}(t) $

To find the derivative $ \mathbf{r}^{\prime}(t) $, differentiate each component of $ r(t) $ with respect to $ t $: \[ \mathbf{r}^{\prime}(t) = \left(\frac{d}{dt}\left(e^{-7t} \cos(7t)\right), \frac{d}{dt}\left(e^{-7t} \sin(7t)\right), \frac{d}{dt}\left(e^{-7t}\right)\right). \] Using the product rule and chain rule: \[ \frac{d}{dt}\left(e^{-7t} \cos(7t)\right) = -7e^{-7t} \cos(7t) - 7e^{-7t} \sin(7t), \] \[ \frac{d}{dt}\left(e^{-7t} \sin(7t)\right) = -7e^{-7t} \sin(7t) + 7e^{-7t} \cos(7t), \] \[ \frac{d}{dt}\left(e^{-7t}\right) = -7e^{-7t}. \] Thus: \[ \mathbf{r}^{\prime}(t) = \left(-7e^{-7t} \cos(7t) - 7e^{-7t} \sin(7t), -7e^{-7t} \sin(7t) + 7e^{-7t} \cos(7t), -7e^{-7t}\right). \]