Calculating Average Distance from Origin to Curve Integral

sibiryk
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How can I find average distance from the origin point x=y=0 to
the points of the curve using curve integral.

Curve given by

x=cos(2t), y=3sin(2t), t at [0,pi]

I looked in books I have but there is no info on this.
 
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sibiryk said:
How can I find average distance from the origin point x=y=0 to
the points of the curve using curve integral.
Curve given by
x=cos(2t), y=3sin(2t), t at [0,pi]
I looked in books I have but there is no info on this.
Find an equation that gives the distance at any value of t. Then integrate it and divide by the interval (def. of average value).
 
I need to find it using curve integral
 
apmcavoy said:
Find an equation that gives the distance at any value of t. Then integrate it and divide by the interval (def. of average value).

Ok. I integrated equation that give the distance.

I got

Integral ((cos^2(2t)+6sin^2(2t))^0.5)*dt integral from 0 to pi

Did I get it right?
 
sibiryk said:
Ok. I integrated equation that give the distance.
I got
Integral ((cos^2(2t)+6sin^2(2t))^0.5)*dt integral from 0 to pi
Did I get it right?

Close. You need to be more careful with your derivatives and algebra.
 
sibiryk said:
Ok. I integrated equation that give the distance.
I got
Integral ((cos^2(2t)+6sin^2(2t))^0.5)*dt integral from 0 to pi
Did I get it right?
3^2\neq 6

For this case:

\text{Average}=\frac{1}{\pi}\int_{0}^{\pi}f\left(x\right)\,dx

Now it's up to you to find f(x).
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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