Fluid dynamics - find velocity of a particle in a vector fie

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Homework Help Overview

The discussion revolves around determining the speed of a fluid particle at specific points within a three-dimensional vector field, specifically at the origin and at the point (1,-2,0) for a given velocity field. The context is fluid dynamics, focusing on the relationship between velocity and speed in vector fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the process of plugging in coordinates into the velocity vector to find speed and question the relationship between velocity and speed. There are attempts to clarify the distinction between these two concepts and the correct method for calculating speed from a velocity vector.

Discussion Status

Some participants have shared their calculations and expressed uncertainty about their results, indicating that they may be missing key steps or concepts. There is an ongoing exploration of the correct approach to finding the magnitude of the velocity vector, with hints about the need for square roots in the calculation of speed.

Contextual Notes

Participants note a lack of examples in their textbooks and confusion stemming from the absence of clear guidance from their professor regarding three-dimensional vector fields. There is also mention of differing answers among participants, suggesting a need for further clarification on the calculations involved.

Tanner Ross
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Homework Statement


Determine the speed of a fluid particle at the origin and at a point (1,-2,0) for each of the following velocity fields when t = 2s. All distances are in meters and t is in seconds.

V = (x+2)i + xtj - zk m/s

Homework Equations


V x dr = 0
dV/dt = u(dV/dx) + v(dV/dy) + w(dV/dz) + dV/dt

The Attempt at a Solution


Started with trying to:

dx/dt = x+2 ==> dx/(x+2) = dt

dy/dt = xt ==> dy/x = tdt

This is as far as I have gotten, my textbook doesn't give a single example when dealing with 3-dimensional vector fields and my professor doesn't give any clarification on this program in his notes nor has he given any examples based on 3-dimensional problems or a vector field problem in general.
 
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Tanner Ross said:

Homework Statement


Determine the speed of a fluid particle at the origin and at a point (1,-2,0) for each of the following velocity fields when t = 2s. All distances are in meters and t is in seconds.

V = (x+2)i + xtj - zk m/s

Homework Equations


V x dr = 0
dV/dt = u(dV/dx) + v(dV/dy) + w(dV/dz) + dV/dt

The Attempt at a Solution


Started with trying to:

dx/dt = x+2 ==> dx/(x+2) = dt

dy/dt = xt ==> dy/x = tdt

This is as far as I have gotten, my textbook doesn't give a single example when dealing with 3-dimensional vector fields and my professor doesn't give any clarification on this program in his notes nor has he given any examples based on 3-dimensional problems or a vector field problem in general.

V is already a velocity vector (Hint: Look at the units!). The velocity at a point depends on the coordinates of the point (x,y,z) and the time t. You are given both in the problem statement.

Velocity is a vector quantity, but the OP asks for the 'speed' of the particle at those two points. How is 'speed' related to 'velocity'?

Computing dV/dt is not giving you a velocity but something else.
 
So then just plugging in the points gives me the velocity, taking the absolute value would give me the speed I am guessing?

My memory of year to year physics classes goes away quickly so I don't remember much from my first year courses.
 
Tanner Ross said:
So then just plugging in the points gives me the velocity, taking the absolute value would give me the speed I am guessing?

My memory of year to year physics classes goes away quickly so I don't remember much from my first year courses.
Sounds like a plan.

Remember, the hardest part of solving a problem is sometimes understanding what you are given to work with.

Except for knowing the difference between velocity and speed, you don't need to know any physics to work this problem.
 
Well just did that and checked one of my answers, and its wrong so I am still missing something.
 
Tanner Ross said:
Well just did that and checked one of my answers, and its wrong so I am still missing something.
Well, post your calculations (and the answer, if you have it.)
 
Well there was a part b which gave a different vector field

b) V = xy i - 2y^(2)j + tyzk m/s

answer is: 0, 8.246 m/s

My answers are 0 and 10m/s
 
Tanner Ross said:
Well just did that and checked one of my answers, and its wrong so I am still missing something.
Show us your work please.

Chet
 
All I did was what was talked about earlier.

Using the points (0,0,0) and (1,-2,0) I plugged them into the vector field b then took the magnitude for the speed:

V(1,-2,0) = {(1)(-2) - 2(-2)^2 - 0} = -10

speed = |-10 m/s| = 10m/s

But I am guessing I am missing something to do with a sqroot
 
  • #10
Tanner Ross said:
All I did was what was talked about earlier.

Using the points (0,0,0) and (1,-2,0) I plugged them into the vector field b then took the magnitude for the speed:

V(1,-2,0) = {(1)(-2) - 2(-2)^2 - 0} = -10

speed = |-10 m/s| = 10m/s

But I am guessing I am missing something to do with a sqroot
Remember, V = xy i - 2y^(2)j + tyzk is a vector. That's what those i, j, k mean. Absolute value has no meaning here. You want to calculate the magnitude.
 
  • #11
Sigh, right. I knew there was supposed to be a sqroot in there.. Haven't done magnitude in a while.
 

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