1. (11 pts) As always, give answers in simplified exact form. (a) Consider the vector function \( \mathrm{r}(t)=\left\langle 6+t, 2 \tan ^{-1}(t), 3 t+e^{t^{2}}\right\rangle \) Find the tangential component of acceleration at \( t=0 \).
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To find the tangential component of acceleration at \( t=0 \), we start by calculating the first and second derivatives of \( \mathbf{r}(t) \). 1. **First Derivative (Velocity)**: \[ \mathbf{r}'(t) = \left\langle 1, \frac{2}{1+t^2}, 3 + 2te^{t^2} \right\rangle \] Evaluating this at \( t=0 \): \[ \mathbf{r}'(0) = \left\langle 1, 2, 3 \right\rangle \] 2. **Second Derivative (Acceleration)**: \[ \mathbf{r}''(t) = \left\langle 0, -\frac{4t}{(1+t^2)^2}, 2e^{t^2} + 4t^2e^{t^2} \right\rangle \] Evaluating this at \( t=0 \): \[ \mathbf{r}''(0) = \left\langle 0, 0, 2 \right\rangle \] 3. **Magnitude of Velocity**: \[ |\mathbf{r}'(0)| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \] 4. **Tangential Component of Acceleration**: The tangential acceleration \( a_{\text{tan}} \) is given by: \[ a_{\text{tan}} = \frac{\mathbf{r}'(0) \cdot \mathbf{r}''(0)}{|\mathbf{r}'(0)|} \] Calculating the dot product: \[ \mathbf{r}'(0) \cdot \mathbf{r}''(0) = 1 \cdot 0 + 2 \cdot 0 + 3 \cdot 2 = 6 \] Thus, \[ a_{\text{tan}} = \frac{6}{\sqrt{14}} = \frac{3\sqrt{14}}{7} \] So, the final answer for the tangential component of acceleration at \( t=0 \) is \( \frac{3\sqrt{14}}{7} \).