morbidaza said:
Okay...no matter how many times I've tried to do this problem I get 0 (which thinking logically makes no sense), but the answer is actually 6 according to the book.
Find the length of the vector...
r(t)= (cos^3(t))i+(sin^3(t))j from 0 to 2pi
Using the formula (with S being integral, and [[]] meaning norm), L=S([[dr/dt]]dt), i get a norm of 3*sint*cost. integrating that i end up with (3/2)sin^2(t). Plugging in 2pi and 0 i get...0 so far as I can tell.
Maybe I'm having a mental lapse but I've done this problem 3 times now and gotten 0 each time. I can also see no errors in my methods...I'm really confused as to what is going on here.
You can write the vector-valued function as
r(t) = x(t)
i + y(t)
j
When you can write it in that form, there is a theorem for arclength (L) which will help you:
L = the definite integral over the interval [a, b] of SQRT[(dx/dt)² + (dy/dt)²] dt
(It looks like crap in text...sorry)
From our definitions of x(t) and y(t), we get
L = int[a, b] of SQRT[(-3 sin t cos²t)² + (3 cos t sin²t)²] dt
L = int[a, b] of SQRT[9 sin²t (cos²t)² + 9 cos²t (sin²t)²] dt
We can factor the 9 out and take the square root:
L = int[a, b] of 3 SQRT[sin²t (cos²t)² + cos²t (sin²t)²] dt
We can also factor out a common factor of sin²t cos²t from each term:
L = int[a, b] of 3 SQRT[(sin²t cos²t)(cos²t + sin²t)] dt
...except that the second factor is 1! (Definition of the unit circle.) It's trivial to take the square root, and now we have an integral we can do:
L = int[a, b] of 3 sin t cos t dt
u = sin t
du = cos t dt
L = int[a, b] of 3u du
L = (3/2)u² over the interval [a, b]
L = (3/2)sin²t over [0, 2pi]
So what gives? This looks like the answer ought to be zero. However, you have to go back and read the fine print on our theorem for arclength. As written, it is only valid when two conditions hold:
1) The curve C does not intersect itself anywhere on the interval [a, b], except possibly at the endpoint.
2) C has a smooth parameterization given by x = f(t) and y = g(t).
When you draw the graph for
r(t), you will see that we're in good shape for #1. However, the graph is NOT smooth at multiples of pi/2 -- the graph forms a sharp point.
You can still solve the problem by noting that the function is symmetric about the x- and y-axes and smooth over the interval [0, pi/2]:
L = 4(3/2)sin²t over [0, pi/2]
L = 6 sin²t over [0, pi/2]
L = 6 sin²(pi/2) - 6 sin²(0)
L = 6(1) - 0
L = 6
That should be the answer you're looking for. And yes, I did spend the last 30 minutes of my birthday with my old calculus book on my desk. I'm pathetic.
Hope this helps...
