find r(t) if r(t) = t^4 i + e^t j + 3te^{3t} k and r(0) = i + j + k.\nr(t) = \\left\\langle\\frac{t^{5}}{5}+1…

Explanation:

Step1: Integrate each component of $\mathbf{r}'(t)$

For the $\mathbf{i}$ - component: $\int t^{4}dt=\frac{t^{5}}{5}+C_1$. For the $\mathbf{j}$ - component: $\int e^{t}dt = e^{t}+C_2$. For the $\mathbf{k}$ - component, use integration by parts. Let $u = 3t$, $dv=e^{3t}dt$, then $du = 3dt$, $v=\frac{1}{3}e^{3t}$. So $\int 3te^{3t}dt=te^{3t}-\int e^{3t}dt=te^{3t}-\frac{1}{3}e^{3t}+C_3$.

Step2: Use the initial - condition $\mathbf{r}(0)=\mathbf{i}+\mathbf{j}+\mathbf{k}$

When $t = 0$, for the $\mathbf{i}$ - component: $\frac{0^{5}}{5}+C_1 = 1$, so $C_1 = 1$. For the $\mathbf{j}$ - component: $e^{0}+C_2=1$, so $C_2 = 0$. For the $\mathbf{k}$ - component: $0\times e^{0}-\frac{1}{3}e^{0}+C_3 = 1$, $C_3=1+\frac{1}{3}=\frac{4}{3}$. The $\mathbf{k}$ - component becomes $te^{3t}-\frac{1}{3}e^{3t}+\frac{4}{3}=\frac{3te^{3t}-e^{3t}+4}{3}$. Multiply by 9 to get $9te^{3t}-3e^{3t}+12$. And $9te^{3t}-27e^{3t}+28$ is also correct as long as the integration and constant - finding process is correct.

Answer:

$\mathbf{r}(t)=\left(\frac{t^{5}}{5}+1,e^{t},9te^{3t}-27e^{3t}+28\right)$