Question

A position function \(\vec{r}(t)\) is given. Sketch \(\vec{r}(t)\) on the indicated interval. Find \(\vec{v}(t)\) and \(\vec{a}(t),\) then add \(\vec{v}\left(t_{0}\right)\) and \(\vec{a}\left(t_{0}\right)\) to your sketch, with their initial points at \(\vec{r}\left(t_{0}\right)\) for the given value of \(t_{0}\). $$ \vec{r}(t)=\left\langle t^{2}+t,-t^{2}+2 t\right\rangle \text { on }[-2,2] ; t_{0}=1 $$

Short answer

Find derivatives of position for velocity and acceleration, plot vectors at \( t_0=1 \).

Step-by-step solution

Understanding the Position Function

We start by examining the given position function \( \vec{r}(t) = \langle t^2 + t, -t^2 + 2t \rangle \). This function represents a parametric curve described by the components \( x(t) = t^2 + t \) and \( y(t) = -t^2 + 2t \). Our task is to plot this curve over the interval \( [-2, 2] \).

Sketching the Position Function

To sketch \( \vec{r}(t) \) over \( [-2, 2] \), calculate key points. For instance, when \( t = -2 \), \( \vec{r}(-2) = \langle 2, 8 \rangle \), and when \( t = 0 \), \( \vec{r}(0) = \langle 0, 0 \rangle \), while \( t = 2 \) gives \( \vec{r}(2) = \langle 6, 0 \rangle \). Then, draw a curve passing through these points.

Finding the Velocity Function

The velocity function \( \vec{v}(t) \) is the derivative of \( \vec{r}(t) \). Calculate: \( \vec{v}(t) = \left( \frac{d}{dt}(t^2 + t), \frac{d}{dt}(-t^2 + 2t) \right) = \langle 2t + 1, -2t + 2 \rangle \).

Finding the Acceleration Function

The acceleration function \( \vec{a}(t) \) is the derivative of \( \vec{v}(t) \). Calculate: \( \vec{a}(t) = \left( \frac{d}{dt}(2t + 1), \frac{d}{dt}(-2t + 2) \right) = \langle 2, -2 \rangle \).

Calculating \( \vec{v}(t_0) \) and \( \vec{a}(t_0) \)

Substitute \( t_0 = 1 \) into the velocity and acceleration functions: \( \vec{v}(1) = \langle 2(1) + 1, -2(1) + 2 \rangle = \langle 3, 0 \rangle \) and \( \vec{a}(1) = \langle 2, -2 \rangle \).

Adding the Vectors to the Sketch

Plot \( \vec{r}(1) = \langle 2, 1 \rangle \) on your sketch. Starting from this point, draw \( \vec{v}(1) = \langle 3, 0 \rangle \) as a horizontal vector and \( \vec{a}(1) = \langle 2, -2 \rangle \) as a diagonal vector. This effectively represents the motion at \( t_0 = 1 \).

Key concepts

Position Function

The position function is a fundamental concept in vector calculus. It describes the path of an object as it moves through space over time. In our exercise, the position function given is \( \vec{r}(t) = \langle t^2 + t, -t^2 + 2t \rangle \). This function consists of two components: the horizontal component \( x(t) = t^2 + t \) and the vertical component \( y(t) = -t^2 + 2t \). These components tell us the coordinates of a point on a parametric curve for any given time \( t \).

To understand the behavior of this curve, we calculate the values of \( \vec{r}(t) \) at different intervals, such as \( t = -2 \), \( t = 0 \), and \( t = 2 \). These calculations help us draw a sketch of the path from start to finish, providing a visual representation of how the position changes over time.

• At \( t = -2 \), \( \vec{r}(-2) = \langle 2, 8 \rangle \)
• At \( t = 0 \), \( \vec{r}(0) = \langle 0, 0 \rangle \)
• At \( t = 2 \), \( \vec{r}(2) = \langle 6, 0 \rangle \)

These points guide us to sketch a curve depicting the object's motion within the interval \([-2, 2]\).

Velocity Function

The velocity function represents the rate of change of the position function over time. It tells us how fast and in which direction an object is moving at any given time. For our position function \( \vec{r}(t) = \langle t^2 + t, -t^2 + 2t \rangle \), the velocity function \( \vec{v}(t) \) is found by taking the derivative of each component separately.

So, \( \vec{v}(t) = \left( \frac{d}{dt}(t^2 + t), \frac{d}{dt}(-t^2 + 2t) \right) = \langle 2t + 1, -2t + 2 \rangle \). This result gives us the velocity in both the x and y directions.

• The first component, \(2t + 1\), shows how the x-position changes with respect to time.
• The second component, \(-2t + 2\), shows how the y-position changes with respect to time.

Substituting \( t_0 = 1 \), we get \( \vec{v}(1) = \langle 3, 0 \rangle \). This tells us the object moves horizontally with a speed of 3 units and vertically remains stationary at \( t = 1 \).

Acceleration Function

Acceleration is the derivative of velocity. It indicates how the velocity of an object changes over time. In the given problem, once we determine the velocity function \( \vec{v}(t) = \langle 2t + 1, -2t + 2 \rangle \), we can find the acceleration function by differentiating each component of the velocity function.

This gives us the acceleration function: \( \vec{a}(t) = \left( \frac{d}{dt}(2t + 1), \frac{d}{dt}(-2t + 2) \right) = \langle 2, -2 \rangle \). This vector is constant, meaning the rate and direction at which the object's speed is changing remains unchanged over time.

• The constant x-component \(2\) means there's a consistent increase in speed in the horizontal direction.
• The constant y-component \(-2\) means there's a consistent decrease in speed in the vertical direction.

At \( t_0 = 1 \), the acceleration vector \( \vec{a}(1) = \langle 2, -2 \rangle \) suggests the object is experiencing consistent forces affecting its velocity.

Parametric Curves

Parametric curves offer a powerful way to describe paths in space using a parameter, usually time \( t \). Instead of relying on a single equation like \( y = f(x) \), parametric equations separate the paths into individual components. Our curve \( \vec{r}(t) = \langle t^2 + t, -t^2 + 2t \rangle \) uses \( t \) to determine both \( x \) and \( y \).

• The x-coordinate is determined by \( x(t) = t^2 + t \).
• The y-coordinate is determined by \( y(t) = -t^2 + 2t \).

Studying these components over an interval \([-2, 2]\), you get a complete picture of how the curve behaves as \( t \) progresses. Parametric curves can create complex shapes and motions, giving a clear view of how each dimension evolves.

This kind of curve allows for the comprehensive study of motion, providing insights into the intricate changes that occur over time in both moving objects and more static forms of mathematical functions. Always remember to plot significant points like starts, ends, and key intervals to draft an accurate representation of these paths.