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Set up, but do not evaluate, an integral that represents the length of the following curve.

$$\left\{\begin{array}{rl} x &= t-t^2\\ y &= \frac{4}{3}t^{3/2}\end{array}\right. \qquad\qquad 1 \leq t \leq 2$$

My solution:

First of all, I find derivative

$$\dfrac{dx}{dt}=1-2t$$

and

$$\dfrac{dy}{dt}=2\sqrt{t}$$

Since parameter $t$ ranges from $t=1$ to $t=2$ we see that

$$(x(1),y(1))=(0,\frac{4}{3})$$

and

$$(x(2),y(2))=(-2, \frac{8\sqrt{2}}{3})$$

It is easy to see in integrating formula we should begin from $x=-2 \ (t=2)$ to $x=0 \ (t=1)$. So length of this parametric curve is $$L=\int \limits_{2}^{1}\sqrt{(1-2t)^2+4t}\ dt$$ But it is obvious that this integral has negative value. What's wrong?

Why should upper limit be 2 and lower limit be 1. I can not grasp this question for quite long period of time. Please help.

ADDITION: Suppose we have an ellipse in parametric equations $x=a\cos \theta, \ y=b\sin \theta, \ \theta \in[0,2\pi]$. And we want to find its area. Is we compute the integral $\int \limits_{0}^{2\pi}{b\sin \theta (a\cos \theta)' d\theta}$ we will get the negative. Here we have ambiguity with arc length.

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  • $\begingroup$ You shoud move your "addition" to a separate question. It has nothing to do with the arc length. $\endgroup$ – Yves Daoust Oct 11 '17 at 12:14
  • $\begingroup$ @YvesDaoust, Possibly you are right. But i attached it here in order to demonstrate what I am not able to understand. $\endgroup$ – ZFR Oct 11 '17 at 12:23
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"It is easy to see in integrating formula we should begin from...": no, this is your mistake.

The values taken by $x$ are irrelevant in setting the integration bounds. (By the way, why worry about the values of $x$ rather than those of $y$, and what in case $x$ wiggles ?) You are probably confusing with an integral on $x$.

An arc length is given by

$$\int_0^s ds=\int_{t_0}^{t_1}\sqrt{\dot x^2+\dot y^2}dt$$ and is positive. As the integrand is positive, you must have $t_0<t_1$.

If you swap the integration bounds so that $t_0>t_1$, you are integrating on decreasing values of $t$ and you must compensate by using the differential $-dt$ instead of $dt$.

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  • $\begingroup$ Why it is irrelevant in arc length? But it is reasonable in area. I have added example in my question. Please take a look $\endgroup$ – ZFR Oct 11 '17 at 12:13
  • $\begingroup$ @RFZ: I commented with your post. Note anyway that an arc length is always positive, by definition, while an area can have both signs and depends on an orientation convention. $\endgroup$ – Yves Daoust Oct 11 '17 at 12:15
  • $\begingroup$ Why in above example with ellipse the lower limit should $2\pi$ instead of $0$ $\endgroup$ – ZFR Oct 11 '17 at 12:32
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Let function $\mathrm x : [1,2] \to \mathbb R^2$ be defined as follows

$$\mathrm x (t) := \begin{bmatrix} t-t^2\\ \frac{4}{3}t^{3/2}\end{bmatrix}$$

We would like to determine the length of the curve $\{ \mathrm x (t) \mid t \in [1,2] \}$, which is given by the integral

$$\ell := \int_1^2 \| \dot{\mathrm x} (t) \|_2 \, \mathrm d t = \int_1^2 \sqrt{(1-2t)^2 + 4 t} \, \mathrm d t$$

Note that the integration is carried out from endpoint $\mathrm a := \mathrm x (1)$ to endpoint $\mathrm b := \mathrm x (2)$. If, instead, we want to integrate from endpoint $\mathrm b$ to endpoint $\mathrm a$, then let $\tau := 2 - t$ and compute the integral

$$\int_0^1 \| \dot{\mathrm x} (2-\tau) \|_2 \, \mathrm d \tau = \int_0^1 \sqrt{(2\tau-3)^2 + 4 (2-\tau)} \, \mathrm d \tau$$

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  • $\begingroup$ Thanks but my question is bit different. I am asking why we are not begining from $t=2$ to $t=1$? because $t=2$ is left endpoint $\endgroup$ – ZFR Oct 11 '17 at 12:00
  • $\begingroup$ The integral is over $t$, not over $x$. You are computing a length, not an area. $\endgroup$ – Rodrigo de Azevedo Oct 11 '17 at 12:02
  • $\begingroup$ Exactly. Because couple of hours ago i was learning also about an area. Why in are we take direction into account? Could you expand it in detail? I would be very grateful for help $\endgroup$ – ZFR Oct 11 '17 at 12:04
  • $\begingroup$ When you compute area, you add slices of infinitesimal area. Here, we are adding infinitesimal bits of the curve. $\endgroup$ – Rodrigo de Azevedo Oct 11 '17 at 12:07
  • $\begingroup$ I will edit my question. Please take a look $\endgroup$ – ZFR Oct 11 '17 at 12:07

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