Question

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

Short answer

\(\mathbf{T}(1) = \frac{\mathbf{i} + 2\mathbf{j} -3\mathbf{k}}{| \mathbf{i} + 2\mathbf{j} -3\mathbf{k} |}\), \(\mathbf{N}(1) = undefined\), \(a_{\mathrm{T}} = 0\), \(a_{\mathrm{N}} = undefined\)

Step-by-step solution

Calculate the derivative of the space curve

First, compute the first derivative of \(\mathbf{r}(t)\), \(\mathbf{r}'(t)\), in order to get the velocity vector, \(\mathbf{v}(t)\). Since \(\mathbf{r}(t) = t\mathbf{i} + 2t\mathbf{j} -3t\mathbf{k}\), its derivative, \(\mathbf{r}'(t)\), is equal to \(\mathbf{i} + 2\mathbf{j} -3\mathbf{k}\). At \(t = 1\), we have: \(\mathbf{v}(1) = \mathbf{i} + 2\mathbf{j} -3\mathbf{k}\)

Find the unit tangent vector

This is just the unit vector of \(\mathbf{v}(t)\). So, normalize \(\mathbf{v}(t)\) to get \(\mathbf{T}(t)=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}\). Note that \(|\mathbf{v}(t)|\) is the magnitude of \(\mathbf{v}(t)\). At \(t = 1\), \(\mathbf{T}(1)=\frac{\mathbf{i} + 2\mathbf{j} -3\mathbf{k}}{| \mathbf{i} + 2\mathbf{j} -3\mathbf{k} |}\)

Calculate the derivative of the unit tangent vector

Find the derivative of \(\mathbf{T}(t)\) to get \(\mathbf{T}'(t)\). This step involves applying the quotient rule to \(\mathbf{T}(t)\). It's a bit complex because it involves taking the derivative of a ratio of vectors. However, at \(t = 1\), because both numerator and denominator of \(\mathbf{T}(1)\) are constants (not depending on \(t\)), we get \(\mathbf{T}'(1) = \mathbf{0}\)

Compute the unit normal vector

The unit normal vector, \(\mathbf{N}(t)\), is just the normalized vector of \(\mathbf{T}'(t)\). But since the derivative of \(\mathbf{T}(t)\) at \(t = 1\) is zero, the principle is undefined and there is no unit normal vector, hence \(\mathbf{N} = undefined\)

Find the tangential acceleration

Tangential acceleration, \(a_{\mathrm{T}}\), is the derivative of the speed (magnitude of velocity). The speed, \(|\mathbf{v}(t)|\), is the magnitude of \(\mathbf{v}(t)\), which is constant as computed earlier. Thus, the derivative is zero. Hence, at \(t = 1\), we have: \(a_{\mathrm{T}} = 0\)

Find the normal acceleration

Normal acceleration, \(a_{\mathrm{N}}\), is the magnitude of acceleration in the direction of the unit normal vector \(\mathbf{N}(t)\). But since \(\mathbf{N}(t)\) is undefined here, \(a_{\mathrm{N}}\) will also be undefined at \(t = 1\)

Key concepts

Tangent and Normal Vectors

When studying space curves in calculus, a key concept is the understanding of tangent and normal vectors as they give important geometrical information about the curve. A tangent vector at any point of a space curve indicates the direction in which the curve is proceeding at that point. It is essentially a 'touching' vector that aligns with the curve but does not cut across it.

To find the tangent vector, we calculate the derivative of the curve's position vector, which gives us the velocity vector. If we denote the position vector by \( \mathbf{r}(t) \), then the velocity vector is \( \mathbf{v}(t) = \mathbf{r}'(t) \). To get the unit tangent vector, \( \mathbf{T}(t) \), we normalize \( \mathbf{v}(t) \) by dividing it by its magnitude.

The normal vector, perpendicularly related to the tangent vector, points towards the center of curvature of the space curve, signifying the instantaneous direction of change. For the curve mentioned in the exercise, since the velocity vector at \( t = 1 \) is constant, its derivative is zero, and therefore, obtaining the unit normal vector is not possible — it is undefined because we do not have a direction of change at that point.

Tangential and Normal Acceleration

The concept of acceleration in space curves is split into tangential and normal components. These components describe how quickly the velocity changes along the curve and how fast the curve deviates from a straight path respectively.

Tangential acceleration, denoted by \( a_{\mathrm{T}} \), is the time rate of change of the speed of an object moving along a curve, or the derivative of the magnitude of the velocity vector. In our case, since the magnitude of the velocity vector is constant, \( a_{\mathrm{T}} \) remains zero at \( t = 1 \). It tells us that the speed along the tangent line is not changing, implying constant motion along the curve at that moment in time.

On the other hand, normal acceleration, denoted by \( a_{\mathrm{N}} \), is associated with the change in direction of the velocity vector and is found by projecting the acceleration vector onto the unit normal vector. However, in the exercise provided, since the unit normal vector is undefined due to a zero derivative of the unit tangent vector, \( a_{\mathrm{N}} \) is also undefined. This signifies that there is no centripetal acceleration pulling the object towards a center of curvature since its path does not deviate—it moves in a straight line.

Vector Derivative

The derivative of a vector function, such as a space curve, brings out rich information on how the vector quantities change over time. If we consider the position vector \( \mathbf{r}(t) \) of a particle moving along a space curve, its derivative with respect to time, \( \mathbf{r}'(t) \), will give us the velocity vector, \( \mathbf{v}(t) \).

This vector derivative is crucial because it forms the basis for understanding motion along the curve, expressing both the rate of change and the direction of motion. Moreover, calculating the derivative of higher-order, such as the derivative of the velocity vector (i.e., the acceleration vector), will provide us insights into the dynamic behavior of the particle in motion. The acceleration can then be decomposed into tangential and normal components, as discussed previously.

For the exercise given, the derivative of the position vector, \( \mathbf{r}'(t) \), is a constant vector, signaling a linear path without any curvature, a case where traditional methods of finding normal vectors and accelerations do not yield meaningful results as they would in curves with changing direction.