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## Q. 3.6

For $\mathbf{f}^{\prime}(t)=t \mathbf{i}+t^{3} \mathbf{j}$ and $\mathbf{f}(0)=\mathbf{i}-2 \mathbf{j}$, calculate $\mathbf{f}(t)$.

## Verified Solution

We first find that

$\mathbf{f}(t)=\int \mathbf{f}^{\prime}(t) d t=\left(\frac{t^{2}}{2}+C_{1}\right) \mathbf{i}+\left(\frac{t^{4}}{4}+C_{2}\right) \mathbf{j}=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{C}$.

To evaluate $\mathbf{C}$, we substitute the value t = 0 into (10) to find that $\mathbf{f}(0)=\mathbf{C}=\mathbf{i}-2 \mathbf{j}$, so that

$\mathbf{f}(t)=\frac{t^{2}}{2} \mathbf{i}+\frac{t^{4}}{4} \mathbf{j}+\mathbf{i}-2 \mathbf{j}=\left(\frac{t^{2}}{2}+1\right) \mathbf{i}+\left(\frac{t^{4}}{4}-2\right) \mathbf{j}$.