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

In summary, Wolfram Alpha had trouble calculating the length of a curve from a to b. The integral can be solved, but if x=-r to r is evaluated it gets undefined (∞-∞ or similar). This is why Wolfram Alpha lagged behind other calculators in this particular calculation.
  • #1
cantRemember
13
0
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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  • #2
Please show how you evaluated your integral. dy/dx is infinite for at least one point in any semicircle.
 
  • #3
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]
 
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  • #4
Studiot said:
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.
 
  • #5
Sayajin said:
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.
 
  • #6
cantRemember said:
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 .
 
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  • #7
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
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]
 
  • #9
Studiot said:
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.
 
  • #10
Sayajin said:
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 "en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating" 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
 
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  • #11
HallsofIvy said:
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.
 
  • #12
HallsofIvy said:
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.
 
  • #13
rollingstein said:
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.
 
  • #14
cantRemember said:
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...
 

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

1. Why is a standard curve length function not working in a semi-circle?

This could be due to the fact that the standard curve length function is designed for linear relationships, and a semi-circle has a curved shape. Therefore, the function may not accurately represent the data and may not provide accurate results.

2. Can the standard curve length function be modified to work in a semi-circle?

It is possible to modify the standard curve length function to work in a semi-circle, but it may require complex mathematical adjustments. Additionally, the accuracy of the results may still be affected due to the inherent differences between linear and curved relationships.

3. Are there any alternative methods to analyze data in a semi-circle shape?

Yes, there are several alternative methods that can be used to analyze data in a semi-circle shape. These include non-linear regression analysis, polynomial curve fitting, and using specific mathematical equations that are suitable for curved relationships.

4. What are the limitations of using a standard curve length function in semi-circle analysis?

The main limitation is the assumption of a linear relationship between variables, which may not hold true in a semi-circle. This can lead to inaccurate results and may not fully capture the complexity of the data. Additionally, it may be difficult to accurately fit a linear function to a curved shape.

5. Can the standard curve length function be used for other non-linear shapes?

The standard curve length function is specifically designed for linear relationships and may not be suitable for non-linear shapes. It is important to carefully consider the underlying relationship between variables and use appropriate methods for analysis. There are other curve fitting techniques that can be used for non-linear shapes, such as exponential or logarithmic functions.

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