Question

Find \(\mathbf{r}(t)\) for the given conditions. $$ \mathbf{r}^{\prime}(t)=\frac{1}{1+t^{2}} \mathbf{i}+\frac{1}{t^{2}} \mathbf{j}+\frac{1}{t} \mathbf{k}, \quad \mathbf{r}(1)=2 \mathbf{i} $$

Short answer

\(\mathbf{r}(t) = (\arctan(t) + 2 - \pi/4 )\mathbf{i} - \frac{1}{t}\mathbf{j} + \ln|t|\mathbf{k}\)

Step-by-step solution

Integrate \(\mathbf{r}^{\prime}(t)\) component-wise

The vector \(\mathbf{r}^{\prime}(t)\) consists of three components, and each of these components should be integrated separately. Integrating the given derivative, we get \(\int \frac{1}{1+t^{2}}dt\mathbf{i}+\int \frac{1}{t^{2}}dt \mathbf{j}+\int \frac{1}{t}dt \mathbf{k}\)

Evaluate the integrals

The antiderivatives for each of the components are as follows: \( \int \frac{1}{1+t^{2}}dt = \arctan(t)\mathbf{i}+const_i\), \( \int \frac{1}{t^{2}}dt = -\frac{1}{t}\mathbf{j}+const_j\), and \( \int \frac{1}{t}dt=\ln |t|\mathbf{k}+const_k\). Here, \(const_i, const_j, const_k\) are constants of integration.

Calculate the constants of integration from \(\mathbf{r}(1)=2 \mathbf{i}\)

We know that the vector function \(\mathbf{r}(t)\) equals \(2 \mathbf{i}\) when \(t=1\). This gives us three equations, one for each component of the vector: For the \(\mathbf{i}\)-component: \(\arctan(1) + const_i = 2\), for the \(\mathbf{j}\)-component: \(\frac{-1}{1} + const_j = 0\), and for the \(\mathbf{k}\)-component: \(\ln(1) + const_k = 0\). Solving these equations gives \(const_i = 2 - \pi/4, const_j = 1,\) and \(const_k = 0\).

Incorporate the constants of integration into \(\mathbf{r}(t)\)

Finally, we can obtain \(\mathbf{r}(t)\) by putting the constants of integration from step 3 into the integral-like function from step 2. The result is \(\mathbf{r}(t) = (\arctan(t) + 2 - \pi/4 )\mathbf{i} - \frac{1}{t}\mathbf{j} + \ln|t|\mathbf{k}\).