# Arc Length Problem

#### 2^Oscar

Hey guys,

Got a bit of a problem with a question I found in a text book. I can do most of it but theres one little part I'm really struggling with:

A curve C is given parametrically by:
x=t-tanht, y=secht, t$$\geq0$$​

The length of arc C measured from the point (0,1) to a general point with parameter t is s. Find s in terms of t and deduce that, for any point on the curve, y=e-s.

I'm happy finding that the arc length is defined as $$\int (tanht)dt$$ between the limits of 0 and s, and i evaluate this integral to be ln(coshs) however after this I am stumped; I am having great trouble getting to y=e-s.

Oscar

Last edited:

#### LCKurtz

Homework Helper
Gold Member
Perhaps you have misunderstood or mis-quoted the problem. Your work looks correct but the reason you can't get to y = e-s is the two aren't equal, as trying s = 0 will show.

#### 2^Oscar

It is a past examination question and I quoted it word for word from what is in my text book. I've doubled checked it and what I have written is definitely what is written down here... so unless there is a typo in the textbook I really don't know what is going on :S

Oscar

#### LCKurtz

Homework Helper
Gold Member
On looking again, perhaps the difficulty is that the problem is asking for y in terms of s, apparently not what you have calculated.

#### LCKurtz

Homework Helper
Gold Member
In fact it's trivial from the equations y = sech(t) and s = ln(cosh(t)).

#### 2^Oscar

Sorry if i'm being stupid here;

Could you please tell me how you got s=ln(cosh(t))? I evaluated the integral as C=ln(cosh(s)), so I am confused as to how you made this step. I'm okay as to how you solved the problem from there on though :)

Thank you for the help,
Oscar

#### LCKurtz

Homework Helper
Gold Member
Your original integral was (or should have been)

$$s(t) = \int_0^t tanh(u)\, du$$

#### 2^Oscar

Ahh i see my error, you were right in your first post, I have misread the question.

I understand now.

Thank you for your help :)
Oscar