Velocity Vector Derivative

Oct 13, 2025 by orovajewels.com 95 views

For ${$ t \geq 0 $}$, a particle moves in the ${$ xy $}$-plane with position vector ${$ \langle x(t), y(t)\rangle$}$, where ${$ x(t) = t e^{-t} $}$ and ${$ y(t) = \ln (2t + 6) $}$.

Introduction

In this article, we will explore the concept of acceleration in the context of a particle moving in the ${$ xy $}$-plane. We will use the given position vector ${$ \langle x(t), y(t)\rangle$}$ to derive the acceleration of the particle. The position vector is defined by the equations ${$ x(t) = t e^{-t} $}$ and ${$ y(t) = \ln (2t + 6) $}$, where ${$ t \geq 0 $}$.

Position Vector

The position vector of the particle is given by ${$ \langle x(t), y(t)\rangle$}$. This vector represents the position of the particle at time ${$ t $}$ in the ${$ xy $}$-plane. The components of the position vector are ${$ x(t) = t e^{-t} $}$ and ${$ y(t) = \ln (2t + 6) $}$.

Velocity Vector

To find the acceleration of the particle, we need to first find the velocity vector. The velocity vector is the derivative of the position vector with respect to time. We can find the derivative of the position vector using the product rule and the chain rule.

Velocity Vector Derivative

The derivative of the position vector is given by:

${$ \frac{d}{dt}\langle x(t), y(t)\rangle = \langle \frac{dx}{dt}, \frac{dy}{dt}\rangle $}$

Using the product rule and the chain rule, we can find the derivative of the position vector:

${$ \frac{dx}{dt} = e^{-t} - t e^{-t} $}$

${$ \frac{dy}{dt} = \frac{2}{2t + 6} $}$

The velocity vector is then given by:

${$ \langle \frac{dx}{dt}, \frac{dy}{dt}\rangle = \langle e^{-t} - t e^{-t}, \frac{2}{2t + 6} \rangle $}$

Acceleration Vector

The acceleration vector is the derivative of the velocity vector with respect to time. We can find the derivative of the velocity vector using the product rule and the chain rule.

Acceleration Vector Derivative

The derivative of the velocity vector is given by:

${$ \frac{d}{dt}\langle \frac{dx}{dt}, \frac{dy}{dt}\rangle = \langle \frac{d^2x}{dt2}, \frac{d^2y}{dt2}\rangle $}$

Using the product rule and the chain rule, we can find the derivative of the velocity vector:

${$ \frac{d^2x}{dt2} = -e^{-t} + (e^{-t} - t e^{-t}) $}$

${$ \frac{d^2y}{dt2} = -\frac{2}{(2t + 6)^2} $}$

The acceleration vector is then given by:

${$ \langle \frac{d^2x}{dt2}, \frac{d^2y}{dt2}\rangle = \langle -e^{-t} + (e^{-t} - t e^{-t}), -\frac{2}{(2t + 6)^2} \rangle $}$

Conclusion

In this article, we have derived the acceleration vector of a particle moving in the ${$ xy $}$-plane. The position vector of the particle is given by ${$ \langle x(t), y(t)\rangle$}$, where ${$ x(t) = t e^{-t} $}$ and ${$ y(t) = \ln (2t + 6) $}$. We have used the product rule and the chain rule to find the derivative of the position vector, which gives us the velocity vector. Finally, we have used the product rule and the chain rule again to find the derivative of the velocity vector, which gives us the acceleration vector.

Mathematical Derivations

The mathematical derivations used in this article are based on the following rules:

Product Rule: If ${$ f(t) = g(t)h(t) $}$, then ${$ \frac{df}{dt} = \frac{dg}{dt}h(t) + g(t)\frac{dh}{dt} $}$
Chain Rule: If ${$ f(t) = g(h(t)) $}$, then ${$ \frac{df}{dt} = \frac{dg}{dh}\frac{dh}{dt} $}$

These rules are used to find the derivative of the position vector, the velocity vector, and the acceleration vector.

References

Calculus: A First Course by Michael Sullivan
Differential Equations: An Introduction to Modern Methods and Applications by James R. Brannan and William E. Boyce

Note: The references provided are for general information and are not specific to the topic of this article.
Q&A: Acceleration of a Particle in the ${$ xy $}$-plane

Introduction

In our previous article, we derived the acceleration vector of a particle moving in the ${$ xy $}$-plane. In this article, we will answer some common questions related to the acceleration of a particle in the ${$ xy $}$-plane.

Q1: What is the acceleration vector of a particle moving in the ${$ xy $}$-plane?

A1: The acceleration vector of a particle moving in the ${$ xy $}$-plane is given by ${$ \langle \frac{d^2x}{dt2}, \frac{d^2y}{dt2}\rangle $}$, where ${$ x(t) = t e^{-t} $}$ and ${$ y(t) = \ln (2t + 6) $}$.

Q2: How do I find the acceleration vector of a particle moving in the ${$ xy $}$-plane?

A2: To find the acceleration vector of a particle moving in the ${$ xy $}$-plane, you need to first find the velocity vector by taking the derivative of the position vector with respect to time. Then, you need to take the derivative of the velocity vector with respect to time to get the acceleration vector.

Q3: What is the relationship between the acceleration vector and the velocity vector?

A3: The acceleration vector is the derivative of the velocity vector with respect to time. This means that the acceleration vector is the rate of change of the velocity vector.

Q4: Can I use the acceleration vector to determine the position of a particle at a given time?

A4: No, the acceleration vector cannot be used to determine the position of a particle at a given time. The acceleration vector only gives information about the rate of change of the velocity vector, not the position of the particle.

Q5: How do I use the acceleration vector to determine the velocity of a particle at a given time?

A5: To use the acceleration vector to determine the velocity of a particle at a given time, you need to integrate the acceleration vector with respect to time. This will give you the velocity vector at the given time.

Q6: Can I use the acceleration vector to determine the position of a particle at a given time if I know the initial velocity?

A6: Yes, if you know the initial velocity of the particle, you can use the acceleration vector to determine the position of the particle at a given time. You can integrate the acceleration vector with respect to time to get the velocity vector, and then integrate the velocity vector with respect to time to get the position vector.

Q7: What is the physical significance of the acceleration vector?

A7: The acceleration vector represents the rate of change of the velocity vector, which is a measure of the force acting on the particle. The acceleration vector can be used to determine the force acting on the particle, which is essential in understanding the motion of the particle.

Q8: Can I use the acceleration vector to determine the force acting on a particle?

A8: Yes, if you know the mass of the particle, you can use the acceleration vector to determine the force acting on the particle. The force acting on the particle is equal to the mass of the particle multiplied by the acceleration vector.

Conclusion

In this article, we have answered some common questions related to the acceleration of a particle in the ${$ xy $}$-plane. We have discussed the relationship between the acceleration vector and the velocity vector, and how to use the acceleration vector to determine the velocity and position of a particle at a given time. We have also discussed the physical significance of the acceleration vector and how to use it to determine the force acting on a particle.

Mathematical Derivations

The mathematical derivations used in this article are based on the following rules:

Product Rule: If ${$ f(t) = g(t)h(t) $}$, then ${$ \frac{df}{dt} = \frac{dg}{dt}h(t) + g(t)\frac{dh}{dt} $}$
Chain Rule: If ${$ f(t) = g(h(t)) $}$, then ${$ \frac{df}{dt} = \frac{dg}{dh}\frac{dh}{dt} $}$

These rules are used to find the derivative of the position vector, the velocity vector, and the acceleration vector.

References

Calculus: A First Course by Michael Sullivan
Differential Equations: An Introduction to Modern Methods and Applications by James R. Brannan and William E. Boyce

Note: The references provided are for general information and are not specific to the topic of this article.