- #1

sanitykey

- 93

- 0

**The Length of a Curve**

Hi, I'm stumped!

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

Last edited by a moderator: