# Arc Length Problem

1. Dec 6, 2011

### bmr676

1. The problem statement, all variables and given/known data
Find the arc length of y=x*e^(x^6) where 0 ≤ x ≤ 3

2. Relevant equations

3. The attempt at a solution
I took the derivative of the equation and squared it to get (e^2(x^6))(1+12(x^6)+36(x^12)) then plugged it into the proper formula to get:
3
∫√(1+(e^2(x^6))(1+12(x^6)+36(x^12))) dx
0

I tried to plug it into my calculator but got the error message "Overload". I then tried Wolfram and got the answer 1.19685...*10^317, which seemed very extraneous.
One problem I thought about was maybe my area of integration was off, but couldn't think of another solution. Any suggestions or see any flaws?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 6, 2011

### ehild

Are you sure you copied the function correctly?

ehild

3. Dec 6, 2011

### bmr676

absolutely sure...

4. Dec 6, 2011

### ehild

If it is so, you can not find the anti-derivative in closed form, and the arc length will be extremely great. It is certainly greater then the distance between the points (1,y(1)) and (3,y(3)) which is about 4˙10^316. Was not it x(e^x)^6 instead?

ehild

Last edited: Dec 6, 2011
5. Dec 6, 2011

### Joffan

The error is lost in the noise... the answer to a high degree of precision is 3*e^(3^6).

Rationale: it is a monotonic function, so the answer is bounded below by the length of the diagonal and above by the Manhattan distance. Both of those are also approx the above number.

6. Dec 6, 2011

### Joffan

Numerical integration on the bottom part of the curve should suggest exactly how much greater than y(3) the arc length is. Hint: it's less than 1.

7. Dec 6, 2011

### bmr676

Then the answer I stated above isn't as out there as I thought it might have been?

8. Dec 6, 2011

### Joffan

The answer Wolfram gave is numerically correct.

It's more interesting to find the answer relative to y(3), since for practical purposes y(3) is the answer anyway.