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

In summary, the length of a curve with the equation x^2/a^2+y^2/b^2=1, for the range 0<x<a, can be calculated by following the steps listed in this example, which uses the variables c^2 and D^2. However, the length of this curve cannot be determined using the general expression for the length of a circumference of an ellipse, because a and b will increase as x increases.
  • #1
sanitykey
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The Length of a Curve

Hi, I'm stumped! :confused:

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

[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]

for the range [tex]0 \leq x \leq a[/tex]


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

[tex]\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\times dx[/tex]

I think...

[tex]y=b\times\sqrt{1-\frac{x^2}{a^2}}[/tex]

[tex]\frac{dy}{dx}=\frac{-b\times x}{a^2\times\sqrt{1-\frac{x^2}{a^2}}}[/tex]

[tex]\left(\frac{dy}{dx}\right)^2=\frac{b^2\times x^2}{a^2\times(a^2-x^2)}[/tex]

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

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...

[tex]\frac{1}{a^2}[/tex] and [tex]\frac{1}{b^2}[/tex] 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 [tex]x^2[/tex] and [tex]y^2[/tex] i'd call my new variables [tex]c^2[/tex] and [tex]D^2[/tex] and then i thought if i did that i'd have to change the range and since [tex]\frac{x^2}{a^2}=c^2[/tex] then the new range would be [tex]0 \leq \sqrt{c^2 \times a^2}\leq a[/tex] which i thought

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

But i think I'm probably wrong any help would be appreciated thanks :smile:
 
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  • #2
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 [tex]\frac{\pi}{2}[/tex] doesn't really make much sense
 
  • #3
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.
 
  • #4
Thanks for your replies, to Office_Shredder ~ yeah i was thinking [tex]\frac{\pi}{2}[/tex] didn't really make sense in fact errm oh yeah how can the length be a constant if I'm finding a general expression :redface: to arildno ~ would you suggest i should just leave my expression as that integral then if that is right?
 

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

1. What is the formula for calculating the length of a curve?

The formula for calculating the length of a curve is given by L = ∫√(1 + (dy/dx)^2)dx, where dy/dx is the derivative of the curve function.

2. How can I find the length of a curve with a specific domain?

To find the length of a curve with a specific domain, you can use the formula L = ∫√(1 + (dy/dx)^2)dx and substitute the given domain values into the curve function before taking the derivative.

3. Can I use the formula for calculating length of curve for any type of curve?

Yes, the formula L = ∫√(1 + (dy/dx)^2)dx can be used for any type of curve as long as its derivative can be determined.

4. How do I calculate the length of a curve with polar coordinates?

To calculate the length of a curve with polar coordinates, you can use the formula L = ∫√(r^2 + (dr/dθ)^2)dθ, where r is the radius and dr/dθ is the derivative of the polar function.

5. Can I use this formula for curves in three-dimensional space?

No, this formula is specifically for calculating the length of a curve in two-dimensional space. For curves in three-dimensional space, a different formula is used.

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