$(a)$ $\vec{r}(t)=A \cos \theta t \hat{i}+B \sin \omega t \hat{j}$ ને

$\vec{r}(t)=x \hat{i}+y \hat{j}$ સાથે સરખાવતાં

$x=A \cos \omega t$ અને $y=B \sin \omega t$

$x$$=A \cos \omega t$ અને $y=B \sin \omega t$

$\frac{x}{ A }$$=\cos \omega t$ અને $\frac{y}{ B }=\sin \omega t$

$\therefore \quad \frac{x^{2}}{ A ^{2}}+\frac{y^{2}}{ B ^{2}}=\cos ^{2} \omega t+\sin ^{2} \omega t$

$\therefore \frac{x^{2}}{ A ^{2}}+\frac{y^{2}}{ B ^{2}}=1$ જેઉપવલયનું પ્રમાણિત સમીકરણ છે.

તેથી કણનો ગતિપથ ઉપવલયાકાર છે.

$(b)$

$\vec{r}(t)=A \cos \omega t \hat{i}+B \sin \omega t \hat{j}$

$\frac{d \vec{r}(t)}{d t}=-A \omega \sin \omega t \hat{i}+B \omega \cos \omega t \hat{j}$

$\therefore \vec{v}(t)=-A \omega \sin \omega t \hat{i}+B \omega \cos \omega t \hat{j}$

$\therefore \frac{d \vec{v}(t)}{d t}=-A \omega^{2} \cos \omega t \hat{i}-B \omega^{2} \sin \omega t \hat{j}$

$\therefore \vec{a}(t)=-\omega^{2}(A \cos \omega t \hat{i}+B \sin \omega t \hat{j})$

$=-\omega^{2} \vec{r}$

હવે $\overrightarrow{ F }$$=m \vec{a}(t)$

$\therefore \overrightarrow{ F }$$= m \omega^{2} \vec{r}$