Arc Length & Parametric Curves

[razored]

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

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?

 

[jhae2.718]

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.

 

[razored]

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

 

[jhae2.718]

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.

 

[razored]

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]

 

[jhae2.718]

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

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 S_a⁰sqrt(1+(dy/dx)²)dx+S₀^bsqrt(1+(dy/dx)²)dx.

Can't believe I missed that.

 

[jhae2.718]

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.