Answer to Question #173400 in Differential Geometry | Topology for Nestor Freeman

Question

A particle moves so that its position vector at time t is given by r =e−t(costi +sintj). Show that at any time t, (a) its velocity v is inclined to the vector r at a constant angle 3 (b) its acceleration vector is at right angles to the vector r.

Answer 1

$\text{r}=(e^{-t}\cos t)\text{i}+(e^{-t}\sin t)\text{j}$

$\text{v}=(-e^{-t}\cos t-e^{-t}\sin t)\text{i}$

$+(-e^{-t}\sin t+e^{-t}\cos t)\text{j}$

$\text{a}=(2e^{-t}\sin t)\text{i}+(-2e^{-t}\cos t)\text{j}$

a)

$\text{r}\cdot \text{v}=(e^{-t}\cos t)(-e^{-t}\cos t-e^{-t}\sin t)$

$+(e^{-t}\sin t)(-e^{-t}\sin t+e^{-t}\cos t)$

$=-e^{-2t}\cos^2t-e^{-2t}\cos t\sin t$

$-e^{-2t}\sin^2t+e^{-2t}\cos t\sin t=-e^{-2t}$

$|\text{r}|=\sqrt{(e^{-t}\cos t)^2+(e^{-t}\sin t)^2 }=e^{-t}$

$(-e^{-t}\cos t-e^{-t}\sin t)^2+(-e^{-t}\sin t+e^{-t}\cos t)^2$

$=e^{-2t}\cos^2t+2e^{-2t}\cos t\sin t+e^{-2t}\sin^2t$

$e^{-2t}\cos^2t-2e^{-2t}\cos t\sin t+e^{-2t}\sin^2t=2e^{-2t}$

$|\text{v}|=\sqrt{2e^{-2t} }=\sqrt{2}e^{-t}$

$\cos\angle(\text{r}, \text{v})=\dfrac{\text{r}\cdot \text{v}}{|\text{r}||\text{v}|}$

$=\dfrac{-e^{-2t}}{e^{-t}\sqrt{2}e^{-t}}=-\dfrac{\sqrt{2}}{2}$

$\angle(\text{r}, \text{v})=\dfrac{3\pi}{4}=\text{const}$

b)

$\text{r}\cdot \text{a}=(e^{-t}\cos t)(2e^{-t}\sin t)$

$+(e^{-t}\sin t)(-2e^{-t}\cos t)=0$

Hence acceleration vector $\text{a}$ is at right angles to the vector $\text{r}$.