# Why won't standard curve length function work in semi-circle?

1. Jan 25, 2013

### cantRemember

Ok, so for a give function f(x) it's curve length from a to b is supposed to be
∫(1+(f '(x))^2)dx evaluated from a to b. However even wolfram alpha had a hard time solving that, plus the results were wrong.
What am I missing?

PS: With f(x)= sqrt(r^2-x^2)

Last edited: Jan 25, 2013
2. Jan 25, 2013

### Studiot

Please show how you evaluated your integral. dy/dx is infinite for at least one point in any semicircle.

3. Jan 25, 2013

### Sayajin

It works actualy :
$\frac{d(\sqrt{r^2-x^2})}{dx}=\frac{x}{\sqrt{r^2-x^2}}$
Now replace this in the length of the curve formula:
$\int_a^b \sqrt{1 + \frac{x^2}{r^2-x^2}} dx =$$\int_a^b \sqrt{\frac{r^2 - x^2 +x^2}{r^2-x^2}}$$=\frac{r}{r}\int_a^b \frac{dx}{\sqrt{1-\frac{x^2}{r^2}}}=$$r \int_a^b \frac{d(\frac{x}{r})}{\sqrt{1 - \frac{x^2}{r^2}}}= r[arcsin \frac{x}{r}]_a^b$

Last edited: Jan 25, 2013
4. Jan 25, 2013

### cantRemember

You are correct. I should test for derivative's continuity first.

5. Jan 25, 2013

### cantRemember

The integral can be solved indeed, but if you try to evaluate it for x=-r to r (i.e. the entire semi-circle) you get something.. ugh... undefined (∞-∞ or similar). This is why (I guess) Wolfram Alpha lagged, having trouble to calculate such a definite integral. Thanks for your assistance.

6. Jan 25, 2013

### Sayajin

The free version of Wolfram Alpha have limited computation time. If it tries to calculate something and it takes too much it does not show the result. This is made so people can buy the pro edition. Finding the antiderivative and them using the second fundamental theorem of calculus to evaluate the integral from -r to r you get r(arcsin(1)-arcsin(-1)) = πr which is just the half of the circle. But you should be careful with a and b because they should not be bigger than r .

Last edited: Jan 25, 2013
7. Jan 25, 2013

### Studiot

Have you thought about what you mean by a semi circle?

If you include both ends of a diameter and use them as your limits you have gone more than half way round the circle and what is left is short by this. If you go all the way round and divide by 2 you don't have this problem.

8. Jan 25, 2013

### HallsofIvy

Staff Emeritus
By the way, another way to find arc length is to write the curve in terms of parametric equations. If x= x(t) and y= y(t) then the arclength, from t0 to t1, is given by
$$\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt$$

And a judicious choice of parameter can avoid problems like the curve being parallel to the y-axis. In particular, x= R cos(t), y= R sin(t), for t= 0 to $\pi$ gives arclength
$$\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi$$

9. Jan 25, 2013

### HallsofIvy

Staff Emeritus
Since the "length" of a single point is 0, this is irrelevant to the question asked.

10. Jan 25, 2013

### cantRemember

The proof of "en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating" [Broken] requires that the f '(x) is continuous in the closed interval [a,b]
And x→r f '(x)→ -∞ so f '(x) is not continuous at x=±r (Plot)
Also see This

Last edited by a moderator: May 6, 2017
11. Jan 25, 2013

### cantRemember

Thanks, this was useful.

12. Jan 25, 2013

### rollingstein

As an aside, do the parametric derivatives have any intuitive meaning? I see that they don't go to ±∞ unlike dy/dx in the Cartesian representation. Is that a general property or specific to this case.

13. Jan 25, 2013

### cantRemember

They can go to ±∞ (i.e. ln(x)''=(1/x)'=-(1/x^2)
and as x→0, f '' (x) → -∞

Besides that, you can think the parametric plots as "trajectories", the point x,y moves as time proceeds. The derivative dx/dt would (probably) be the x-coordinate "speed", while d^2x/dt^2 would be it's x-axis acceleration.

14. Jan 25, 2013

### rollingstein

Sorry I should have framed better. I meant dx/dt and dy/dt alone. Could they go to ±∞ anywhere on a curve whose length is to be measured?

I can't come up with a case.....