Problem 1
Question
\( \mathbf{r}(t)\) is the position vector of a moving particle. Graph the curve and the velocity and acceleration vectors at the indicated time. Find the speed at that time. $$ \mathbf{r}(t)=t^{2} \mathbf{i}+\frac{1}{4} t^{4} \mathbf{j} ; t=1 $$
Step-by-Step Solution
Verified Answer
The speed at \( t = 1 \) is \( \sqrt{5} \).
1Step 1: Understand the position vector
The position vector of the particle is given by \( \mathbf{r}(t) = t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j} \). This represents the coordinates of the particle as a function of time \( t \). At \( t = 1 \), the position is \( \mathbf{r}(1) = 1^2 \mathbf{i} + \frac{1}{4} \times 1^4 \mathbf{j} = \mathbf{i} + \frac{1}{4} \mathbf{j} \).
2Step 2: Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector with respect to time \( t \). \[ \mathbf{v}(t) = \frac{d}{dt}[t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j}] = 2t \mathbf{i} + t^3 \mathbf{j}. \] At \( t = 1 \), the velocity vector is \( \mathbf{v}(1) = 2 \times 1 \mathbf{i} + 1^3 \mathbf{j} = 2 \mathbf{i} + \mathbf{j} \).
3Step 3: Find the acceleration vector
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector with respect to time \( t \). \[ \mathbf{a}(t) = \frac{d}{dt}[2t \mathbf{i} + t^3 \mathbf{j}] = 2 \mathbf{i} + 3t^2 \mathbf{j}. \] At \( t = 1 \), the acceleration vector is \( \mathbf{a}(1) = 2 \mathbf{i} + 3 \times 1^2 \mathbf{j} = 2 \mathbf{i} + 3 \mathbf{j} \).
4Step 4: Calculate the speed at \( t = 1 \)
The speed is the magnitude of the velocity vector at \( t = 1 \). \[ \text{Speed} = ||\mathbf{v}(1)|| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}. \]
5Step 5: Graph the curve and vectors
Graph the trajectory of \( \mathbf{r}(t) = t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j} \) for the range of \( t \) values around 1, and draw the velocity vector \( 2 \mathbf{i} + \mathbf{j} \) and acceleration vector \( 2 \mathbf{i} + 3 \mathbf{j} \) at the point \( (1, \frac{1}{4}) \). This visual represents the path, the direction of the velocity, and the acceleration.
Key Concepts
Position VectorVelocity VectorAcceleration VectorSpeed Calculation
Position Vector
A position vector is a fundamental concept in multivariable calculus. It represents the location of a particle in space as a function of time. In this exercise, the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j} \) tells us that the x-coordinate is \( t^2 \) and the y-coordinate is \( \frac{1}{4} t^4 \). This vector not only shows where the particle is at any time \( t \), but also provides the path the particle traces out in space.
For example, when \( t = 1 \), substitute into the function to get the particle's position as \( \mathbf{r}(1) = \mathbf{i} + \frac{1}{4} \mathbf{j} \). This means the particle resides at coordinates \((1, \frac{1}{4})\) in this time frame. Visualizing position vectors can help in understanding how a particle moves through space over time.
For example, when \( t = 1 \), substitute into the function to get the particle's position as \( \mathbf{r}(1) = \mathbf{i} + \frac{1}{4} \mathbf{j} \). This means the particle resides at coordinates \((1, \frac{1}{4})\) in this time frame. Visualizing position vectors can help in understanding how a particle moves through space over time.
Velocity Vector
The velocity vector is derived by taking the derivative of the position vector with respect to time. This vector shows both the speed and direction of a particle's movement at any given moment.
In our problem, the velocity vector is given by \( \mathbf{v}(t) = \frac{d}{dt}[t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j}] = 2t \mathbf{i} + t^3 \mathbf{j} \). At \( t = 1 \), we calculate \( \mathbf{v}(1) = 2 \mathbf{i} + \mathbf{j} \).
This indicates that at \( t = 1 \), the particle is moving with a velocity of 2 units in the x-direction and 1 unit in the y-direction. Velocity vectors are crucial for understanding not just how fast a particle is moving, but precisely where it is headed.
In our problem, the velocity vector is given by \( \mathbf{v}(t) = \frac{d}{dt}[t^2 \mathbf{i} + \frac{1}{4} t^4 \mathbf{j}] = 2t \mathbf{i} + t^3 \mathbf{j} \). At \( t = 1 \), we calculate \( \mathbf{v}(1) = 2 \mathbf{i} + \mathbf{j} \).
This indicates that at \( t = 1 \), the particle is moving with a velocity of 2 units in the x-direction and 1 unit in the y-direction. Velocity vectors are crucial for understanding not just how fast a particle is moving, but precisely where it is headed.
Acceleration Vector
Acceleration vectors are derived from the velocity vector. They show how a particle's velocity changes over time. Think of it as measuring how quickly the particle speeds up or slows down, and in what direction this change occurs.
In the exercise, the acceleration vector is calculated by differentiating the velocity vector: \( \mathbf{a}(t) = \frac{d}{dt}[2t \mathbf{i} + t^3 \mathbf{j}] = 2 \mathbf{i} + 3t^2 \mathbf{j} \). At time \( t = 1 \), the acceleration vector becomes \( \mathbf{a}(1) = 2 \mathbf{i} + 3 \mathbf{j} \).
This tells us that the rate of change of the velocity is consistent as it maintains a 2 unit increase in the x-direction, and speeds up by 3 units in the y-direction when \( t = 1 \). Understanding acceleration is essential in predicting future motion of particles, especially in more complex paths.
In the exercise, the acceleration vector is calculated by differentiating the velocity vector: \( \mathbf{a}(t) = \frac{d}{dt}[2t \mathbf{i} + t^3 \mathbf{j}] = 2 \mathbf{i} + 3t^2 \mathbf{j} \). At time \( t = 1 \), the acceleration vector becomes \( \mathbf{a}(1) = 2 \mathbf{i} + 3 \mathbf{j} \).
This tells us that the rate of change of the velocity is consistent as it maintains a 2 unit increase in the x-direction, and speeds up by 3 units in the y-direction when \( t = 1 \). Understanding acceleration is essential in predicting future motion of particles, especially in more complex paths.
Speed Calculation
Speed is the magnitude of the velocity vector. It illustrates how fast a particle is moving, regardless of direction. To find speed, you compute the norm of the velocity vector.
Here, for \( t = 1 \), we have \( \mathbf{v}(1) = 2 \mathbf{i} + \mathbf{j} \). Calculate its magnitude as: \[ ||\mathbf{v}(1)|| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}. \]
The speed at \( t = 1 \) is therefore \( \sqrt{5} \), which encompasses the total motion in both the x and y directions. By focusing on speed, we focus on the total rate of movement which helps in comparing different stages of motion or different particles.
Here, for \( t = 1 \), we have \( \mathbf{v}(1) = 2 \mathbf{i} + \mathbf{j} \). Calculate its magnitude as: \[ ||\mathbf{v}(1)|| = \sqrt{(2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5}. \]
The speed at \( t = 1 \) is therefore \( \sqrt{5} \), which encompasses the total motion in both the x and y directions. By focusing on speed, we focus on the total rate of movement which helps in comparing different stages of motion or different particles.
Other exercises in this chapter
Problem 1
Sketch some of the level curves associated with the given function. $$ f(x, y)=x+2 y $$
View solution Problem 1
In Problems, compute the gradient for the given function. $$ f(x, y)=x^{2}-x^{3} y^{2}+y^{4} $$
View solution Problem 1
In Problems, graph the curve traced by the given vector function. \(\mathbf{r}(t)=2 \sin t \mathbf{i}+4 \cos t \mathbf{j}+t \mathbf{k} ; t \geq 0\)
View solution Problem 1
Verify the divergence theorem. \(\mathbf{F}=x y \mathbf{i}+y z \mathbf{j}+x z \mathbf{k} ; D\) the region bounded by the unit cube defined by \(0 \leq x \leq 1,
View solution