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202408221017 Exercise 5.68

A particle moves along the space curve

\mathbf{r}=e^{-t}\cos t\,\hat{\mathbf{i}}+e^{-t}\sin t\,\hat{\mathbf{j}}+e^{-t}\,\hat{\mathbf{k}}.

Find the magnitude of the (a) velocity and (b) acceleration at any time t.

Extracted from Spiegel, M. R. (1971). Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists.

Roughwork.

Let displacement

\mathbf{r}=r_x\,\hat{\mathbf{i}}+r_y\,\hat{\mathbf{j}}+r_z\,\hat{\mathbf{k}}

where

\begin{aligned} r_x(t) & = e^{-t}\cos t \\ r_y(t) & = e^{-t}\sin t \\ r_z(t) & = e^{-t} \\ \end{aligned}

then velocity

\mathbf{v}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}}+v_z\,\hat{\mathbf{k}}

where

\begin{aligned} v_x & = \frac{\mathrm{d}r_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t) \\ v_x(t)& = -e^{-t}\sin t-e^{-t}\cos t \\ v_y & = \frac{\mathrm{d}r_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\sin t) \\ v_y(t)& = e^{-t}\cos t-e^{-t}\sin t \\ v_z & = \frac{\mathrm{d}r_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}) \\ v_z(t)& = -e^{-t} \\ \end{aligned}

then acceleration

\mathbf{a}=a_x\,\hat{\mathbf{i}}+a_y\,\hat{\mathbf{j}}+a_z\,\hat{\mathbf{k}}

where

\begin{aligned} a_x & = \frac{\mathrm{d}v_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}\sin t-e^{-t}\cos t) \\ a_x(t)& = 2e^{-t}\sin t\\ a_y & = \frac{\mathrm{d}v_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t-e^{-t}\sin t) \\ a_y(t)& = -2e^{-t}\cos t\\ a_z & = \frac{\mathrm{d}v_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}) \\ a_z(t)& = e^{-t} \\ \end{aligned}

Cheat.

\mathbf{r}(t) = \underbrace{e^{-t}}_{\textrm{(1)}}\underbrace{(\cos t,\sin t, 1)}_{\textrm{(2)}}

where

\begin{aligned} \textrm{(1):}&\enspace \textrm{radial centripetal} \\ \textrm{(2):}&\enspace \textrm{circumferential anticlockwise} \\ \end{aligned}

By courtesy of WolframAlpha

In a cylindrical coordinate system, the position of a particle can be written as

\mathbf{r}=\rho\,\hat{\boldsymbol{\rho}}+z\,\hat{\mathbf{z}}.

The velocity of the particle is the time derivative of its position,

\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}}=\dot{\rho}\,\hat{\boldsymbol{\rho}}+\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}+\dot{z}\,\hat{\mathbf{z}},

where the term \rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}} comes from the Poisson formula

\displaystyle{\frac{\mathrm{d}\hat{\boldsymbol{\varphi}}}{\mathrm{d}t}}=\dot{\varphi}\,\hat{\mathbf{z}}\times\hat{\boldsymbol{\rho}}.

Its acceleration is

\mathbf{a}=\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=(\ddot{\rho}-\rho\dot{\varphi}^2)\,\hat{\boldsymbol{\rho}}+(2\dot{\rho}\dot{\varphi}+\rho\ddot{\varphi})\,\hat{\boldsymbol{\varphi}}+\ddot{z}\,\hat{\mathbf{z}}.

Kinematics in Wikipedia on Cylindrical coordinate system

Stop wandering! You are facing up to

\begin{aligned} v=|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2} & = \cdots \\ a=|\mathbf{a}|=\sqrt{a_x^2+a_y^2+a_z^2} & = \cdots \\ \end{aligned}

But then, I’m fond of opting out.

This problem is not to be attempted.