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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
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
Last edited by a moderator:
to arildno ~ would you suggest i should just leave my expression as that integral then if that is right?