## Calculus (3rd Edition)

$$\frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89}$$
Since \begin{aligned} \frac{d s}{d t} &=\sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}} \\ &=\sqrt{9 t^{4}-24 t^{2}+16+4 t^{2}} \\ &=\sqrt{9 t^{4}-20 t^{2}+16} \end{aligned} Now, we will find the minimal speed, $$f(t) =\sqrt{9 t^{4}-20 t^{2}+16}$$ Then $$f'(t) =\frac{36 t^{3}-40 t }{2\sqrt{9 t^{4}-20 t^{2}+16} }$$ Hence $f'(t) =0$ at $t=0$ and $t= \sqrt{\frac{10}{9}}$ then $$\frac{d s}{d t}\bigg|_{t=0}= 4,\ \ \frac{d s}{d t}\bigg|_{t=0}=\sqrt{4.89}$$ Hence $$\frac{d s}{d t}\bigg|_{ \text{min}}=\sqrt{4.89}$$