Derivative of Position Vector at Specified Time

In summary, the student is struggling with the concept of derivatives of vectors at a specified time in a proof in orbital mechanics. They question whether the derivatives of position and velocity vectors at time 0 should be 0, but understand that this may not be the case due to the analogy with scalar quantities. They ask for clarification on how to approach taking the derivative in a specific problem.
  • #1
bghommy
2
0

Homework Statement


My homework problem is a proof in orbital mechanics, but I'm not looking for specific help on that just yet, I'd like to work through it on my own. In doing so however, I'm having a hard time conceptualizing the idea of derivatives of vectors at a specified time. If r is a general position vector, and r0 is the position vector at time t0, and the same applies for velocity vectors v and v0, it seems to me that the derivatives of each of the vectors specified at time 0 should be 0, because these values are constant. But I also don't see how that can be true because v0 should be the derivative of r0.

Homework Equations


r=ar0+bv0 where a and b are scalar functions of time.

The Attempt at a Solution


If I attempt to take the derivative of the above equation, I'm not sure whether or not I can say the derivatives of r0 and v0 are 0, leaving me with v=a'r0+b'v0 or not.

Thanks for any help guys, sorry if this is a bit of a dumb question but it's really messing with my head right now. Cheers
 
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  • #2
Vectors are analogous to their scalar cousins. Just because an initial position is fixed doesn't imply the initial velocity must be zero, or that the initial acceleration must be zero. Consider throwing a ball from a rooftop. The initial position is fixed but not zero. The initial velocity is not zero because it's thrown, and the initial acceleration is not zero because gravity doesn't go away :smile:
 

Related to Derivative of Position Vector at Specified Time

1. What is the purpose of finding the derivative of a position vector at a specified time?

The derivative of a position vector at a specified time is used to calculate the instantaneous velocity of an object at that specific time. This can help in understanding the motion and speed of an object at a particular moment.

2. How is the derivative of a position vector calculated?

The derivative of a position vector is calculated by taking the derivative of each component of the vector with respect to time. This can be done using the chain rule or by finding the slope of the tangent line to the curve at the specified time.

3. What is the difference between average velocity and instantaneous velocity?

The average velocity of an object is the total displacement divided by the total time taken. On the other hand, instantaneous velocity is the velocity of an object at a specific moment in time, which can be calculated by finding the derivative of the position vector at that time.

4. Can the derivative of a position vector be negative?

Yes, the derivative of a position vector can be negative. This indicates that the object is moving in the negative direction at the specified time. A positive derivative indicates movement in the positive direction and a zero derivative indicates that the object is at rest at that moment.

5. How can the derivative of a position vector be used to find the acceleration of an object?

The second derivative of a position vector is the acceleration of an object. This can be calculated by taking the derivative of the velocity vector. The sign of the second derivative can indicate whether the object is accelerating or decelerating at the specified time.

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