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Question Number 20819 by ajfour last updated on 03/Sep/17

Commented by ajfour last updated on 03/Sep/17

$$Aparticlemovesalongthespiral$$$$shown.Determinethemagnitude$$$$ofthevelocityoftheparticlein$$$$termsof\mathit{b},\mathit{\theta },\stackrel{.}{\mathit{\theta }}.$$$$$$

Answered by dioph last updated on 04/Sep/17

$$\frac{d}{dt}r{\theta }^2=0$$$$\stackrel{.}{r}{\theta }^2+2r\theta \stackrel{.}{\theta }=0$$$$\stackrel{.}{r}=-\frac{2r\theta \stackrel{.}{\theta }}{{\theta }^2}=-\frac{2r\stackrel{.}{\theta }}{\theta }$$$$\stackrel{.}{r}=-\frac{2b\stackrel{.}{\theta }}{{\theta }^3}$$$$v=\sqrt{{\left(\stackrel{.}{r}\right)}^2+r^2{\left(\stackrel{.}{\theta }\right)}^2}$$$$v=\sqrt{\frac{4b^2}{{\theta }^6}{\left(\stackrel{.}{\theta }\right)}^2+\frac{b^2}{{\theta }^4}{\left(\stackrel{.}{\theta }\right)}^2}$$$$v=\frac{b\stackrel{.}{\theta }}{{\theta }^2}\sqrt{\frac{4}{{\theta }^2}+1}$$

Commented by ajfour last updated on 04/Sep/17

$$yessir,soitis.$$

Commented by ajfour last updated on 04/Sep/17

$$whatifthecurveis\mathit{r}={\mathit{b}\mathit{e}}^{{\mathit{\theta }}^2/2}?$$

Commented by dioph last updated on 04/Sep/17

$$\stackrel{.}{r}={be}^{{\theta }^2/2}\theta \stackrel{.}{\theta }$$$$v=\sqrt{{\left(\stackrel{.}{r}\right)}^2+r^2{\left(\stackrel{.}{\theta }\right)}^2}$$$$v=\sqrt{b^2{\theta }^2{e}^{{\theta }^2}{\left(\stackrel{.}{\theta }\right)}^2+b^2{e}^{{\theta }^2}{\left(\stackrel{.}{\theta }\right)}^2}$$$$v={be}^{{\theta }^2/2}\stackrel{.}{\theta }\sqrt{{\theta }^2+1}$$

Commented by dioph last updated on 04/Sep/17

$$\mathrm{indeed},\mathrm{thanks}\mathrm{for}\mathrm{pointing}\mathrm{out}.$$$$\mathrm{it}\mathrm{is}\mathrm{now}\mathrm{corrected}$$

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