Total distance of a particle

  • #1
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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.
 
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Answers and Replies

  • #2
LCKurtz
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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##.
 
  • #3
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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?
 
  • #4
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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?
 
  • #5
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Actually, how would you even define net distance for a vector situation like this? I kind of confused myself now
 
  • #6
LCKurtz
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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.
 
  • #7
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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?
 
  • #8
LCKurtz
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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.
 

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