# Consider an object moving along the parametrized curve with equations: x(t) = e^t + e^{-t} and y(t) = e^{-t}

• Answer the following:
• Find the maximum speed of the object and the time it takes.
• What is the minimum speed of the object along with the time it takes?
• t is the time interval $[0,4]$ in seconds.

This problem aims to find the maximum speed of an object that covers a distance in the shape of a parameterized curve whose equations are given.

To better understand the problem, you must be familiar with the parameterized curve in a plane, terminal, and initial velocities. A parametrized curve is a trail in the $xy$ plane outlined by the point $x(t), y(t)$ as the parameter $t$ spans over an interval $I$.

The set builder notation for curve will be:

$c = \{ (x(t), y(t)) \colon t \in I \}$

## Expert Answer

We are given the following two equations of the object that is moving along a parametrized curve:

$x(t) = e^t + e^{-t}$

$y(t) = e^{-t}$

$[0, 4]$ is the time interval $t$.

Position vector at time $t$ will be:

$R(t) = <x(t),y(t)> = <e^t + e^{-t}, e^{-t} >$

Velocity vector at time $t$ is:

$v(t) = \dfrac{d}{dt} R(t)$

$= \dfrac{d}{d_t} < e^t + e^{-t}, e^{-t} >$

$v(t) = < e^t – e^{-t}, – e^{-t} >$

Scalar speed at time $t$ comes out to be:

$v(t) = |v(t)| = |< e^t – e^{-t}, – e^{-t} >|$

$= \sqrt{(e^t – e^{-t})^2 + e^{-2t}}$

$= \sqrt{e^{2t} + e^{2t} -2 + e^{-2t}}$

$v(t) = \sqrt{e^{2t} + 2e^{-2t} -2 }$

Consider the function,

$f(t) = \sqrt{e^{2t} + 2e^{-2t} -2 }$

$f'(t) = \dfrac{e^{2t}-2e^{-2t}} {\sqrt{e^{2t} + 2e^{-2t} -2 }}$

For minima or maxima,

$f'(t) = 0$

$\dfrac{e^{2t}-2e^{-2t}} {\sqrt{e^{2t} + 2e^{-2t} -2 }} = 0$

$e^{2t}-2e^{-2t} = 0$

$e^{4t} = 2$

$4t = ln(2)$

$t = \dfrac{1}{4}ln(2)$

$\dfrac{1}{4}ln(2)$ is the critical point of $f$.

End points and critical points are found as follows:

$f(t) = \sqrt{e^{2t} + 2e^{-2t} -2 }$

$f(0) = \sqrt{e^{2(0)} + 2e^{-2(0)} -2 } = 1$

$f(4) = \sqrt{e^{2(4)} + 2e^{-2(4)} -2 } = 54.58$

$f(\dfrac{1}{4}ln(2)) = \sqrt{\sqrt{2} + 2 \left( \dfrac{\sqrt{2}}{2} \right) -2 }$

$= \sqrt{2\sqrt{2} -2 } = 0.91$

Thus, the Maximum speed at interval $4$ is $54.58$,

Whereas the Minimum speed at interval $f(\dfrac{1}{4}ln(2))$ is $0.91$.

## Numerical Result

The maximum speed of the object on the time interval is $54.58$ at time $t=4$.
The minimum speed of the object on the time interval is $0.91$  at time $t=f(\dfrac{1}{4}ln(2))$.

## Example

We are given the following two equations of the object that is moving along a parametrized curve:

$x(t) = e^t + e^{-t}$

$y(t) = e^{-t}$

Finding the speed on the interval $t=2$:

$f(t) = \sqrt{e^{2t} + 2e^{-2t} -2 }$

$f(2) = \sqrt{e^{2(2)} + 2e^{-2(2)} -2 } = 7.25$

The speed of the object on the time interval is $7.25$ at time $t=2$.

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