The curves \( \vec{r}_{1}(t)=\left\langle-4 t, t^{2},-t^{6}\right\rangle \) and \( \vec{r}_{2}(t)=\langle\sin (5 t), \sin (4 t), t\rangle \) intersect at the origin. Find the acute angle of intersection \( (\theta) \) in degrees. \( \theta=\square \)

To find the acute angle of intersection between the two curves at the origin, we first need to compute the tangent vectors of the curves at the point of intersection, which is the origin \((0,0,0)\). 1. **Find the tangent vectors \(\vec{r}'_1(t)\) and \(\vec{r}'_2(t)\)**: - For \(\vec{r}_1(t) = \langle -4t, t^2, -t^6 \rangle\): \[ \vec{r}'_1(t) = \langle -4, 2t, -6t^5 \rangle \] - When \(t=0\): \[ \vec{r}'_1(0) = \langle -4, 0, 0 \rangle \] - For \(\vec{r}_2(t) = \langle \sin(5t), \sin(4t), t \rangle\): \[ \vec{r}'_2(t) = \langle 5 \cos(5t), 4 \cos(4t), 1 \rangle \] - When \(t=0\): \[ \vec{r}'_2(0) = \langle 5, 4, 1 \rangle \] 2. **Calculate the dot product of the tangent vectors**: \[ \vec{a} = \vec{r}'_1(0) = \langle -4, 0, 0 \rangle, \quad \vec{b} = \vec{r}'_2(0) = \langle 5, 4, 1 \rangle \] \[ \vec{a} \cdot \vec{b} = (-4)(5) + (0)(4) + (0)(1) = -20 \] 3. **Find the magnitudes of the tangent vectors**: \[ |\vec{a}| = \sqrt{(-4)^2 + 0^2 + 0^2} = 4 \] \[ |\vec{b}| = \sqrt{5^2 + 4^2 + 1^2} = \sqrt{25 + 16 + 1} = \sqrt{42} \] 4. **Use the dot product to find the cosine of the angle \(\theta\)**: \[ \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{-20}{4 \cdot \sqrt{42}} = \frac{-5}{\sqrt{42}} \] Since we are interested in the acute angle, we take the absolute value: \[ \cos(\theta) = \frac{5}{\sqrt{42}} \] 5. **Find \(\theta\)**: \[ \theta = \cos^{-1}\left(\frac{5}{\sqrt{42}}\right) \] Calculating this value: \[ \theta \approx 49.4^\circ \] Thus, the acute angle of intersection is approximately: \[ \theta \approx 49.4 \text{ degrees} \] So the final answer is: \[ \theta \approx 49.4^\circ \]