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 need to calculate the derivative of the position function \(\bar{r}(t)\) to obtain the velocity function \(\bar{v}(t)\). The velocity \(\bar{v}(t)\) is given by: \[ \bar{v}(t) = \frac{d}{dt}\bar{r}(t) = \left\langle \frac{d}{dt}(5 t^{2}+1), \frac{d}{dt}(6 t+6), \frac{d}{dt}(t^{3}+2 t) \right\rangle. \] Calculating the derivatives: 1. \(\frac{d}{dt}(5 t^{2}+1) = 10 t\), 2. \(\frac{d}{dt}(6 t+6) = 6\), 3. \(\frac{d}{dt}(t^{3}+2 t) = 3 t^{2}+2\). Thus, the velocity function is: \[ \bar{v}(t) = \left\langle 10 t, 6, 3 t^{2} + 2 \right\rangle. \] Now, let's evaluate the velocity at \(t = 4\): \[ \bar{v}(4) = \left\langle 10 \cdot 4, 6, 3 \cdot 4^{2} + 2 \right\rangle = \left\langle 40, 6, 50 \right\rangle. \] Next, we compute the speed, which is the magnitude of the velocity vector: \[ \text{Speed} = \|\bar{v}(4)\| = \sqrt{40^{2} + 6^{2} + 50^{2}}. \] Calculating each term: - \(40^{2} = 1600\), - \(6^{2} = 36\), - \(50^{2} = 2500\). Now summing them: \[ \text{Speed} = \sqrt{1600 + 36 + 2500} = \sqrt{4136}. \] Finally, calculating the square root: \[ \text{Speed} \approx 64.3083. \] Thus, the speed of the object at \(t = 4\) is: \[ \text{Speed} \approx 64.3083. \]