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

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

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

 Quote by Studiot 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.

 Quote by 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$
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.

 Quote by 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.
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 .

 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.
 Recognitions: Gold Member Science Advisor 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 $$\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi$$

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

 Quote by 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 .

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

 Quote by HallsofIvy 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 $$\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi$$
Thanks, this was useful.

 Quote by HallsofIvy 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 $$\int_0^{\pi}\sqrt{R^2sin^2(t)+ R^2cos^2(t)}dt= R\int_0^{\pi} dt= R\pi$$
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.

 Quote by 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.
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.

 Quote by cantRemember 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.....

 Tags calculus 1, integral, length of a curve