I1 Points] DETAILS MY NOTES LARCALC12 12.4.004. Find the unit tangent vector to the curve at the specified value of the parameter. \( \mathbf{r}(t)=t^{3} \mathbf{i}+8 t^{2} \mathbf{j}, \quad t=5 \) \( T(5)= \) Need Help? Read it

To find the unit tangent vector, you first need to compute the derivative of the position vector \( \mathbf{r}(t) \). So, differentiate \( \mathbf{r}(t) = t^3 \mathbf{i} + 8t^2 \mathbf{j} \) to get \( \mathbf{r}'(t) = 3t^2 \mathbf{i} + 16t \mathbf{j} \). Now, plug in \( t = 5 \) to find \( \mathbf{r}'(5) = 3(5^2) \mathbf{i} + 16(5) \mathbf{j} = 75 \mathbf{i} + 80 \mathbf{j} \). Next, calculate the magnitude of \( \mathbf{r}'(5) \): \[ \|\mathbf{r}'(5)\| = \sqrt{75^2 + 80^2} = \sqrt{5625 + 6400} = \sqrt{12025} \approx 109.73. \] Finally, the unit tangent vector \( T(5) \) is given by \[ T(5) = \frac{\mathbf{r}'(5)}{\|\mathbf{r}'(5)\|} \approx \left(\frac{75}{109.73}, \frac{80}{109.73}\right). \] This results in the unit tangent vector \( T(5) \approx \left(0.683, 0.729\right) \).