Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the space curve \(\mathbf{r}(t) .\) $$ \mathbf{r}(t)=4 t \mathbf{i}-4 t \mathbf{j}+2 t \mathbf{k} \quad t=2 $$

Short answer

The tangential unit vector \(\mathbf{T}(t)\) at \(t=2\) is \(\left(\frac{4}{3}\right)\mathbf{i} - \left(\frac{4}{3}\right)\mathbf{j} + \left(\frac{1}{3}\right)\mathbf{k}\), while the unit normal vector \( \mathbf{N}(t)\) does not exist at \(t=2\). The tangential acceleration \(a_{\mathrm{T}} = 0\), likewise the normal acceleration \(a_{\mathrm{N}}=0\).

Step-by-step solution

Find velocity and acceleration vectors

Find the derivative of \(\mathbf{r}(t)\) to obtain the velocity vector, \(\mathbf{v}(t)\), and the second derivative of \(\mathbf{r}(t)\) to obtain the acceleration vector, \(\mathbf{a}(t)\). In this case, \(\mathbf{r}(t) = 4t\mathbf{i} - 4t\mathbf{j} + 2t\mathbf{k}\), meaning \(\mathbf{v}(t) = \mathbf{r}'(t) = 4\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\) and \(\mathbf{a}(t) = \mathbf{r}''(t) = \mathbf{0}\).

Calculate the magnitudes

The magnitude of \(\mathbf{v}(t)\) at \(t=2\) is \(\|\mathbf{v}(2)\| = \sqrt{(4^2) + (-4^2) + (2^2)} = 6\). Since the acceleration vector is a zero vector, its magnitude will be 0.

Finding Tangential Unit Vector

The unit tangential vector is found by dividing the velocity vector by its magnitude. So, \(\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}\). Thus, at \(t=2\), \(\mathbf{T}(2) = \frac{\mathbf{v}(2)}{\|\mathbf{v}(2)\|} = \left(\frac{4}{3}\right)\mathbf{i} - \left(\frac{4}{3}\right)\mathbf{j} + \left(\frac{1}{3}\right)\mathbf{k}\).

Finding Normal Unit Vector and Accelerations

The unit normal vector \(\mathbf{N}(t)\) can be found as the derivative of the unit tangent vector divided by its magnitude. However, since our acceleration vector is zero, the normal unit vector will not exist at \(t =2 \). As for the tangential acceleration \(a_{\mathrm{T}}\), it can be found by taking the dot product of the acceleration vector at \(t=2\) and the unit tangent vector at \(t=2\), i.e., \(a_{\mathrm{T}} = \mathbf{a}(2) \cdot \mathbf{T}(2)\). Since the acceleration vector at \(t=2\) is a zero vector, \(a_{\mathrm{T}} = 0\). The normal acceleration, \(a_{\mathrm{N}}\), is the magnitude of the component of the acceleration vector orthogonal to the tangent vector. This, however, also computes to \(a_{\mathrm{N}} = 0\) as the acceleration vector is zero.

Key concepts

Tangent and Normal Vectors

Understanding tangent and normal vectors is crucial for interpreting the direction and orientation of a space curve at a given point.

Tangent vectors, denoted by \( \mathbf{T}(t) \) in calculus, represent the direction of the curve at a particular point, much like the direction in which an object is moving. To find this vector, you typically calculate the derivative of the position vector \( \mathbf{r}(t) \) to obtain the velocity vector, and then normalize it to get the unit tangent vector.

Normal vectors, written as \( \mathbf{N}(t) \) in calculus, are perpendicular to the tangent vectors and point towards the center of the curve's osculating circle at the point of interest. In the given problem, the acceleration vector was found to be zero. This unique scenario means that the curve is moving at a constant speed and there's no change in the direction of motion, leading to no defined normal vector at \( t=2 \).

To recall, the steps involve taking the derivative of the position vector to get the velocity vector \( \mathbf{v}(t) \) and normalizing it by its magnitude to obtain \( \mathbf{T}(t) \), and then differentiating \( \mathbf{T}(t) \) again and normalizing to find \( \mathbf{N}(t) \), with consideration for the unique zero acceleration case here.

Tangential and Normal Acceleration

Acceleration can be decomposed into components that are tangential and normal to the trajectory of a moving body on a path.

Tangential acceleration \( a_{\mathrm{T}} \) is the component of acceleration that is parallel to the velocity vector, indicating how quickly the speed of the object is changing along the curve. It is computed by taking the dot product of the acceleration vector and the unit tangent vector. For our exercise, since the acceleration vector is zero, the tangential acceleration at \( t=2 \) is also zero, signifying no change in the speed.

Normal acceleration \( a_{\mathrm{N}} \) relates to the change in the direction of the velocity vector, and is perpendicular to the tangent vector. This component is significant when considering how sharply a curve bends. To find it, one would calculate the magnitude of the component of the acceleration vector that is orthogonal to the tangent vector. Again, due to the unique case of zero acceleration in our problem, \( a_{\mathrm{N}} = 0 \) at \( t=2 \) just as well, meaning there is no change in direction and no centripetal force acting on the body.

Magnitude of Vectors

The magnitude of a vector is a measure of its length or size, which can be visually interpreted as the distance from the origin to the point the vector represents in space.

To calculate the magnitude of a vector, a common method involves taking the square root of the sum of the squares of its components. In our example, the magnitude of the velocity vector \( \mathbf{v}(t) \) was calculated using the formula \[ \|\mathbf{v}(t)\| = \sqrt{v_x^2 + v_y^2 + v_z^2} \], where \( v_x, v_y, \) and \( v_z \) are the components of the velocity vector along the \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) directions, respectively.

In the given space curve example, after computing the velocities at \( t=2 \), the magnitude was found to be 6. It's important to recognize that this scalar quantity represents the speed of the object moving along the curve at that specific instant. The magnitude of vectors is essential in fields ranging from physics to engineering, illustrating the scope and essence of motion, forces, and other vector-related phenomena.