 #1
 141
 2
 Homework Statement:
 See Attachment
 Relevant Equations:

ds=sqrt(dx^2+dy^2+dz^2)
Arc Length = integral on C of ds
Problem: See Attachment. Parts (a) & (b) are clear, but my confusion arises in (c) I feel like there is a much simpler form. While technically my answer is correct, there must be something I'm missing.
I parameterized the curve C=(t, e^2t, e^2t) and got c'(t)=(1,2e^2t,2e^2t), which should be correct since I got the answer for (b). I said ds=sqrt(c'(t))
So ds=sqrt(1+4e^2t+4e^4t) =sqrt((1+2e^2t)^2)=1 + 2e^2t. The limits are from t=0 to t=x_0.
Integral(1+2e^2t) dt = t+e^2t, and I'll evaluate at the limits to get:
x_0 + e^(2x_0)  1. So if x_0=ln(W+1)^(1/3), then e^(2x_0)= (W+1)^2/3. But still, why ask for it in this crazy form? Or is it just a silly excuse to
I parameterized the curve C=(t, e^2t, e^2t) and got c'(t)=(1,2e^2t,2e^2t), which should be correct since I got the answer for (b). I said ds=sqrt(c'(t))
So ds=sqrt(1+4e^2t+4e^4t) =sqrt((1+2e^2t)^2)=1 + 2e^2t. The limits are from t=0 to t=x_0.
Integral(1+2e^2t) dt = t+e^2t, and I'll evaluate at the limits to get:
x_0 + e^(2x_0)  1. So if x_0=ln(W+1)^(1/3), then e^(2x_0)= (W+1)^2/3. But still, why ask for it in this crazy form? Or is it just a silly excuse to