Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the plane curve \(\mathbf{r}(t)\) $$ \mathbf{r}(t)=t \mathbf{i}+\frac{1}{t} \mathbf{j}, \quad t=1 $$

Short answer

The unit tangent vector at \(t=1\) is \(\mathbf{T}(1)=\frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\) and the unit normal vector is the normalized version of \(2\mathbf{j} \times \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\). The tangential acceleration \(a_{\mathrm{T}}\) is 0 and normal acceleration \(\mathbf{a}_{\mathrm{N}}\) is the magnitude of \(2\mathbf{j} \times \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\).

Step-by-step solution

Find The Velocity Vector

Start by differentiating the position vector \(\mathbf{r}(t)\) to get the velocity vector \(\mathbf{v}(t)\):\n\(\mathbf{v}(t)=\mathbf{r}'(t)=\mathbf{i}-\frac{1}{t^2}\mathbf{j}\). Then, substituting \(t=1\) we find: \(\mathbf{v}(1)=\mathbf{i}-\mathbf{j}\).

Compute The Tangent Vector

The unit tangent vector \(\mathbf{T}(t)\) is obtained from the velocity vector \(\mathbf{v}(t)\) by normalizing it. This is done by dividing \(\mathbf{v}(t)\) by its magnitude:\n\(\mathbf{T}(t) = \frac{\mathbf{v}(t)}{ ||\mathbf{v}(t)|| }\), \nHence, we find \(\mathbf{T}(1)=\frac{\mathbf{i}-\mathbf{j}}{ ||\mathbf{i}-\mathbf{j}||} = \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\).

Determine The Acceleration Vector

The acceleration vector is found by differentiating the velocity vector: \(\mathbf{a}(t) = \mathbf{v}'(t)\). When we differentiate we get: \(\mathbf{a}(t)=2t^{-3}\mathbf{j}\). Substituting \(t=1\) yields: \(\mathbf{a}(1)=2\mathbf{j}\).

Compute The Normal and Tangential Components of Acceleration

First, we get the tangential component of the acceleration \(\mathbf{a}_{\mathrm{T}}\) by calculating the dot product of the acceleration and the unit tangent vector: \(\mathbf{a}_{\mathrm{T}}=\mathbf{a}(t)\cdot\mathbf{T}(t)\). After evaluation we get \(\mathbf{a}_{\mathrm{T}}=0\). The normal component of acceleration \(\mathbf{a}_{\mathrm{N}}\) is found by taking the cross product of the acceleration and the unit tangent vector: \(\mathbf{a}_{\mathrm{N}}=\mathbf{a}(t)\times\mathbf{T}(t)\) which is equal to \(2\mathbf{j} \times \frac{\mathbf{i}-\mathbf{j}}{\sqrt{2}}\). Normalizing \(\mathbf{a}_{\mathrm{N}}\) yields the final unit normal vector \(\mathbf{N}(t)\).