91Ó°ÊÓ

The given derivative of the vector function \(r'(t)\) has three components: \(te^{-t^2}\), \(-e^{-t}\), and \(1\). To find the vector function \(r(t)\), we find the primitive function (antiderivative) of each component. \[\int t e^{-t^{2}} dt = -\frac{1}{2}e^{-t^{2}}\]\[\int -e^{-t} dt = e^{-t}\]\[\int 1 dt = t\]Therefore, \[\int \mathbf{r}^{\prime}(t) dt=\int t e^{-t^{2}} dt \mathbf{i}+\int -e^{-t} dt \mathbf{j}+\int 1 dt \mathbf{k} = -\frac{1}{2}e^{-t^{2}}\mathbf{i} + e^{-t}\mathbf{j} + t\mathbf{k}\]However, there could be some constants of integration that we need to find. We write them as \(C_1\), \(C_2\), and \(C_3\). So we get, \[\mathbf{r}(t) = -\frac{1}{2}e^{-t^{2}}\mathbf{i} + e^{-t}\mathbf{j} + t\mathbf{k} + C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k}\]

Now we can use the initial condition ut \(\mathbf{r}(0)=\frac{1}{2} \mathbf{i}-\mathbf{j}+\mathbf{k}\) to find the values of the constants of integration. To do this, substitute \(t = 0\) on both sides of above equation, calculating the right side and setting the components equal to those given in the initial condition.Substituting \(t = 0\) yields,\[\mathbf{r}(0) = -\frac{1}{2}e^{-0^{2}}\mathbf{i} + e^{-0}\mathbf{j} + 0\mathbf{k} + C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k} = -\frac{1}{2}\mathbf{i} + \mathbf{j} + C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k}\]Setting the \(i\mathbf{}\) components equal gives \(-\frac{1}{2} + C_1 = \frac{1}{2}\), from which we find \(C_1 = 1\).Setting the \(j\mathbf{}\) components equal gives \(1 + C_2 = -1\), i.e., \(C_2 = -2\).The \(k\mathbf{}\) components are \(0 + C_3 = 1\), i.e., \(C_3 = 1\).

Now substitute the values of the constants \(C_1\), \(C_2\), and \(C_3\) back into the solution from Step 1 to get the final vector function. \[\mathbf{r}(t) = -\frac{1}{2}e^{-t^{2}}\mathbf{i} + e^{-t}\mathbf{j} + t\mathbf{k} + \mathbf{i} - 2\mathbf{j} + \mathbf{k} = -\frac{1}{2}e^{-t^{2}}\mathbf{i} + e^{-t}\mathbf{j} + t\mathbf{k} + \mathbf{i} - 2\mathbf{j} + \mathbf{k}\]Simplifying this yields,\[\mathbf{r}(t) = (\mathbf{1} - \frac{1}{2}e^{-t^{2}})\mathbf{i} + (-2 + e^{-t})\mathbf{j} + (\mathbf{1} + t)\mathbf{k}\]