Calculus/Arc length

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Arc length

Suppose that we are given a function that is continuous on an interval and we want to calculate the length of the curve drawn out by the graph of from to . If the graph were a straight line this would be easy — the formula for the length of the line is given by Pythagoras' theorem. And if the graph were a piecewise linear function we can calculate the length by adding up the length of each piece.

The problem is that most graphs are not linear. Nevertheless we can estimate the length of the curve by approximating it with straight lines. Suppose the curve is given by the formula for . We divide the interval into subintervals with equal width and endpoints . Now let so is the point on the curve above . The length of the straight line between and is

So an estimate of the length of the curve is the sum

As we divide the interval into more pieces this gives a better estimate for the length of . In fact we make that a definition.

Length of a Curve

The length of the curve for is defined to be

The Arclength Formula

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Suppose that   is continuous on   . Then the length of the curve given by   between   and   is given by

 

And in Leibniz notation

 

Proof: Consider   . By the Mean Value Theorem there is a point   in   such that

 

So

   
 
 
 

Putting this into the definition of the length of   gives

 

Now this is the definition of the integral of the function   between   and   (notice that   is continuous because we are assuming that   is continuous). Hence

 

as claimed.

Example: Length of the curve   from   to  

As a sanity check of our formula, let's calculate the length of the "curve"   from   to   . First let's find the answer using the Pythagorean Theorem.

 

and

 

so the length of the curve,   , is

 

Now let's use the formula

 

Exercises

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1. Find the length of the curve   from   to   .
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2. Find the length of the curve   from   to   .
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Solutions

Arclength of a parametric curve

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For a parametric curve, that is, a curve defined by   and   , the formula is slightly different:

 

Proof: The proof is analogous to the previous one: Consider   and   .

By the Mean Value Theorem there are points   and   in   such that

 

and

 

So

   
 
 
 

Putting this into the definition of the length of the curve gives

 

This is equivalent to:

 

Exercises

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3. Find the circumference of the circle given by the parametric equations   ,   , with   running from   to   .
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4. Find the length of one arch of the cycloid given by the parametric equations   ,   , with   running from   to   .
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Solutions

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Arc length