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

In summary: The problem is asking for y in terms of s, apparently not what you have calculated.In fact it's trivial from the equations y = sech(t) and s = ln(cosh(t)).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 :)
  • #1
2^Oscar
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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[tex]\geq0[/tex]​

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 [tex]\int (tanht)dt[/tex] 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
 
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  • #2
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.
 
  • #3
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
 
  • #4
On looking again, perhaps the difficulty is that the problem is asking for y in terms of s, apparently not what you have calculated.
 
  • #5
In fact it's trivial from the equations y = sech(t) and s = ln(cosh(t)).
 
  • #6
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
 
  • #7
Your original integral was (or should have been)

[tex] s(t) = \int_0^t tanh(u)\, du [/tex]
 
  • #8
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
 

1. What is the Arc Length Problem?

The Arc Length Problem is a mathematical problem that involves finding the length of a curve on a graph. It is commonly used in calculus and is important in understanding the behavior of functions.

2. How do you find the value of s in the Arc Length Problem?

To find the value of s in the Arc Length Problem, you can use the formula s = ∫√(1+(dy/dx)^2) dx, where dy/dx represents the derivative of the function. This formula is derived from the Pythagorean theorem and can be used to find the length of any curve on a graph.

3. What is the significance of the curve y=e-s in the Arc Length Problem?

The curve y=e-s is significant in the Arc Length Problem because it represents the exponential function, which is commonly used in many scientific and mathematical applications. It is also used as an example in many textbooks and problems related to the Arc Length Problem.

4. How can you deduce the value of y=e-s in the Arc Length Problem?

To deduce the value of y=e-s in the Arc Length Problem, you can use the formula y=e-s in combination with the given value of s. By plugging in the value of s, you can solve for y and find the corresponding point on the curve.

5. Who is Oscar in the Arc Length Problem?

Oscar is not a specific person in the Arc Length Problem. It is simply used as a placeholder name for the given function y=e-s. The name Oscar has no mathematical significance and is often used in practice problems as a way to identify a specific function or variable.

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