Find the tangential and normal components of the acceleration vector for the curve \( \vec{r}(t)=\left\langle-4 t, 3 t^{3},-t^{2}\right\rangle \) at the point \( t=-1 \) \( \vec{a}(-1)=\vec{T}+\square \vec{N} \) Give your answers to two decimal places Submit Question

To find the tangential and normal components of the acceleration vector for the curve \(\vec{r}(t)=\left\langle-4 t, 3 t^{3},-t^{2}\right\rangle \) at the point \( t=-1 \), we will first need to calculate the velocity and acceleration vectors, as well as the unit tangent and normal vectors. 1. **Calculate Velocity and Acceleration**: The velocity vector \(\vec{v}(t)\) is the derivative of \(\vec{r}(t)\): \[ \vec{v}(t) = \frac{d\vec{r}}{dt} = \left\langle -4, 9t^2, -2t \right\rangle \] At \(t = -1\): \[ \vec{v}(-1) = \left\langle -4, 9(-1)^2, -2(-1) \right\rangle = \left\langle -4, 9, 2 \right\rangle \] The acceleration vector \(\vec{a}(t)\) is the derivative of the velocity vector: \[ \vec{a}(t) = \frac{d\vec{v}}{dt} = \left\langle 0, 18t, -2 \right\rangle \] At \(t = -1\): \[ \vec{a}(-1) = \left\langle 0, 18(-1), -2 \right\rangle = \left\langle 0, -18, -2 \right\rangle \] 2. **Calculate the Magnitude of the Velocity**: The magnitude of the velocity vector at \(t = -1\): \[ |\vec{v}(-1)| = \sqrt{(-4)^2 + 9^2 + 2^2} = \sqrt{16 + 81 + 4} = \sqrt{101} \] 3. **Find the Unit Tangent Vector**: The unit tangent vector \(\vec{T}\) is given by: \[ \vec{T} = \frac{\vec{v}(-1)}{|\vec{v}(-1)|} = \frac{1}{\sqrt{101}} \left\langle -4, 9, 2 \right\rangle \] 4. **Find Tangential Acceleration**: The tangential component of the acceleration \(a_T\) is given by: \[ a_T = \frac{\vec{a}(-1) \cdot \vec{T}}{|\vec{T}|} = \vec{a}(-1) \cdot \vec{T} \] Calculating the dot product: \[ \vec{a}(-1) \cdot \vec{T} = \left\langle 0, -18, -2 \right\rangle \cdot \frac{1}{\sqrt{101}} \left\langle -4, 9, 2 \right\rangle = \frac{1}{\sqrt{101}} \left(0(-4) + -18(9) + -2(2)\right) = \frac{-162 - 4}{\sqrt{101}} = \frac{-166}{\sqrt{101}} \] Approximate \( a_T \): \[ a_T \approx \frac{-166}{10.05} \approx -16.52 \] 5. **Calculate Normal Acceleration**: The normal component \(a_N\) is given by: \[ a_N = |\vec{a}(-1)| - a_T \] The magnitude of \(\vec{a}(-1)\): \[ |\vec{a}(-1)| = \sqrt{0^2 + (-18)^2 + (-2)^2} = \sqrt{324 + 4} = \sqrt{328} \] Approximate \( a_N \): \[ a_N \approx \sqrt{328} + 16.52 \approx 18.13 + 16.52 \approx 34.65 \] Thus, we find: \[ \vec{a}(-1) = \vec{T} + a_N \vec{N} \] with \(a_T \approx -16.52\) and \( a_N \approx 34.65\).