A-particle-P-moves-on-the-curve-with-polar-equation-r-e-k-where-r-are-polar-coordinates-referred-to-a-fixed-pole-and-k-is-a-positive-constant-Given-that-the-radial-velocity-of-P-is-k-

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Question Number 63298 by Rio Michael last updated on 02/Jul/19

$$AparticleP,movesonthecurvewithpolarequation$$$$r=e^{k\theta },where(r,\theta )arepolarcoordinatesreferredtoafixed$$$$poleandkisapositiveconstant.Giventhattheradialvelocity$$$$ofPis\frac{k}{r}showthatthetransverseaccelerationofthparticle$$$$iszero.$$$$$$

Commented by Prithwish sen last updated on 02/Jul/19

$$\mathrm{r}={\mathrm{e}}^{\mathrm{k}\theta }$$$$\mathrm{taking}\log \mathrm{both}\mathrm{sides}$$$$\mathrm{logr}=\mathrm{k}\theta$$$$\frac{1}{\mathrm{r}}\frac{\mathrm{dr}}{\mathrm{d}\theta }=\mathrm{k}$$$$\because \mathrm{Radial}\mathrm{vel}.$$$$\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{\mathrm{k}}{\mathrm{r}}$$$$\therefore \frac{\mathrm{d}\theta }{\mathrm{dt}}=\frac{\mathrm{d}\theta }{\mathrm{dr}}.\frac{\mathrm{dr}}{\mathrm{dt}}=\frac{1}{{\mathrm{r}}^2}\Rightarrow {\mathrm{r}}^2\frac{\mathrm{d}\theta }{\mathrm{dt}}=1$$$$\therefore \mathrm{Transverse}\mathrm{accl}.$$$$\frac{1}{\mathrm{r}}\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{r}}^2\frac{\mathrm{d}\theta }{\mathrm{dt}}\right)=0\mathrm{proved}$$$$\mathrm{please}\mathrm{check}.$$

Commented by Rio Michael last updated on 02/Jul/19

$$perfect!$$

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