the position function of a particle is given by ( mathbf{r}(t)=leftlangle t^{2}, 5 t, t^{2}-16…

the position function of a particle is given by ( mathbf{r}(t)=leftlangle t^{2}, 5 t, t^{2}-16 t\right\rangle ). when is the speed a minimum? ( t= )

the position function of a particle is given by ( mathbf{r}(t)=leftlangle t^{2}, 5 t, t^{2}-16 t\right\rangle ). when is the speed a minimum? ( t= )

Answer

Explanation:

Step1: Find the velocity vector

The velocity vector ( \mathbf{v}(t)) is the derivative of the position vector ( \mathbf{r}(t)). If ( \mathbf{r}(t)=\langle t^{2},5t,t^{2}-16t\rangle), then ( \mathbf{v}(t)=\mathbf{r}'(t)=\langle 2t,5,2t - 16\rangle).

Step2: Find the speed function

The speed ( s(t)) is the magnitude of the velocity vector. [ \begin{align*} s(t)&=\sqrt{(2t)^{2}+5^{2}+(2t - 16)^{2}}\ &=\sqrt{4t^{2}+25+4t^{2}-64t + 256}\ &=\sqrt{8t^{2}-64t + 281} \end{align*} ] To find the minimum of ( s(t)), we can instead find the minimum of ( f(t)=8t^{2}-64t + 281) (since ( y = \sqrt{u}) and (u = f(t)) is a non - negative function and the square root function is increasing for (u\geq0)).

Step3: Use the formula for the vertex of a quadratic function

For a quadratic function (y = ax^{2}+bx + c) ((a = 8), (b=-64), (c = 281)), the (x) - coordinate of the vertex (where the function is minimized since (a>0)) is given by (t=-\frac{b}{2a}). [ t=-\frac{-64}{2\times8}=\frac{64}{16} = 4 ]

Answer:

(4)