Calculating Length of Curve: 0 ≤ x ≤ a | x2/a2 + y2/b2 = 1

[sanitykey]

The Length of a Curve

Hi, I'm stumped!

I've been asked to calculate the length of a curve with the equation...

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

for the range $$0 \leq x \leq a$$


I've been trying to apply the equation...

$$\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\times dx$$

I think...

$$y=b\times\sqrt{1-\frac{x^2}{a^2}}$$

$$\frac{dy}{dx}=\frac{-b\times x}{a^2\times\sqrt{1-\frac{x^2}{a^2}}}$$

$$\left(\frac{dy}{dx}\right)^2=\frac{b^2\times x^2}{a^2\times(a^2-x^2)}$$

$$Length=\int\sqrt{1+\frac{b^2\times x^2}{a^2\times(a^2-x^2)}}\times dx$$

and then if that is right i get stuck and don't know how to do that integral with the limits above.

I tried to say that since...

$$\frac{1}{a^2}$$ and $$\frac{1}{b^2}$$ are constants i can ignore them and follow the method here:

http://www.maths.abdn.ac.uk/~igc/tch/ma1002/int/node21.html (example 2.8 the last one on the page)

but instead of $$x^2$$ and $$y^2$$ i'd call my new variables $$c^2$$ and $$D^2$$ and then i thought if i did that i'd have to change the range and since $$\frac{x^2}{a^2}=c^2$$ then the new range would be $$0 \leq \sqrt{c^2 \times a^2}\leq a$$ which i thought

might be the same as $$0 \leq c \leq 1$$ which i think gives an answer of $$\frac{\pi}{2}$$.

But i think I'm probably wrong any help would be appreciated thanks

 

[Office_Shredder]

As a quick check to why your answer is wrong:

The curve you're looking at is an ellipse. Surely you can agree that if a and b increase, the length will increase. So $$\frac{\pi}{2}$$ doesn't really make much sense

 

[arildno]

sanitykey:
Just forget about about trying to find the general expression for the length of the circumference of an ellipse.

It can be done, but you'll only end up with a rather pointless series solution.

 

[sanitykey]

Thanks for your replies, to Office_Shredder ~ yeah i was thinking $$\frac{\pi}{2}$$ didn't really make sense in fact errm oh yeah how can the length be a constant if I'm finding a general expression to arildno ~ would you suggest i should just leave my expression as that integral then if that is right?