For the curve defined by \( \vec{r}(t)=\left\langle e^{-t} \cos (t), e^{t} \sin (t)\right\rangle \) find the unit tangent vector, unit normal vector, normal acceleration, and tangential acceleration at \( t=\frac{5 \pi}{6} \) \( \vec{T}\left(\frac{5 \pi}{6}\right)= \) \( \vec{N}\left(\frac{5 \pi}{6}\right)= \) \( a_{T}= \) \( a_{N}= \)

To find the unit tangent vector, unit normal vector, normal acceleration, and tangential acceleration for the curve defined by \[ \vec{r}(t) = \left\langle e^{-t} \cos(t), e^{t} \sin(t) \right\rangle \] at \( t = \frac{5\pi}{6} \), we will follow these steps: 1. **Calculate the derivative \(\vec{r}'(t)\)** to find the velocity vector. 2. **Find the unit tangent vector \(\vec{T}(t)\)**. 3. **Calculate the second derivative \(\vec{r}''(t)\)** to find the acceleration vector. 4. **Find the unit normal vector \(\vec{N}(t)\)**. 5. **Calculate the tangential acceleration \(a_T\)** and normal acceleration \(a_N\). ### Step 1: Calculate \(\vec{r}'(t)\) The derivative of \(\vec{r}(t)\) is given by: \[ \vec{r}'(t) = \left\langle \frac{d}{dt}(e^{-t} \cos(t)), \frac{d}{dt}(e^{t} \sin(t)) \right\rangle \] Using the product rule: \[ \frac{d}{dt}(e^{-t} \cos(t)) = -e^{-t} \cos(t) + e^{-t}(-\sin(t)) = e^{-t}(-\cos(t) - \sin(t)) \] \[ \frac{d}{dt}(e^{t} \sin(t)) = e^{t} \sin(t) + e^{t} \cos(t) = e^{t}(\sin(t) + \cos(t)) \] Thus, \[ \vec{r}'(t) = \left\langle e^{-t}(-\cos(t) - \sin(t)), e^{t}(\sin(t) + \cos(t)) \right\rangle \] ### Step 2: Find the unit tangent vector \(\vec{T}(t)\) The unit tangent vector is given by: \[ \vec{T}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} \] We need to calculate \(|\vec{r}'(t)|\): \[ |\vec{r}'(t)| = \sqrt{(e^{-t}(-\cos(t) - \sin(t)))^2 + (e^{t}(\sin(t) + \cos(t)))^2} \] ### Step 3: Calculate \(\vec{r}''(t)\) Next, we differentiate \(\vec{r}'(t)\) to find \(\vec{r}''(t)\): \[ \vec{r}''(t) = \left\langle \frac{d}{dt} \left( e^{-t}(-\cos(t) - \sin(t)) \right), \frac{d}{dt} \left( e^{t}(\sin(t) + \cos(t)) \right) \right\rangle \] Calculating these derivatives will give us the acceleration vector. ### Step 4: Find the unit normal vector \(\vec{N}(t)\) The unit normal vector is given by: \[ \vec{N}(t) = \frac{\vec{T}'(t)}{|\vec{T}'(t)|} \] ### Step 5: Calculate \(a_T\) and \(a_N\) The tangential acceleration \(a_T\) is given by: \[ a_T = \vec{T}(t) \cdot \vec{r}''(t) \] The normal acceleration \(a_N\) is given by: \[ a_N = \frac{|\vec{r}'(t)|^2}{R} \] where \(R\) is the radius of curvature. Now, let's perform the calculations for \(t = \frac{5\pi}{6}\). ### Performing the calculations 1. **Calculate \(\vec{r}'\left(\frac{5\pi}{6}\right)\)** and \(\vec{r}''\left(\frac{5\pi}{6}\right)\). 2. **Calculate \(|\vec{r}'\left(\frac{5\pi}{6}\right)|\)**. 3. **Find \(\vec{T}\left(\frac{5\pi}{6}\right)\)**. 4. **Find \(\vec{N}\left(\frac{5\pi}{6}\right)\)**. 5. **Calculate \(a_T\) and \(a_N\)**. Let's compute these values. ### Step 1: Calculate \(\vec{r}'\left(\frac{5\pi}{6}\right)\) Substituting \(t = \frac{5\pi}{6}\): \[ \vec{r}'\left(\frac{5\pi}{6}\right) = \left\langle e^{-\frac{5\pi}{6}}(-\cos(\frac{5\pi}{6}) - \sin(\frac{5\pi}{6})), e^{\frac{5\pi}{6}}(\sin(\frac{5\pi}{6}) + \cos(\frac{5\pi}{6})) \right\rangle \] Calculating \(\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}\) and \(\sin\left(\frac{5\pi}{6}\right) = \frac{1}{2}\): \[ \vec{r}'\left(\frac{5\pi}{6}\right) = \left\langle e^{-\frac{5\pi}{6}}\left(-\left(-\frac{\sqrt{3}}{2}\right) - \frac{1}{2}\right), e^{\frac{5\pi}{6}}\left(\frac{1}{2} - \frac{\sqrt{3}}{2}\right) \right\rangle \] Calculating the components: \[ \vec{r}'\left(\frac{5\pi}{6}\right) = \left\langle e^{-\frac{5\pi}{6}}\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right), e^{\frac{5\pi}{6}}\left(\frac{1 - \sqrt{3}}{2}\right) \right\rangle \] Now, let's calculate the magnitude \(|\vec{r}'\left(\frac{5\pi}{6}\right)|\) and then find \(\vec{T}\left(\frac{5\pi}{6}\right)\). ### Step 2: Calculate \(|\vec{r}'\left(\frac{5\pi}{6}\right