Finding The Length of a Curve

  • #1

Homework Statement


Find the length of the curve r(t)=<e^(t) , e^(t)sin(t) , e^(t)cos(t)> between points (1,0,1) and (e^(2pi) , 0 , e^(2pi))

Homework Equations



Length of curve=∫(llv(t)ll Where the limits of integration are the distance between the given points.

The Attempt at a Solution



First, to know what my limits are. I would like to assume they're from 1 to e^(2pi), but something tells me I have to calculate the distance somehow. Would that distance be D=√(Δx^(2)+Δy^(2)+Δz^(2)). Is this correct?

That's the only part I'm confused on. The integral itself isn't so bad.
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement


Find the length of the curve r(t)=<e^(t) , e^(t)sin(t) , e^(t)cos(t)> between points (1,0,1) and (e^(2pi) , 0 , e^(2pi))

Homework Equations



Length of curve=∫(llv(t)ll Where the limits of integration are the distance between the given points.

The Attempt at a Solution



First, to know what my limits are. I would like to assume they're from 1 to e^(2pi), but something tells me I have to calculate the distance somehow. Would that distance be D=√(Δx^(2)+Δy^(2)+Δz^(2)). Is this correct?

That's the only part I'm confused on. The integral itself isn't so bad.

The limits are NOT the distance between the given points. For example, if we calculate the length of the curve (cos(t),sin(t)) once around the circle (say from (1,0) to (1,0) again), the distance between the given points is zero! In the circle case the limits would be t=0 and t=2*pi. To find them in your case, ask yourself: what t gives (x,y,z) = (1,0,1)? What t gives (x,y,z) = (e^(2pi),0,e^(2pi))?

RGV
 
  • #3
The limits are NOT the distance between the given points. For example, if we calculate the length of the curve (cos(t),sin(t)) once around the circle (say from (1,0) to (1,0) again), the distance between the given points is zero! In the circle case the limits would be t=0 and t=2*pi. To find them in your case, ask yourself: what t gives (x,y,z) = (1,0,1)? What t gives (x,y,z) = (e^(2pi),0,e^(2pi))?

RGV

Oh! I see. Well then, my limits would be from 0 to 2pi, right? Because e^0, e^(0)sin(0), e^(0)cos(0) give the point (1,0,1). And e^(2pi), e^(2pi)sin(2pi), e^(2pi)cos(2pi) give the point (e^(2pi), 0, e^(2pi)). Correct?
 
  • #4
Ray Vickson
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Oh! I see. Well then, my limits would be from 0 to 2pi, right? Because e^0, e^(0)sin(0), e^(0)cos(0) give the point (1,0,1). And e^(2pi), e^(2pi)sin(2pi), e^(2pi)cos(2pi) give the point (e^(2pi), 0, e^(2pi)). Correct?

What do YOU think?

RGV
 
  • #6
HallsofIvy
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Homework Statement


Find the length of the curve r(t)=<e^(t) , e^(t)sin(t) , e^(t)cos(t)> between points (1,0,1) and (e^(2pi) , 0 , e^(2pi))

Homework Equations



Length of curve=∫(llv(t)ll Where the limits of integration are the distance between the given points.

The Attempt at a Solution



First, to know what my limits are. I would like to assume they're from 1 to e^(2pi), but something tells me I have to calculate the distance somehow.
No, those are only the two x values. You will be integrating with respect to the parameter t so your limits of integration will be the values of t that give those points. What is t when [itex]x= e^t= 1[/itex], [itex]y= e^t sin(t)= 0[/itex], and [itex]z= e^tcos(t)= 1[/itex]? (Of course, any one of those equations will give the correct answer- check that the others are correct to show that (1, 0, 1) really is on that curve.) What is t when [itex]x= e^t= e^{2\pi}[/itex], [itex]y= e^tsin(t)= 0[/itex], and [itex]z= e^tcos(t)= e^{2\pi}[/itex]? Again, the first equation is enough to find t, but you the others to verify that the point really is on the curve.

Would that distance be D=√(Δx^(2)+Δy^(2)+Δz^(2)). Is this correct?
Well, no, that is the straight line distance between the two points. You may be thinking this- if you take [itex]\Delta x[/itex], [itex]\Delta y[/itex], [itex]\Delta z[/itex] to be very small, that will approximate the length of a short segment of the curve. You can add those for many such short segments to get the "Riemann sum" that approximates the length of the curve: [itex]\sum \sqrt{\Delta x^2+ \Delta y^2+ \Delta z^2}[/itex]. Of course, those will correspond to small [itex]\Delta t[/itex] so that you can "multiply and divide" by that to get [itex]\sum\sqrt{\left(\Delta x/\Delta t\right)^2+ \left(\Delta y\Delta t\right)^2+ \left(\Delta z/\Delta t\right)^2}\Delta t[/itex] which, after taking the limit process, becomes the integral
[tex]\int \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2+ \left(\frac{dz}{dt}\right)^2} dt[/tex]

That's the only part I'm confused on. The integral itself isn't so bad.
 

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