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Arc Length & Parametric Curves

  • Thread starter razored
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  • #1
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Homework Statement


Find the length of the curve y=x^2-4|x|-x from x=-4 to x=4.

The Attempt at a Solution



I realized there is a corner at x=0 so i tried to get around this by pluggin in x for x>=0 and -x for x<0. However, my integrals don't match the answer.

http://texify.com/img/%5CLARGE%5C%21%5CLARGE%5C%21%5Cint_%7B-4%7D%5E%7B0%7D%20%5Csqrt%7B1%2B%282x-3%29%5E2%7D%20dx%20%2B%20%5Cint_0%5E%7B4%7D%20%5Csqrt%7B1%2B%282x-5%29%5E2%7D%20dx%20.gif [Broken]

Which does not match the answer 19.56, what did I do wrong?


Also, when I do try to find arc length in parametric curves and often more that not, a curve repeats itself in a given interval. How do I detect this and avoid it to just find length of one cycle?
 
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Answers and Replies

  • #2
jhae2.718
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How is this curve parametric? x and y are not functions of t.

Arc length for a curve on Cartesian coordinates is the integral from a to b of sqrt(1+(dy/dx)2) dx.

How are x2-3x and x2-5x derivatives of y=x2-4|x|-x? You are finding the arc length of a different curve: x3/3-3/2x2+C on [-4,0) and x3/3-5/2x2+C on (0,4].

If you were operating with parametric equations, Arc length for a curve is the integral from a to b of sqrt((dx/dt)2+(dy/dt)2) dt.
 
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  • #3
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my mistake I was so intent on writing it up that i forgot to take derivative. I did not do this on the actual work when trying to solve this problem.


About hte parametric curves(this one isn't), when i'm given an x=f(t) and y=g(t) to find it in an interval say [a,b] where it repeats in the interval [a,b] say a few times, how will i know when to break it up to find one length of the cycle?


sorry about the confusion
 
  • #4
jhae2.718
Gold Member
1,161
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my mistake I was so intent on writing it up that i forgot to take derivative. I did not do this on the actual work when trying to solve this problem.


About hte parametric curves(this one isn't), when i'm given an x=f(t) and y=g(t) to find it in an interval say [a,b] where it repeats in the interval [a,b] say a few times, how will i know when to break it up to find one length of the cycle?


sorry about the confusion
No problem.

I would look at the behavior of the curve and look for the point where it begins to repeat itself. I would try to find an approximation of where it begins to do so and then solve the equations for a value of t where it does so.

Could you post an example of such a curve? That would help me in understanding exactly what you want, and it would give me something to work with to get you a better answer on what to do.

If the case is a parametric function that moves to a value and then repeats along itself, find the endpoint. It may be a max min; try finding a value for which dy/dx=0.
 
  • #5
172
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No problem.

I would look at the behavior of the curve and look for the point where it begins to repeat itself. I would try to find an approximation of where it begins to do so and then solve the equations for a value of t where it does so.

Could you post an example of such a curve? That would help me in understanding exactly what you want, and it would give me something to work with to get you a better answer on what to do.

If the case is a parametric function that moves to a value and then repeats along itself, find the endpoint. It may be a max min; try finding a value for which dy/dx=0.
x=cos(3t)
y=sin(3t)

x interval [-1,1]
 
  • #6
jhae2.718
Gold Member
1,161
20

Homework Statement


Find the length of the curve y=x^2-4|x|-x from x=-4 to x=4.

The Attempt at a Solution



I realized there is a corner at x=0 so i tried to get around this by pluggin in x for x>=0 and -x for x<0. However, my integrals don't match the answer.

http://texify.com/img/%5CLARGE%5C%21%5CLARGE%5C%21%5Cint_%7B-4%7D%5E%7B0%7D%20%5Csqrt%7B1%2B%282x-3%29%5E2%7D%20dx%20%2B%20%5Cint_0%5E%7B4%7D%20%5Csqrt%7B1%2B%282x-5%29%5E2%7D%20dx%20.gif [Broken]

Which does not match the answer 19.56, what did I do wrong?
To deal with the corner, we split y=x2-4|x|-x into a piecewise function:
y=x2+4x-x=x2+3x on (-inf,0)
and
y=x2-4x-x=x2-5x on (0,inf)

Remember, |x|=-x, -inf<x<0 and x, 0<x<inf

You incorrectly removed the absolute value term. This should give you the correct answer when you take Sa0sqrt(1+(dy/dx)2)dx+S0bsqrt(1+(dy/dx)2)dx.

Can't believe I missed that.
 
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  • #7
jhae2.718
Gold Member
1,161
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x=cos(3t)
y=sin(3t)

x interval [-1,1]
Looking at the graph of x=cos(3t) and y-sin(3t) on a t interval beginning at 0 and ending at some c such that the graph traces over itself, we notice that at t=0 the graph is located at (1,0) in Cartesian coordinates. Thus, the graph will repeat itself the next time x=1.

Thus, we set x=cos(3t)=1. Solving for t, the first solution we get greater than 0 is the upper limit of integration. Then we take find the arc length from 0 to that value.
 

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