Question

A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.​

Answer 1

Answer:

The speed of this particle is constantly $c$.

Explanation:

Position vector of this particle at time $t$:

$\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}$.

Write $\mathbf{r}(t)$ as a column vector to distinguish between the components:

$\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}$.

Both $b$ and $Q$ are constants. Therefore, $b\, \cos(Q)$ and $b \sin (Q)$ would also be constants with respect to $t$. Hence, $\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0$ and $\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0$.

Differentiate $\mathbf{r}(t)$ (component-wise) with respect to time $t$ to find the velocity vector of this particle at time $t\!$:

$\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}$.

The speed $v$ (a scalar) of a particle is the magnitude of its velocity :

$\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}$.

Therefore, the speed of this particle is constantly $c$ (a constant.)