Question

Find \(v(t), a(t), \mathbf{T}(t),\) and \(N(t)\) (if it exists) for an object moving along the path given by the vector-valued function \(\mathrm{r}(t) .\) Use the results to determine the form of the path. Is the speed of the object constant or changing? $$ \mathbf{r}(t)=4 t \mathbf{i}-2 t \mathbf{j} $$

Short answer

The velocity is \(v(t) = 4\mathbf{i} - 2\mathbf{j}\), acceleration is \(a(t) = 0\), the unit tangent vector is \(\mathbf{T}(t) = \frac{1}{\sqrt{20}}(4\mathbf{i} - 2\mathbf{j})\), and the unit normal vector \(N(t)\) doesn't exist. The speed is constant.

Step-by-step solution

Find the Velocity

The velocity function \(v(t)\) is the derivative of the position function \(\mathbf{r}(t)\). For the given \(\mathbf{r}(t) = 4t\mathbf{i} - 2t\mathbf{j}\), differentiation yields \(v(t) = 4\mathbf{i} - 2\mathbf{j}\).

Find the Acceleration

The acceleration function \(a(t)\) is the derivative of the velocity function \(v(t)\). Differentiating \(v(t) = 4\mathbf{i} - 2\mathbf{j}\) provides \(a(t) = 0\).

Calculate the Speed

Speed is the magnitude of the velocity vector. The magnitude of \(v(t) = 4\mathbf{i} - 2\mathbf{j}\) is calculated as \(\|v(t)\| = \sqrt{(4^2) + (-2)^2} = \sqrt{20} \), this is a constant, hence the speed is constant.

Find the Unit Tangent Vector

The unit tangent vector \(\mathbf{T}(t)\) is the velocity vector \(v(t)\) divided by the speed. With, \(v(t) = 4\mathbf{i} - 2\mathbf{j}\) and \(\|v(t)\| = \sqrt{20}\), we have \(\mathbf{T}(t) = \frac{v(t)}{\|v(t)\|} = \frac{1}{\sqrt{20}}(4\mathbf{i} - 2\mathbf{j})\).

Find the Unit Normal Vector

Since the acceleration is zero, finding the unit normal vector becomes a bit tricky. In this case, acceleration vector \(a(t) = 0\), so the unit normal vector \(N(t)\) does not exist. In general, \(N(t)\) doesn’t exist when the speed is constant.