Suppose that $\vec{r}^{\prime}(t)=<10 t, e^{0.2 t}, \sqrt{t}>$ and $\vec{r}(0)=<5,1,1>$. Find $\vec{r}(t)$ \[ \vec{r}(t)=< \]

Step 1 ：We are given the derivative function of a vector function, \(\vec{r}^\prime(t)=<10t, e^{0.2t}, \sqrt{t}>\), and the initial condition \(\vec{r}(0)=<5,1,1>\). We need to find the original vector function \(\vec{r}(t)\).

Step 2 ：To find \(\vec{r}(t)\), we integrate each component of \(\vec{r}^\prime(t)\) with respect to \(t\).

Step 3 ：The integral of \(10t\) with respect to \(t\) is \(5t^2\).

Step 4 ：The integral of \(e^{0.2t}\) with respect to \(t\) is \(5e^{0.2t}\).

Step 5 ：The integral of \(\sqrt{t}\) with respect to \(t\) is \(\frac{2}{3}t^{3/2}\).

Step 6 ：We also need to consider the initial condition \(\vec{r}(0)=<5,1,1>\). This means that when \(t=0\), \(\vec{r}(t)=<5,1,1>\). We can use this information to find the constants of integration for each component of \(\vec{r}(t)\).

Step 7 ：For the first component, the constant of integration is 5, so the first component of \(\vec{r}(t)\) is \(5t^2 + 5\).

Step 8 ：For the second component, the constant of integration is -4, so the second component of \(\vec{r}(t)\) is \(5e^{0.2t} - 4\).

Step 9 ：For the third component, the constant of integration is 1, so the third component of \(\vec{r}(t)\) is \(\frac{2}{3}t^{3/2} + 1\).

Step 10 ：Thus, the function \(\vec{r}(t)\) is given by \(\vec{r}(t)=<5t^2 + 5, 5e^{0.2t} - 4, \frac{2}{3}t^{3/2} + 1>\).

Step 11 ：\(\boxed{\vec{r}(t)=<5t^2 + 5, 5e^{0.2t} - 4, \frac{2}{3}t^{3/2} + 1>}\)