Given y = 156 - (x - 40)2/60. x = 0 and x = 85 Find distance traveled
Arc Length S = integral of square root of ( y' )2
The Attempt at a Solution
Doing this I get the trig sub tan(t) = (x - 40)/ 30 (Told by teacher to use this instead of -(x - 40)/30 )
so x = 30tan(t) + 40
and dx = 30sec2(t)
So then under the radical I get 1 + ( - (tan (t) )2
So is this equal to the square root of sec2(t)?
Assuming it is, I get the integral of sec2(t) times 30sec2(t) or sec3(t)
I know how to do this (I think) but I'm really not sure (I can't use what's written in the back of my book).
Doing the work I get:
S(x) = 15sec(t)tan(t) + (1/2) ln[ (sec(t) + tan(t) ]
Substitution x back in for t gives
tan(t) = (x-40)/30
sec(t) = square root [ 1 + ( ( -x + 40)/(30) )2
But evaluating this from 0 to 85 gives me an arc length of ~ 75.05, which I know can't be right, but I can't figure out where I went wrong.