Question

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

Explanation:

Step1: Find the velocity vector

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

Step2: Find the speed function

The speed \( s(t)\) is the magnitude of the velocity vector.

$$
LATEXBLOCK0
$$

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:

\(4\)