Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the plane curve \(\mathbf{r}(t)\) $$ \mathbf{r}(t)=\left(t-t^{3}\right) \mathbf{i}+2 t^{2} \mathbf{j}, \quad t=1 $$

Short answer

The unit tangent vector at \(t=1\) is \(-\frac{1}{2}\mathbf{i}+\mathbf{j}\), the normal unit vector at \(t=1\) is \(-\mathbf{j}\), the tangential acceleration at \(t=1\) is 7, and the normal acceleration at \(t=1\) is \(\sqrt{13}\).

Step-by-step solution

Calculate the velocity vector

Start by taking the derivative of \(\mathbf{r}(t)\) to get the velocity vector \(\mathbf{v}(t)\).\n\[\mathbf v(t) = \mathbf r'(t) = (1 - 3t^2) \mathbf i + 4t \mathbf j.\]

Find the unit tangent vector

Normalize the velocity vector from the previous step to get the unit tangent vector. At time \(t = 1\), the velocity vector is \(\mathbf v(1) = -2\mathbf i + 4\mathbf j\). Therefore, we find that\n\[ \mathbf T(1) = \frac{\mathbf v(1)}{||\mathbf v(1)||} = \frac{-2\mathbf i + 4\mathbf j}{\sqrt{(-2)^2 + 4^2}} = -\frac{1}{2}\mathbf i + \mathbf j. \]

Calculate the acceleration vector

Now we find the acceleration vector by taking the derivative of the velocity vector. Calculating \(\mathbf a(t) = \mathbf v'(t)\) results in\n\[\mathbf a(t) = -6t\mathbf i + 4\mathbf j.\] At \(t=1\) we find \(\mathbf a(1) = -6\mathbf i + 4\mathbf j.\)

Find the tangential acceleration

Now find the tangential acceleration at \(t = 1\) by taking the scalar product of the acceleration and unit tangent vectors, \(a_T = \mathbf a \cdot \mathbf T\). This gives\n\[a_T(1) = \mathbf a(1) \cdot \mathbf T(1) = (-6\mathbf i + 4\mathbf j) \cdot (-\frac{1}{2}\mathbf i + \mathbf j) = 3 + 4 = 7.\]

Find the normal unit vector

First, calculate the derivative of the tangent vector, \(\mathbf T '(t)\). At time \(t=1\), this is \((1 - 3t) \mathbf i\). This gives\n\[\mathbf B(t) = \frac{\mathbf T '(1)}{||\mathbf T '(1)||} =\frac{2 \mathbf{i}}{||2 \mathbf{i}||} = \mathbf{i}.\] For a two dimensional case, the binormal vector, \(\mathbf B(t)\), is a rotation of the tangent vector, \(\mathbf T (t)\), by 90°. By the right hand rule, the normal vector is \(\mathbf N(t) = \mathbf B(t) \times \mathbf T(t)\). At \(t=1\), we get\n\[\mathbf N(1) = \mathbf B(1) \times \mathbf T(1) = \mathbf{i} \times (-\frac{1}{2}\mathbf i + \mathbf j) = -\mathbf j.\]

Find the normal acceleration

We can find the normal acceleration at \(t=1\) as \(a_{\mathrm{N}} = || \mathbf a \times \mathbf T ||\). We then find that\n\[a_{\mathrm{N}}=||(-6\mathbf i + 4\mathbf j) \times (-\frac{1}{2}\mathbf i + \mathbf j)|| = ||3\mathbf i + 2\mathbf j|| = \sqrt{3^2 + 2^2} = \sqrt{13}.\]