1, x = 2+t

^{2}, y = t^{2}+ t^{3 }-
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1, x = 2+t^{2} , y = t^{2} + t^{3 }

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When finding total length of a curve, you must first find an expression for the differential vector along your path of interest, ds.

You know from the pythagorean theorem that the length of any vector in your space is going to be the sqrt(x^2+y^2+z^2....+...)

In your case, you have differential elements dx and dy in your measure space.

So the differential arc length ds = sqrt(dx^2 + dy^2)

= sqrt(1 + (dy/dx)^2)dx or sqrt(1+(dx/dy)^2)dy whichever you prefer

By the chain rule you know that dx = f'(t)dt and dy = g'(t)dt.

Algebraically, you know that dy/dx = f'(t)dt/g'(t)dt

leaving us with ds = sqrt(1+f'(t)/g'(t))f'(t)dt or sqrt(1+g'(t)/f'(t))g'(t)dt

the length then is simply the integral of ds from a to b.

Or, in a more clear form,

Integrate sqrt((dx/dt)^2 + (dy/dt)^2) with respect to t from 0 to 2p.

I will leave the actual arithmetic to you.

Cheers,

--Tim

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