Find the speed of the object whose position function is given by \( \bar{r}(t)=\left\langle 5 t^{2}+1,6 t+6, t^{3}+2 t\right\rangle \) at \( t=4 \) Please show your answer to at least 4 decimal places. Speed \( =\square \)

To find the speed of the object, we first compute the velocity vector \( \vec{v}(t) \) by differentiating the position function \[ \bar{r}(t)=\langle 5t^2+1,\; 6t+6,\; t^3+2t \rangle \] with respect to \( t \). 1. Differentiate each component: - For the \( x \)-component: \[ \frac{d}{dt}(5t^2+1) = 10t \] - For the \( y \)-component: \[ \frac{d}{dt}(6t+6) = 6 \] - For the \( z \)-component: \[ \frac{d}{dt}(t^3+2t) = 3t^2+2 \] 2. Thus, the velocity vector is: \[ \vec{v}(t)=\langle 10t,\; 6,\; 3t^2+2 \rangle \] 3. The speed is the magnitude of the velocity vector: \[ \text{Speed} = |\vec{v}(t)| = \sqrt{(10t)^2 + 6^2 + (3t^2+2)^2} \] 4. Evaluate the speed at \( t=4 \): - Substitute \( t=4 \) into each component: \[ 10t = 10(4) = 40, \] \[ 6 \text{ remains } 6, \] \[ 3t^2+2 = 3(4^2)+2 = 3(16)+2 = 48+2 = 50. \] 5. Now, compute the magnitude: \[ \text{Speed} = \sqrt{(40)^2 + (6)^2 + (50)^2} = \sqrt{1600 + 36 + 2500} = \sqrt{4136}. \] 6. Finally, compute the numerical value to at least 4 decimal places: \[ \sqrt{4136} \approx 64.3102. \] Thus, the speed of the object at \( t=4 \) is \[ \text{Speed} = 64.3102. \]