Question

Find \(\mathbf{T}(t), \mathbf{N}(t), a_{\mathrm{T}},\) and \(a_{\mathrm{N}}\) at the given time \(t\) for the plane curve \(\mathbf{r}(t)\) $$ \mathbf{r}(t)=e^{t} \cos t \mathbf{i}+e^{t} \sin t \mathbf{j}, \quad t=\frac{\pi}{2} $$

Short answer

After performing the calculations, you'll obtain the unit tangent vector \(\mathbf{T}(t)\), unit normal vector \(\mathbf{N}(t)\) and the tangential and normal accelerations \(a_{T}\) and \(a_{N}\) at the given time \(t=\frac{\pi}{2}\). To confirm your solution is correct, you can check that \(\mathbf{T}\) and \(\mathbf{N}\) should be unit vectors, and \(\mathbf{T}\) and \(\mathbf{N}\) should be perpendicular at \(t=\frac{\pi}{2}\).

Step-by-step solution

Find velocity vector

Calculate the derivative of \(\mathbf{r}(t)\) to find the velocity vector \(\mathbf{v}(t)\).\n\[\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=(e^{t} \cos t - e^{t} \sin t) \mathbf{i} + (e^{t} \cos t + e^{t} \sin t) \mathbf{j}\]\nThen, substitute \(t=\frac{\pi}{2}\) into \(\mathbf{v}(t)\) to get the velocity vector at the given time.

Find acceleration vector

Calculate the derivative of \(\mathbf{v}(t)\) to find the acceleration vector \(\mathbf{a}(t)\).\n\[\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=(-e^{t} \sin t - 2e^{t} \cos t) \mathbf{i} + (e^{t} \cos t - 2e^{t} \sin t) \mathbf{j}\]\nThen, substitute \(t=\frac{\pi}{2}\) into \(\mathbf{a}(t)\) to get the acceleration vector at the given time.

Find unit tangent vector

\(\mathbf{T}(t)\) can be obtained by normalizing the velocity vector:\n\[\mathbf{T}(t)=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}\]\nCalculate |v(t)|, the magnitude of v(t), and divide v(t) by it to get T(t). Substitute \(t=\frac{\pi}{2}\) to get \(\mathbf{T}(\frac{\pi}{2})\)

Find unit normal vector and tangential and normal accelerations

First, find \(\mathbf{N}(t)\) which is normal to the unit tangent vector. It is given by\n\[\mathbf{N}(t)=\frac{\frac{d\mathbf{T}}{dt}}{\left|\frac{d\mathbf{T}}{dt}\right|}\]\nTo find the tangential and normal accelerations: \(a_{T}\) is the derivative of the speed along the path and \(a_{N}\) is the change of direction of the velocity vector.\n\[a_{T} = \mathbf{v}'(t) . \mathbf{T}(t) \quad and \quad a_{N} = || \mathbf{v}'(t) . \mathbf{N}(t)||\]\nSubstitute \(t=\frac{\pi}{2}\) into the equations to get \(a_{T}(\frac{\pi}{2})\) and \(a_{N}(\frac{\pi}{2})\).

Key concepts

Velocity Vector Calculus

Understanding the velocity vector is fundamental in the study of plane curves in calculus. The velocity vector, denoted as \(\mathbf{v}(t)\), represents the rate of change of position with respect to time. It is obtained by differentiating the position vector \(\mathbf{r}(t)\) with respect to \(t\).

In the context of the given problem, the velocity vector at any time \(t\) is calculated by differentiating each component of the position vector. This results in \[\mathbf{v}(t)=\frac{d\mathbf{r}}{dt}=(e^{t} \cos t - e^{t} \sin t) \mathbf{i} + (e^{t} \cos t + e^{t} \sin t) \mathbf{j}\]. To find the specific velocity at \(t=\frac{\pi}{2}\), we substitute this value into our expression for \(\mathbf{v}(t)\), which gives us the velocity at the given time. The velocity vector captures the direction and speed of the particle moving along the curve at any instant.

Acceleration Vector Calculus

Acceleration is the rate at which velocity changes with time, and in vector calculus, it's denoted by \(\mathbf{a}(t)\). The acceleration vector provides insights into how the speed and direction of a particle's motion change over time.

To find the acceleration vector for a plane curve, we differentiate the velocity vector \(\mathbf{v}(t)\) with respect to time. For the given plane curve, this yields \[\mathbf{a}(t)=\frac{d\mathbf{v}}{dt}=(-e^{t} \sin t - 2e^{t} \cos t) \mathbf{i} + (e^{t} \cos t - 2e^{t} \sin t) \mathbf{j}\]. When \(t=\frac{\pi}{2}\) is substituted, we obtain the acceleration vector at that specific moment. The acceleration not only describes how the velocity changes in magnitude but also in direction along the plane curve.

Tangent and Normal Vectors

In the study of plane curves, tangent and normal vectors play crucial roles in describing the geometry and motion along the curve. A tangent vector, \(\mathbf{T}(t)\), points in the direction of the curve's movement and is a normalized version of the velocity vector. The normal vector, \(\mathbf{N}(t)\), is perpendicular to the tangent vector and points towards the curve's concavity.

To find these vectors for a given curve, \(\mathbf{T}(t)\) is calculated by normalizing the velocity vector \(\mathbf{v}(t)\): \[\mathbf{T}(t)=\frac{\mathbf{v}(t)}{|\mathbf{v}(t)|}\]. Meanwhile, the normal vector \(\mathbf{N}(t)\) can be found by differentiating \(\mathbf{T}(t)\) and normalizing it. These vectors are essential for understanding how a curve bends and for calculating other important quantities like curvature and torsion.

Tangential and Normal Accelerations

Diving deeper into the dynamics of a particle moving along a curve, we have the tangential acceleration, \(a_{\mathrm{T}}\), and the normal acceleration, \(a_{\mathrm{N}}\). Tangential acceleration corresponds to the change in the speed of the particle along the curve and is given by the projection of the acceleration vector onto the tangent vector. The normal acceleration measures how quickly the direction of the velocity vector is changing and is associated with the centripetal force experienced by the particle.

Mathematically, these are determined by: \[a_{\mathrm{T}} = \mathbf{a}(t) \cdot \mathbf{T}(t)\] and \[a_{\mathrm{N}} = ||\mathbf{a}(t) \times \mathbf{T}(t)||\], where \(\cdot\) represents the dot product and \(\times\) represents the cross product. By substituting \(t=\frac{\pi}{2}\) into these formulas, we can find the magnitudes of tangential and normal accelerations at that instant, offering a complete picture of the acceleration's effect on the curve's motion.