Calculating the Total Distance Traveled by a Particle using Parametric Equations

  • Thread starter Thread starter member 508213
  • Start date Start date
  • Tags Tags
    Particle
Click For Summary
To calculate the total distance traveled by a particle defined by the parametric equations x=t^2-3 and y=(2/3)t^3 from t=0 to 5, the arc length formula is applied. The correct approach involves integrating the square root of the sum of the squares of the derivatives, specifically ∫ from 0 to 5 of √((dx/dt)² + (dy/dt)²) dt. After reevaluating, one participant confirmed the total distance as 87.72. The discussion also clarified that in this context, the total distance and net distance are equivalent, as the integral of the velocity vector magnitude is always positive. The importance of showing work for verification was emphasized, as discrepancies in answers can occur.
member 508213

Homework Statement


a particle's position is represented parametrically by x=t^2-3 and y=(2/3)t^3

Find the total distance traveled by the particle from t= 0 to 5

Homework Equations


Can't think of any

The Attempt at a Solution


I cannot think of a way to do it keeping it in terms of t. All I could do was convert the original equations to Cartesian and evaluate the difference from t=0 to 5 which in terms of x is -3 to 22.
 
Last edited by a moderator:
Physics news on Phys.org
Austin said:

Homework Statement


a particle's position is represented parametrically by x=t^2-3 and y=(2/3)t^3

Find the total distance traveled by the particle from t= 0 to 5

Homework Equations


Can't think of any

The Attempt at a Solution


I cannot think of a way to do it keeping it in terms of t. All I could do was convert the original equation to Cartesian and evaluate the difference from t=0 to 5 which in terms of x is -3 to 22.
Look in your text for the formula for arc length of a parametric curve ##\vec R(t) = \langle x(t),y(t)\rangle##.
 
Ok thanks I was not thinking of arc length for some reason. I got 87.72 as the total distance can that be verified as correct?
 
Also, just as a side question, in this problem what if it asked for net distance instead of total distance, I was just trying to conceptualize that. In this case the net distance would be equal to the total distance correct? When would there be a vector situation where the net distance would not equal the total distance?
 
Actually, how would you even define net distance for a vector situation like this? I kind of confused myself now
 
You will have to show your work to get it checked. I get a different answer so one of us is wrong. Also note that ##\int_a^b|\vec V(t)|~dt## is always positive as long as ##a<b## so there is no need to talk about "net" distance.
 
LCKurtz said:
You will have to show your work to get it checked. I get a different answer so one of us is wrong. Also note that ##\int_a^b|\vec V(t)|~dt## is always positive as long as ##a<b## so there is no need to talk about "net" distance.
It would be the integral from 0 to 5 of squareroot of (dx/dt)^2+(dy/dt)^2 right? So integral from 0 to 5 of squareroot of ( (2t)^2 + (2t^2)^2) dt right?
 
Austin said:
It would be the integral from 0 to 5 of squareroot of (dx/dt)^2+(dy/dt)^2 right? So integral from 0 to 5 of squareroot of ( (2t)^2 + (2t^2)^2) dt right?

That's right. And this morning, I agree with your answer. Must have had a mistake last night.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
9K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K