Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the space curve \(\mathbf{r}(t) .\) $$ \mathbf{r}(t)=e^{t} \sin t \mathbf{i}+e^{t} \cos t \mathbf{j}+e^{t} \mathbf{k} \quad t=0 $$

Short answer

The unit tangent vector at \(t=0\) is \(\mathbf{T}(0) = \mathbf{i}/\sqrt{2} + \mathbf{k}/\sqrt{2}\), the unit normal vector is \(\mathbf{N}(0) = \mathbf{0}\), the tangential component of acceleration is \(a_{\mathrm{T}} = 0\), and the normal component of acceleration is given by \(a_{\mathrm{N}} = ||\mathbf{a}(t)||\).

Step-by-step solution

Calculate the derivative of \(\mathbf{r}(t)\)

First, we need to find the derivative of \(\mathbf{r}(t)\) at time \(t=0\). This is given by \(\mathbf{r}'(t)=e^{t} \cos(t) \mathbf{i} - e^{t} \sin(t) \mathbf{j} + e^{t} \mathbf{k}\). Evaluated at \(t=0\), we get \(\mathbf{r}'(0) = \mathbf{i} + \mathbf{k}\).

Find the unit tangent vector \(\mathbf{T}(t)\)

The unit tangent vector \(\mathbf{T}(t)\) is given by the normalization of \(\mathbf{r}'(t)\). The magnitude ||\(\mathbf{r}'(t)\)|| at \(t=0\) is \(\sqrt{2}\), so \(\mathbf{T}(0) = (\mathbf{i} + \mathbf{k})/\sqrt{2}\).

Calculate \(\mathbf{T}'(t)\) and Normalize to get \(\mathbf{N}(t)\)

First calculate \(\mathbf{T}'(t)\). Considering \(\mathbf{T}(t)\) does not depend on \(t\), \(\mathbf{T}'(t)\) is \(\mathbf{0}\), and thus \(\mathbf{N}(t)\) is also \(\mathbf{0}\).

Calculate the Acceleration Components \(a_{\mathrm{T}}\) and \(a_{\mathrm{N}}\)

The tangential component of acceleration is given by \(a_{\mathrm{T}} = \mathbf{a}(t) \cdot \mathbf{T}(t)\) and the normal component by \(a_{\mathrm{N}} = ||\mathbf{a}(t) - a_{\mathrm{T}} \mathbf{T}(t)||\). But considering that \(\mathbf{N}(0) = \mathbf{0}\), we have \(a_{\mathrm{T}} = 0\) and \(a_{\mathrm{N}} = ||\mathbf{a}(t)||\).