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

by cantRemember
Tags: calculus 1, integral, length of a curve
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cantRemember
#1
Jan25-13, 03:48 AM
P: 13
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)
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Studiot
#2
Jan25-13, 04:37 AM
P: 5,462
Please show how you evaluated your integral. dy/dx is infinite for at least one point in any semicircle.
Sayajin
#3
Jan25-13, 05:30 AM
P: 16
It works actualy :
[itex]\frac{d(\sqrt{r^2-x^2})}{dx}=\frac{x}{\sqrt{r^2-x^2}}[/itex]
Now replace this in the length of the curve formula:
[itex]\int_a^b \sqrt{1 + \frac{x^2}{r^2-x^2}} dx =[/itex][itex] \int_a^b \sqrt{\frac{r^2 - x^2 +x^2}{r^2-x^2}}[/itex][itex]=\frac{r}{r}\int_a^b \frac{dx}{\sqrt{1-\frac{x^2}{r^2}}}=[/itex][itex]r \int_a^b \frac{d(\frac{x}{r})}{\sqrt{1 - \frac{x^2}{r^2}}}= r[arcsin \frac{x}{r}]_a^b[/itex]

cantRemember
#4
Jan25-13, 05:51 AM
P: 13
Why won't standard curve length function work in semi-circle?

Quote Quote by Studiot View Post
Please show how you evaluated your integral. dy/dx is infinite for at least one point in any semicircle.
You are correct. I should test for derivative's continuity first.
cantRemember
#5
Jan25-13, 05:55 AM
P: 13
Quote Quote by Sayajin View Post
It works actualy :
[itex]\frac{d(\sqrt{r^2-x^2})}{dx}=\frac{x}{\sqrt{r^2-x^2}}[/itex]
Now replace this in the length of the curve formula:
[itex]\int_a^b \sqrt{1 + \frac{x^2}{r^2-x^2}} dx =[/itex][itex] \int_a^b \sqrt{\frac{r^2 - x^2 +x^2}{r^2-x^2}}[/itex][itex]=\frac{r}{r}\int_a^b \frac{dx}{\sqrt{1-\frac{x^2}{r^2}}}=[/itex][itex]r \int_a^b \frac{d(\frac{x}{r})}{\sqrt{1 - \frac{x^2}{r^2}}}= r[arcsin \frac{x}{r}]_a^b[/itex]
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.
Sayajin
#6
Jan25-13, 06:08 AM
P: 16
Quote Quote by cantRemember View Post
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.
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 .
Studiot
#7
Jan25-13, 07:06 AM
P: 5,462
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.
HallsofIvy
#8
Jan25-13, 08:56 AM
Math
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P: 39,565
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
[tex]\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt[/tex]

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 [itex]\pi[/itex] gives arclength
[tex]\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi[/tex]
HallsofIvy
#9
Jan25-13, 08:57 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,565
Quote Quote by Studiot View Post
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.
Since the "length" of a single point is 0, this is irrelevant to the question asked.
cantRemember
#10
Jan25-13, 09:42 AM
P: 13
Quote Quote by Sayajin View Post
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 .

The proof of this formula 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
cantRemember
#11
Jan25-13, 09:44 AM
P: 13
Quote Quote by HallsofIvy View Post
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
[tex]\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt[/tex]

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 [itex]\pi[/itex] gives arclength
[tex]\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi[/tex]
Thanks, this was useful.
rollingstein
#12
Jan25-13, 10:09 AM
PF Gold
P: 330
Quote Quote by HallsofIvy View Post
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
[tex]\int_{t_0}^{t_1}\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2} dt[/tex]

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 [itex]\pi[/itex] gives arclength
[tex]\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi[/tex]
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.
cantRemember
#13
Jan25-13, 10:34 AM
P: 13
Quote Quote by rollingstein View Post
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.
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.
rollingstein
#14
Jan25-13, 10:56 AM
PF Gold
P: 330
Quote Quote by cantRemember View Post
They can go to ∞ (i.e. ln(x)''=(1/x)'=-(1/x^2)
and as x→0, f '' (x) → -∞
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.....


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