Question

Find the unit tangent vector to the curve at the specified value of the parameter. $$ \mathbf{r}(t)=\ln t \mathbf{i}+2 t \mathbf{j}, \quad t=e $$

Short answer

The unit tangent vector to the curve at \(t=e\) is \(\frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}\)

Step-by-step solution

Differentiate the curve

The first step is to differentiate the components of the curve. The derivative of \(\ln t\) with respect to \(t\) is \({1/t}\) and the derivative of \(2t\) with respect to \(t\) is \(2\). So, \(\mathbf{r}'(t) = {1/t}\mathbf{i} + 2\mathbf{j}\) .

Evaluate the derivative at the specified parameter

Next, evaluate the derivative of the curve at the specified parameter value, \(t=e\), by substituting \(e\) for \(t\). \(\mathbf{r}'(e) = {1/e}\mathbf{i} + 2\mathbf{j} = {1}\mathbf{i} + 2\mathbf{j}\) .

Find the normalized tangent vector

The final step is to normalize the derivative to get the unit tangent vector \(\mathbf{T} = \mathbf{r}'/||\mathbf{r}'||\). First, find the magnitude of the derivative, \(||\mathbf{r}'|| = \sqrt{(1)^{2} + (2)^{2}} = \sqrt{5}\). Then divide the derivative by its magnitude to get the unit tangent vector, \(\mathbf{T} = \frac{1}{\sqrt{5}}\mathbf{i} + \frac{2}{\sqrt{5}}\mathbf{j}\) .