Arc length for these parametric equations

[Chiborino]

Homework Statement

Find the arc length of the curve described by the parametric equations: x=2e^t & y=3e^3t/2 ln3≤t≤2ln3

Homework Equations

S = ∫(a->b) √[(dy/dt)^2 + (dx/dt)^2]dt

The Attempt at a Solution

Differentiated the two parametrics:
dy/dt = 2e^t
dx/dt = (3/2)*3e^3t/2 = (9/2)e^3t/2

Plugged it all in and got:
∫(ln3->2ln3) √[(2e^t)^2 + ((9/2)e^3t/2)^2]dt
= ∫(ln3->2ln3)√[(4e^2t + (81/4)e^3t]dt

I'm stuck at this integral, I don't see any viable choice for u, and I don't think I'm allowed to approximate it.

*sorry about the sloppy bound notation.

 

[zcd]

Try factoring out a $$e^{2t}$$ from inside the square root and then doing a u substitution. Keep in mind that $$\sqrt{e^{2t}}=e^{t}$$

 

[Chiborino]

Try factoring out a $$e^{2t}$$ from inside the square root and then doing a u substitution. Keep in mind that $$\sqrt{e^{2t}}=e^{t}$$

Wow! Didn't even think of that.

So the integrand simplifies to sqrt([e^2t](4+(81/4)e^t]) = (e^t)*sqrt(4 + 81/4)*e^t)). And u = e^t, du = e^t*dt, right? :D

If so, I can totally handle it from there, thanks!

Edit: meant e^t*dt, not du

 

[Chiborino]

I did a u-sub, then a v-sub (v=[81/4]u+4), and got (2/3e^t + 32/243)|(ln2->3ln2)

And then I got 9, is this answer correct?

 

[zcd]

That's not what I'm getting.
$$\int_{\ln(2)}^{2\ln(3)} \sqrt{4e^{2t}+\frac{81}{4}e^{3t}}\,dt=\frac{9}{2}\int_{\ln(2)}^{2\ln(3)} e^{t}\sqrt{\frac{16}{81}+e^{t}}\,dt$$

Simplifies to
$$\left[ 3(\frac{16}{81}+e^{t})^{\frac{3}{2}}\right]_{\ln(3)}^{2\ln{3}}$$

definitely not 9