Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. A particle moves along a path modeled by \(\mathbf{r}(t)=\cosh (b t) \mathbf{i}+\sinh (b t) \mathbf{j}\) where \(b\) is a positive constant. (a) Show that the path of the particle is a hyperbola. (b) Show that \(\mathbf{a}(t)=b^{2} \mathbf{r}(t)\)

Short answer

The path of the particle is a hyperbola and the acceleration at time t is equal to \(b^{2}\) times the position vector at time t

Step-by-step solution

Understanding the problem

Here, the problem presents a particle's motion along a path modelled by \(\mathbf{r}(t)=\cosh (b t) \mathbf{i}+\sinh (b t) \mathbf{j}\), where \(b\) is a positive constant. It asks if the path of the particle is a hyperbola and if the acceleration of this particle is a multiple of its position vector.

Convert the parametric equation into Cartesian form

First of all, we know that \( \cosh^2(t) - \sinh^2(t) = 1\). Thus, for our particle's motion, we obtain btx: \( (\frac{\mathbf{i}}{b})^{2} - (\frac{\mathbf{j}}{b})^{2} = 1 \) which satisfies the standard equation for a hyperbola with a = b and b = 1/b, with center at the origin.

Differentiate the position vector

To find the acceleration, we first determine the velocity, which is the first derivative of the position vector. The derivative of \(\cosh(bt)\) is \(b\sinh(bt)\) and the derivative of \(\sinh(bt)\) is \(b\cosh(bt)\), thus \(\mathbf{v}(t) = b\sinh (b t) \mathbf{i} +b\cosh (b t) \mathbf{j}\).

Further differentiate to find acceleration

The acceleration is the first derivative of the velocity, which in our case simplifies to \( \mathbf{a}(t) = b^{2}\cosh (b t) \mathbf{i}+b^{2}\sinh (b t) \mathbf{j}\). When comparing this to \(\mathbf{r}(t)\), we find that indeed \(\mathbf{a}(t)=b^{2} \mathbf{r}(t)\).