Question

Find a set of parametric equations for the tangent line to the graph at \(t=t_{0}\) and use the equations for the line to approximate \(\mathbf{r}\left(t_{0}+\mathbf{0 . 1}\right)\). $$ \mathbf{r}(t)=\langle t, \ln t, \sqrt{t}\rangle, \quad t_{0}=1 $$

Short answer

The parametric equations for the tangent line are \( \mathbf{r}(t) = \langle 1, 0, 1 \rangle + (t - 1)\langle 1, 1, 0.5 \rangle \) and the approximate value of \( \mathbf{r}(t_{0}+0.1) \) is \( \mathbf{r}(1.1) = \langle 1.1, 0.1, 1.05 \rangle \)

Step-by-step solution

Find the derivative of the vector function

The derivative of a vector function \(\mathbf{r}(t)\) is given by \( \mathbf{r}'(t) = \langle \frac{d}{dt}(t), \frac{d}{dt}(\ln t), \frac{d}{dt}(\sqrt{t}) \rangle \) which simplifies to \( \mathbf{r}'(t) = \langle 1, \frac{1}{t}, \frac{1}{2\sqrt{t}} \rangle \)

Substitute for \(t_{0}\) to find the slope

Substitute \(t_{0}=1\) into the derivative \( \mathbf{r}'(t) = \langle 1, \frac{1}{t}, \frac{1}{2\sqrt{t}} \rangle \) to get \( \mathbf{r}'(1) = \langle 1, 1, 0.5 \rangle \)

Write the equations for the tangent line

The parametric equations for the tangent line to the graph at \(t_{0}\) are given by \( \mathbf{r}(t) = \mathbf{r}(t_{0}) + (t-t_{0})\mathbf{r}'(t_{0}) \). Substituting for \( \mathbf{r}(t_{0}) = \langle 1, 0, 1 \rangle \) and \( \mathbf{r}'(t_{0}) = \langle 1, 1, 0.5 \rangle \), we get \( \mathbf{r}(t) = \langle 1, 0, 1 \rangle + (t - 1)\langle 1, 1, 0.5 \rangle \)

Approximate \( \mathbf{r}(t_{0}+0.1) \)

Substitute \(t = t_{0} + 0.1 = 1.1\) into the parametric equations \( \mathbf{r}(t) = \langle 1, 0, 1 \rangle + (t - 1)\langle 1, 1, 0.5 \rangle \) to get \( \mathbf{r}(1.1) = \langle 1.1, 0.1, 1.05 \rangle \)