91Ó°ÊÓ

Parametric equations are a way to define a set of functions that describe a curve in terms of one or more variables known as parameters. In this particular problem, we have the equations \(x = \sinh t\), \(y = \cosh t\), and \(z = t\). Each of these equations describes the respective coordinates in space—\(x\), \(y\), and \(z\)—in terms of the parameter \(t\). These equations help us understand how the position of a point on the curve changes as \(t\) varies. In our exercise, we checked whether the point \((0, 1, 0)\) lies on the curve when \(t = 0\), and indeed, it does. This verification is crucial as it assures us that our interest is at a valid point on the curve. Understanding parametric equations is key to analyzing motion and geometry in calculus.

The velocity vector describes the rate at which the position of a point on a curve changes over time. To find this, we need to differentiate the parametric equations with respect to \(t\). In our solution, the velocity vector \(\mathbf{r}'(t)\) was found by taking derivatives:

• \(\frac{d}{dt}(\sinh t) = \cosh t\)
• \(\frac{d}{dt}(\cosh t) = \sinh t\)
• \(\frac{d}{dt}(t) = 1\)

This gives the velocity vector \(\mathbf{r}'(t) = (\cosh t, \sinh t, 1)\). At \(t = 0\), it becomes \((1, 0, 1)\). The velocity vector is important because it indicates the direction of motion and speed at any point on the curve.

The acceleration vector represents how the velocity vector changes with time. To determine this, we differentiate the velocity vector with respect to \(t\). Here, the calculation gives us:

• \(\frac{d}{dt}(\cosh t) = \sinh t\)
• \(\frac{d}{dt}(\sinh t) = \cosh t\)
• \(\frac{d}{dt}(1) = 0\)

Thus, the acceleration vector is \(\mathbf{r}''(t) = (\sinh t, \cosh t, 0)\). Evaluating at \(t = 0\), it turns into \((0, 1, 0)\). The acceleration vector indicates how the velocity is changing, which is essential for understanding the dynamics of motion on the curve.

The cross product is a vector multiplication operation that gives a vector perpendicular to two original vectors in three-dimensional space. It’s crucial in calculating curvature, as it provides a sense of how two vectors diverge in space. For the curvature calculation, we needed the cross product of the velocity and acceleration vectors, \(\mathbf{r}'(0) \times \mathbf{r}''(0)\).The formula is represented as a determinant:\[\mathbf{r}'(0) \times \mathbf{r}''(0) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 1 \ 0 & 1 & 0 \end{vmatrix} = (-1, 0, 1)\] The magnitude of this cross product plays a role in determining the curvature \(\kappa\), indicating how sharply the curve bends at a point. The cross product is a foundational tool in vector calculus, particularly when analyzing physical phenomena such as forces or motion in three-dimensional space.