Question

A decomposition of the acceleration vector: Find $com{p}_{\mathbf{v}\mathbf{\left(}\mathbf{t}\mathbf{\right)}}\mathit{a}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, where v and a are the velocity and acceleration vectors, respectively, of the following functions.

Short answer

$Thecomponentofa\left(t\right)alongv\left(t\right)is\sqrt{3}{e}^t.$

Step-by-step solution

Step 1. Given Information.

$$\begin{gathered} Given, \\ x-coordinate:e^t\sin t, \\ y-coordinate:e^t\cos t, \\ z-coordinate:e^t. \end{gathered}$$

Step 2. Calculating the velocity and acceleration vector.

Step 3. Finally calculating the component of a(t) along v(t).

$$\begin{gathered} Fromstep2wegetv\left(t\right)anda\left(t\right). \\ Nowcom{p}_{v\left(t\right)}a\left(t\right)=\frac{\left|a\left(t\right).v\left(t\right)\right|}{\left|\left|v\left(t\right)\right|\right|} \\ Nowfrom\left(i\right)and\left(ii\right)weget \\ com{p}_{v\left(t\right)}a\left(t\right)=\frac{2e^t\cos t\left(e^t\cos t+e^t\sin t\right)+\left(-2e^t\sin t\left(e^t\cos t-e^t\sin t\right)\right)+e^t{e}^t}{\sqrt{{\left(e^t\cos t+e^t\sin t\right)}^2+{\left(e^t\cos t-e^t\sin t\right)}^2+{\left(e^t\right)}^2}} \\ Solvingtheaboveexpressionweget, \\ =\frac{e^{2t}\left(2{\cos }^{2}t+2\cos t\sin t-2\cos t\sin t+2{\sin }^{2}t+1\right)}{e^t\sqrt{{\cos }^{2}t+{\sin }^{2}t+2\cos t\sin t+{\cos }^{2}t+{\sin }^{2}t-2\cos t\sin t+1}} \\ =\frac{3e^{2t}}{e^t\sqrt{3}}=\sqrt{3}{e}^t. \\ Sothecomponentofa\left(t\right)alongthedirectionofv\left(t\right)willbe\sqrt{3}{e}^t. \end{gathered}$$