find the length of the curve x=t^3, y=3t^2/2, 0

Question

find the length of the curve x=t^3, y=3t^2/2, 0<t<sqrt(3)

anonymous · Answer

Formula for length of curve is:
$$\int_a^b \sqrt{1 + (\frac{\mathrm{d}y}{\mathrm{d}x})^2} \, \mathrm{d}x$$
That won't format correctly, but basically:
Integral from a to b of sqrt(1 + (dy/dx)^2) with respect to x.

anonymous · Answer

hmm... this is for a parametrized curve..  are you sure that formula will work?

anonymous · Answer

No, it will not work, but it gives us something to build on. Since I'm tired of my notation not showing up properly, read this: http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

anonymous · Answer

ahh wait.. i see my error.  I calculated dy/dt wrong... integrated instead of taking derivative.

anonymous · Answer

Aw man, eashy, spoilin' my fun.

anonymous · Answer

i used L=int(0,sqrt(3))sqrt( (dx/dx)^2+(dy/dt)) dt

anonymous · Answer

its 7?

anonymous · Answer

Yes, indeed it is.

anonymous · Answer

woo thanks! now to move on to polar coordinates .

anonymous · Answer

That is still one irritating integral.

anonymous · Answer

i agree.  at least this class is easier than calc 2

anonymous · Answer

At the end of the day, it's all really fun.