Arc Length Problem Find 0 ≤ x ≤ 3

[bmr676]

Homework Statement

Find the arc length of y=x*e^(x^6) where 0 ≤ x ≤ 3

Homework Equations

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?

 

[ehild]

Are you sure you copied the function correctly?

ehild

 

[bmr676]

absolutely sure...

 

[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

 

[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.

 

[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.

 

[bmr676]

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

 

[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.