Problem 160 - Calculus 3

Question

Find the velocity, acceleration, and speed of a particle with the given position function. $$ \mathbf{r}(t)=\left\langle t^{2}-1, t\right\rangle $$

Step-by-Step Solution

Answer

Velocity: \(\langle 2t, 1 \rangle\), Acceleration: \(\langle 2, 0 \rangle\), Speed: \(\sqrt{4t^2 + 1}\).

Step 1: Find the Velocity

The velocity of the particle is the derivative of the position function \( \mathbf{r}(t) \) with respect to time \( t \). The position function is \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle \).Take the derivative:\[ \mathbf{v}(t) = \frac{d}{dt} \langle t^2 - 1, t \rangle = \langle \frac{d}{dt}(t^2 - 1), \frac{d}{dt}(t) \rangle = \langle 2t, 1 \rangle \]So the velocity function is \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).

Step 2: Find the Acceleration

The acceleration of the particle is the derivative of the velocity function \( \mathbf{v}(t) = \langle 2t, 1 \rangle \) with respect to time \( t \).Take the derivative:\[ \mathbf{a}(t) = \frac{d}{dt} \langle 2t, 1 \rangle = \langle \frac{d}{dt}(2t), \frac{d}{dt}(1) \rangle = \langle 2, 0 \rangle \]So the acceleration function is \( \mathbf{a}(t) = \langle 2, 0 \rangle \).

Step 3: Find the Speed

The speed of the particle is the magnitude of the velocity vector \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).Find the magnitude:\[ \text{speed} = \| \mathbf{v}(t) \| = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1} \] Thus, the speed of the particle is \( \sqrt{4t^2 + 1} \).

Key Concepts

Velocity

To understand the concept of velocity, it's helpful to think about how quickly something is moving in a specific direction. Velocity is not just about speed; it also includes the direction of motion. It is a vector quantity, which means it has both magnitude (how much) and direction (where to). In this problem, the velocity of a particle \( \mathbf{v}(t) \) is given by the derivative of the position function \( \mathbf{r}(t)\) with respect to time \( t \).

• The position function is like a map of where the particle is in space at any given time.
• Taking the derivative helps us find how the position changes over time, which is precisely what velocity tells us.

Here, the given position function is \( \mathbf{r}(t) = \langle t^2 - 1, t \rangle\). When we calculate the derivative, we find that the velocity \( \mathbf{v}(t) = \langle 2t, 1 \rangle\). This equation means that for every second that passes, the velocity in the x-direction changes by \(2t\), and in the y-direction, it remains constant at 1. Understanding and identifying velocity precisely allows you to predict how an object's speed and direction will change over time.

Acceleration

Acceleration gives us information about how the velocity changes over time. This is crucial because it tells us not just how fast something is moving, but how much faster or slower it is getting. Like velocity, acceleration is also a vector, meaning it has both direction and magnitude. In this exercise, to find acceleration, we took the derivative of the velocity function \( \mathbf{v}(t) = \langle 2t, 1 \rangle\).

• By calculating this derivative, we find the acceleration function \( \mathbf{a}(t) = \langle 2, 0 \rangle \).
• Here, the acceleration in the x-direction is 2.
• The acceleration in the y-direction is 0, meaning there is no change in speed or direction along the y-axis.

This tells us that, over time, the particle's velocity continues to increase in the x-direction by a fixed amount (2 units per unit of time), while staying constant vertically. Hence, acceleration is a powerful tool in forecasting how an object's motion evolves due to forces acting on it.

Speed

Speed is a measure of how fast something is moving, without regard for its direction. Unlike velocity or acceleration, speed is a scalar quantity, meaning it does not include direction. It is simply how fast the particle is traveling at any given moment. In this exercise, speed is determined by taking the magnitude of the velocity vector \( \mathbf{v}(t) = \langle 2t, 1 \rangle \).

• To find this magnitude, we use the formula for the magnitude of a vector: \(\| \mathbf{v}(t) \| = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}\).
• This calculation shows us the particle's speed at any point in time.

The expression \( \sqrt{4t^2 + 1} \) helps us understand how the speed of the particle changes as time progresses. It means as time increases, the speed will also increase, but not linearly due to the square term. Using these calculations, you can better anticipate the dynamics of a particle's journey.