Arc Length Problem: Find s & Deduce y=e-s | Oscar

[2^Oscar]

Hey guys,

Got a bit of a problem with a question I found in a textbook. I can do most of it but there's 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.Can anyone please help me out?Oscar

 

[LCKurtz]

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 textbook. 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


Thanks for the speedy reply,
Oscar

 

[LCKurtz]

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]

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]

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