Question

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34

$32.\mathrm{r}\left(t\right)=⟨\sec t,\frac{1}{t},e^t\ln t⟩$

Short answer

The velocity and acceleration vectors are;

$\begin{array}{rl}& v\left(t\right)=⟨\sec t\tan t,\frac{-1}{t^2},e^t(\frac{1}{t}+\ln t)⟩\\ & a\left(t\right)=⟨{\sec }^3+\sec t{\tan }^{2}t,\frac{2}{t^3},e^t(\frac{-1}{t^2}+\frac{2}{t}+\ln t)⟩\end{array}$

Step-by-step solution

Step 1. Given data

The given position vector is$\mathbf{r}\left(t\right)=⟨\sec t,\frac{1}{t},e^t\ln t⟩$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

$v\left(t\right)=\frac{d}{dt}r\left(t\right)=⟨\frac{d}{dt}x\left(t\right),\frac{d}{dt}y\left(t\right),\frac{d}{dt}z\left(t\right)⟩$

Therefore,

$\begin{array}{rl}v\left(t\right)& =\frac{d}{dt}r\left(t\right)\\ & =\frac{d}{dt}⟨\sec t,\frac{1}{t},e^t\ln t⟩\\ & =⟨\frac{d}{dt}\left(\sec t\right),\frac{d}{dt}\frac{1}{t},\frac{d}{dt}{e}^t\ln t⟩\\ & =⟨\sec t\tan t,\frac{-1}{t^2},\frac{e^t}{t}\left(\ln t\right)\left(e^t\right)⟩\\ & =⟨\sec t\tan t,\frac{-1}{t^2},e^t(\frac{1}{t}+\ln t)⟩\end{array}$

Hence the velocity vector$v\left(t\right)=⟨\sec t\tan t,\frac{-1}{t^2},e^t(\frac{1}{t}+\ln t)⟩$

Step 3. Acceleration vector

The acceleration vector a(t) is given by

$a\left(t\right)=\frac{d}{dt}v\left(t\right)$

Therefore,

$\begin{array}{rl}a\left(t\right)& =\frac{d}{dt}v\left(t\right)\\ & =\frac{d}{dt}⟨\sec t\tan t,\frac{-1}{t^2},e^t(\frac{1}{t}+\ln t)⟩\\ & =⟨{\sec }^{3}t+\sec t{\tan }^{2}t,\frac{2}{t^3},e^t(\frac{-1}{t^2}+\frac{1}{t})+(\frac{1}{t}+\ln t)e^t⟩\\ & =⟨{\sec }^{3}t+\sec t{\tan }^{2}t,\frac{2}{t^3},e^t(\frac{-1}{t^2}+\frac{2}{t}+\ln t)⟩\end{array}$

Hence the acceleration vector$a\left(t\right)=⟨{\sec }^{3}t+\sec t{\tan }^{2}t,\frac{2}{t^3},e^t(\frac{-1}{t^2}+\frac{2}{t}+\ln t)⟩$