Question

Position functions \(\vec{r}_{1}(t)\) and \(\vec{r}_{2}(s)\) for two objects are given that follow the same path on the respective intervals. (a) Show that the positions are the same at the indicated \(t_{0}\) and \(s_{0}\) values; i.e., show \(\vec{r}_{1}\left(t_{0}\right)=\vec{r}_{2}\left(s_{0}\right) .\) (b) Find the velocity, speed and acceleration of the two objects at \(t_{0}\) and \(s_{0},\) respectively. $$ \begin{array}{l} \vec{r}_{1}(t)=\langle t, \sqrt{t}\rangle \text { on }[0,1] ; t_{0}=1 \\ \vec{r}_{2}(s)=\langle\sin t, \sqrt{\sin t}\rangle \text { on }[0, \pi / 2] ; s_{0}=\pi / 2 \end{array} $$

Short answer

Positions are the same: \(\langle 1, 1 \rangle\). Velocities differ: \(\vec{v}_1(1) = \langle 1, \frac{1}{2} \rangle\) and \(\vec{v}_2(\pi/2) = \langle 0, 0 \rangle\). Accelarations: \(\vec{a}_1(1) = \langle 0, -\frac{1}{4} \rangle\) and \(\vec{a}_2(\pi/2) = \langle 0, -\frac{1}{2} \rangle\).

Step-by-step solution

Evaluate positions at given values

Substitute \(t_0 = 1\) into \(\vec{r}_{1}(t)\) to find the position of the first object: \[\vec{r}_{1}(1) = \langle 1, \sqrt{1} \rangle = \langle 1, 1 \rangle.\]Substitute \(s_0 = \pi / 2\) into \(\vec{r}_{2}(s)\) to find the position of the second object: \[\vec{r}_{2}(\pi/2) = \langle \sin(\pi/2), \sqrt{\sin(\pi/2)} \rangle = \langle 1, 1 \rangle.\]Thus, \(\vec{r}_{1}(1) = \vec{r}_{2}(\pi/2)\), confirming the positions are the same at these values.

Find velocity functions

Find the derivative of each position function to find the velocity functions.For \(\vec{r}_{1}(t) = \langle t, \sqrt{t} \rangle\), the velocity is \[\vec{v}_{1}(t) = \frac{d}{dt}\langle t, \sqrt{t} \rangle = \langle 1, \frac{1}{2\sqrt{t}} \rangle.\]For \(\vec{r}_{2}(s) = \langle \sin s, \sqrt{\sin s} \rangle\), the velocity is \[\vec{v}_{2}(s) = \frac{d}{ds}\langle \sin s, \sqrt{\sin s} \rangle = \langle \cos s, \frac{1}{2\sqrt{\sin s}} \cos s \rangle.\]

Evaluate velocities at given values

Substitute \(t_0 = 1\) into \(\vec{v}_{1}(t)\) to find the velocity at \(t_0\):\[\vec{v}_{1}(1) = \langle 1, \frac{1}{2\sqrt{1}} \rangle = \langle 1, \frac{1}{2} \rangle.\]Substitute \(s_0 = \pi/2\) into \(\vec{v}_{2}(s)\) to find the velocity at \(s_0\):\[\vec{v}_{2}(\pi/2) = \langle \cos(\pi/2), \frac{\cos(\pi/2)}{2\sqrt{\sin(\pi/2)}} \rangle = \langle 0, 0 \rangle.\]

Calculate speeds

The speed is the magnitude of the velocity vector for each object at their respective positions.For \(\vec{v}_{1}(1) = \langle 1, \frac{1}{2} \rangle\), calculate the speed:\[||\vec{v}_1(1)|| = \sqrt{1^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}.\]For \(\vec{v}_{2}(\pi/2) = \langle 0, 0 \rangle\), the speed is:\[||\vec{v}_2(\pi/2)|| = \sqrt{0^2 + 0^2} = 0.\]

Find acceleration functions

The acceleration is the derivative of the velocity function.For \(\vec{v}_{1}(t) = \langle 1, \frac{1}{2\sqrt{t}} \rangle\), the acceleration is:\[\vec{a}_{1}(t) = \frac{d}{dt}\langle 1, \frac{1}{2\sqrt{t}} \rangle = \langle 0, -\frac{1}{4t^{3/2}} \rangle.\]For \(\vec{v}_{2}(s) = \langle \cos s, \frac{1}{2\sqrt{\sin s}} \cos s \rangle\), the acceleration is:\[\vec{a}_{2}(s) = \frac{d}{ds}\langle \cos s, \frac{\cos s}{2\sqrt{\sin s}} \rangle.\] Use product and chain rule to differentiate.

Evaluate accelerations at given values

Substitute \(t_0 = 1\) into \(\vec{a}_{1}(t)\) to find the acceleration at \(t_0\):\[\vec{a}_{1}(1) = \langle 0, -\frac{1}{4} \rangle.\]Substitute \(s_0 = \pi/2\) into \(\vec{a}_{2}(s)\) using the derivative result for a clearer expression:\[\vec{a}_{2}(\pi/2) = \langle 0, -\frac{1}{2} \rangle.\]

Key concepts

Vector Functions

In calculus, vector functions are vital because they allow us to describe motion and paths in space. These functions take a scalar input and return a vector output. This is immensely useful when dealing with quantities in physics that require both a direction and magnitude, such as position, velocity, and acceleration.

In our exercise, we have two different vector functions, \(\vec{r}_{1}(t) = \langle t, \sqrt{t} \rangle\) and \(\vec{r}_{2}(s) = \langle \sin s, \sqrt{\sin s} \rangle\).
This means each of these functions inputs a real number and returns a position in two-dimensional space, represented by a vector.

Understanding these vector paths is crucial since they represent the trajectories or positions of objects moving through space. Analyzing them helps in understanding how at certain points these paths coincide, as seen where \( t_0 \) and \( s_0 \) give them the same vector output. In practice, this means the objects are at the same point in space at these particular values.

Derivatives

The derivative of a vector function, just like with scalar functions, gives us critical information about rates of change.
When applied to vector functions, these derivatives give us physical quantities.

For example, when you find the derivative of \(\vec{r}_{1}(t) = \langle t, \sqrt{t} \rangle\), denoted as \(\vec{v}_{1}(t) = \langle 1, \frac{1}{2\sqrt{t}} \rangle\), you are essentially determining the velocity of the object.
Velocity tells you how fast and in what direction the object is moving. Similarly, calculating the derivative of \(\vec{r}_{2}(s)\) gives \(\vec{v}_{2}(s) = \langle \cos s, \frac{1}{2\sqrt{\sin s}} \cos s \rangle\), which is the velocity function of the second object.

The derivative can go further; by taking the derivative of the velocity functions, we find the acceleration functions \(\vec{a}_{1}(t)\) and \(\vec{a}_{2}(s)\).
These functions describe how the object is speeding up or slowing down, which is another critical aspect of motion.

Physics Applications

This exercise beautifully demonstrates the relationship between calculus and physics. Calculus helps us understand the motion of objects through its fundamental concepts, such as derivatives and vector functions.

• Position: Calculating the position vector at specific times, \( t_0 \) or \( s_0 \), allows us to pinpoint where an object is in space.
• Velocity: Using the derivative of the position vector, we derive velocity. This tells us the speed and direction of an object as it moves along its path, which is crucial for track prediction.
• Acceleration: Acceleration, the derivative of velocity, tells us about changes in speed over time and direction alterations.

Understanding these concepts allows physicists and engineers to model real-world problems and predict future motion, helping in designing efficient systems, maximizing safety, and optimizing performance.