Question Number 63298 by Rio Michael last updated on 02/Jul/19
$${A}\:{particle}\:{P},\:{moves}\:{on}\:{the}\:{curve}\:{with}\:{polar}\:{equation}\:\: \\ $$$${r}\:=\:{e}^{{k}\theta} \:,\:{where}\:\left({r},\theta\right)\:{are}\:{polar}\:{coordinates}\:{referred}\:{to}\:{a}\:{fixed} \\ $$$${pole}\:{and}\:{k}\:{is}\:{a}\:{positive}\:{constant}.\:{Given}\:{that}\:{the}\:{radial}\:{velocity} \\ $$$${of}\:{P}\:{is}\:\frac{{k}}{{r}}\:\:{show}\:{that}\:{the}\:{transverse}\:{acceleration}\:{of}\:{th}\:{particle} \\ $$$${is}\:{zero}. \\ $$$$ \\ $$
Commented by Prithwish sen last updated on 02/Jul/19
$$\mathrm{r}=\mathrm{e}^{\mathrm{k}\theta} \\ $$$$\mathrm{taking}\:\mathrm{log}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{logr}=\mathrm{k}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{r}}\:\frac{\mathrm{dr}}{\mathrm{d}\theta}\:=\:\mathrm{k} \\ $$$$\because\mathrm{Radial}\:\mathrm{vel}. \\ $$$$\frac{\mathrm{dr}}{\mathrm{dt}}\:=\:\frac{\mathrm{k}}{\mathrm{r}} \\ $$$$\therefore\:\frac{\mathrm{d}\theta}{\mathrm{dt}}\:=\:\frac{\mathrm{d}\theta}{\mathrm{dr}}.\frac{\mathrm{dr}}{\mathrm{dt}}\:=\:\frac{\mathrm{1}}{\mathrm{r}^{\mathrm{2}} }\:\Rightarrow\:\mathrm{r}^{\mathrm{2}} \frac{\mathrm{d}\theta}{\mathrm{dt}}\:=\:\mathrm{1} \\ $$$$\therefore\:\mathrm{Transverse}\:\mathrm{accl}. \\ $$$$\frac{\mathrm{1}}{\mathrm{r}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{r}^{\mathrm{2}} \frac{\mathrm{d}\theta}{\mathrm{dt}}\right)\:=\:\mathrm{0}\:\mathrm{proved} \\ $$$$\mathrm{please}\:\mathrm{check}. \\ $$
Commented by Rio Michael last updated on 02/Jul/19
$${perfect}! \\ $$