## Thomas' Calculus 13th Edition

$$\mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$
Since \begin{align*} \frac{d \mathbf{r}}{d t}&=\int-(\mathbf{i}+\mathbf{j}+\mathbf{k}) d t\\ &=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k})+\mathbf{C}_{1} \end{align*} and $$\frac{d \mathbf{r}}{d t}(0)=\mathbf{0} \Rightarrow-(0 \mathbf{i}+0 \mathbf{j}+0 \mathbf{k})+\mathbf{C}_{1}=\mathbf{0} \Rightarrow \mathbf{C}_{1}=\mathbf{0}$$ Then $$\frac{d \mathbf{r}}{d t}=-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k})$$ and \begin{align*} \mathbf{r}&=\int-(t \mathbf{i}+t \mathbf{j}+t \mathbf{k}) d t\\ &=-\left(\frac{t^{2}}{2} \mathbf{i}+\frac{t}{2} \mathbf{j}+\frac{t^{2}}{2} \mathbf{k}\right)+\mathbf{C}_{2} \end{align*} $$\mathbf{r}(0)=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$ Then $$\mathbf{C}_{2}=10 \mathbf{i}+10 \mathbf{j}+10 \mathbf{k}$$ and $$\mathbf{r}=\left(-\frac{t^{2}}{2}+10\right) \mathbf{i}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{j}+\left(-\frac{t^{2}}{2}+10\right) \mathbf{k}$$