Why won't standard curve length function work in semicircle?by cantRemember Tags: calculus 1, integral, length of a curve 

#1
Jan2513, 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^2x^2) 



#2
Jan2513, 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.




#3
Jan2513, 05:30 AM

P: 16

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



#4
Jan2513, 05:51 AM

P: 13

Why won't standard curve length function work in semicircle? 



#5
Jan2513, 05:55 AM

P: 13





#6
Jan2513, 06:08 AM

P: 16





#7
Jan2513, 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. 



#8
Jan2513, 08:56 AM

Math
Emeritus
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Thanks
PF Gold
P: 38,900

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 t_{0} to t_{1}, 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 yaxis. 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
Jan2513, 08:57 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900





#10
Jan2513, 09:42 AM

P: 13

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 



#11
Jan2513, 09:44 AM

P: 13





#12
Jan2513, 10:09 AM

P: 305





#13
Jan2513, 10:34 AM

P: 13

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 xcoordinate "speed", while d^2x/dt^2 would be it's xaxis acceleration. 



#14
Jan2513, 10:56 AM

P: 305

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


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