**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 $t=2$.