parametric curve/arc length & simpson's rule

**harold** · February 27th 2010, 12:26 PM

This one has been giving me a hard time:

Consider the parametric curve $c(t) = (t cos(t), t sin(t)), 0 \le t \le 2\pi$ (a spiral). Recall from calculus that the arclength formula of a parametric curve $c(t) = (x(t), y(t))$ with $a\le t\le b$ is given by $l(c)=\int_a^b\sqrt{x'(t)^2 + y'(t)^2} \, dt$ .

What I need to figure out is:
1) How to calculate the arclength of the spiral by hand.
2) Use Simpson's Rule to approximate the integral. Find the error bound using the appropriate error formula, and compare it to the actual error.
3) Determine the number of subintervals $n$ required to approximate the integral to within $10^{-6}$ by using the composite trapezoidal rule.

**Aryth** · February 27th 2010, 12:34 PM

Originally Posted by harold

This one has been giving me a hard time:

Consider the parametric curve $c(t) = (t cos(t), t sin(t)), 0 \le t \le 2\pi$ (a spiral). Recall from calculus that the arclength formula of a parametric curve $c(t) = (x(t), y(t))$ with $a\le t\le b$ is given by $l(c)=\int_a^b\sqrt{x'(t)^2 + y'(t)^2} \, dt$ .

What I need to figure out is:
1) How to calculate the arclength of the spiral by hand.
2) Use Simpson's Rule to approximate the integral. Find the error bound using the appropriate error formula, and compare it to the actual error.
3) Determine the number of subintervals $n$ required to approximate the integral to within $10^-6$ by using the composite trapezoidal rule.

Surprisingly, calculating this by hand is the easiest method. First you differentiate the components:

$x'(t) = cos(t) - tsin(t)$

$y'(t) = sin(t) + tcos(t)$

Square them both:

$(x'(t))^2 = cos^2(t) - 2tcos(t)sin(t) + t^2sin^2(t)$

$(y'(t))^2 = sin^2(t) + 2tcos(t)sin(t) + t^2cos^2(t)$

Now add them together to get:

$(x'(t))^2 + (y'(t))^2 = cos^2(t) + sin^2(t) + t^2(sin^2(t) + cos^2(t))$

Recalling the identity $cos^2(t) + sin^2(t) = 1$ , you get:

$(x'(t))^2 + (y'(t))^2 = 1 + t^2$

Now, take the square root and throw it in the integral:

$l(c) = \int_0^{2\pi} \sqrt{1 + t^2} ~dt$

Integrate that and you're all set.

**harold** · February 27th 2010, 12:47 PM

Thanks Aryth! That makes a lot of sense! I still seem stuck on parts 2 and 3 for some reason. Also for part 1 (just the curve itself) how would I graph it in maple or matlab? Not the code, I mean would I graph the end resulting integral you found or is their a way to graph the oringinal (spiral?) curve itself?

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