- #1

- 8

- 0

## Homework Statement

Given y = 156 - (x - 40)

^{2}/60. x = 0 and x = 85 Find distance traveled

## Homework Equations

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 = 30sec

^{2}(t)

So then under the radical I get 1 + ( - (tan (t) )

^{2}

So is this equal to the square root of sec

^{2}(t)?

Assuming it is, I get the integral of sec

^{2}(t) times 30sec

^{2}(t) or sec

^{3}(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

and

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.