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We can summarize this method in the following theorem.

Arc length of a parametric curve

Consider the plane curve defined by the parametric equations

x = x ( t ) , y = y ( t ) , t 1 t t 2

and assume that x ( t ) and y ( t ) are differentiable functions of t. Then the arc length of this curve is given by

s = t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t .

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function y = F ( x ) . Then y ( t ) = F ( x ( t ) ) and the Chain Rule gives y ( t ) = F ( x ( t ) ) x ( t ) . Substituting this into [link] gives

s = t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t = t 1 t 2 ( d x d t ) 2 + ( F ( x ) d x d t ) 2 d t = t 1 t 2 ( d x d t ) 2 ( 1 + ( F ( x ) ) 2 ) d t = t 1 t 2 x ( t ) 1 + ( d y d x ) 2 d t .

Here we have assumed that x ( t ) > 0 , which is a reasonable assumption. The Chain Rule gives d x = x ( t ) d t , and letting a = x ( t 1 ) and b = x ( t 2 ) we obtain the formula

s = a b 1 + ( d y d x ) 2 d x ,

which is the formula for arc length obtained in the Introduction to the Applications of Integration .

Finding the arc length of a parametric curve

Find the arc length of the semicircle defined by the equations

x ( t ) = 3 cos t , y ( t ) = 3 sin t , 0 t π .

The values t = 0 to t = π trace out the red curve in [link] . To determine its length, use [link] :

s = t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t = 0 π ( −3 sin t ) 2 + ( 3 cos t ) 2 d t = 0 π 9 sin 2 t + 9 cos 2 t d t = 0 π 9 ( sin 2 t + cos 2 t ) d t = 0 π 3 d t = 3 t | 0 π = 3 π .

Note that the formula for the arc length of a semicircle is π r and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.

A semicircle is drawn with radius 3. There is an arrow pointing counterclockwise. On the graph there are also written three equations: x(t) = 3 cos(t), y(t) = 3 sin(t), and 0 ≤ t ≤ π.
The arc length of the semicircle is equal to its radius times π.
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Find the arc length of the curve defined by the equations

x ( t ) = 3 t 2 , y ( t ) = 2 t 3 , 1 t 3 .

s = 2 ( 10 3 / 2 2 3 / 2 ) 57.589

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We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x -axis, the parametric equations for this curve can be written as

x ( t ) = 140 t , y ( t ) = −16 t 2 + 2 t

where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions x ( t ) and y ( t ) using v as an independent variable, so as to eliminate any confusion with the parameter t :

x ( v ) = 140 v , y ( v ) = −16 v 2 + 2 v .

Then we write the arc length formula as follows:

s ( t ) = 0 t ( d x d v ) 2 + ( d y d v ) 2 d v = 0 t 140 2 + ( −32 v + 2 ) 2 d v .

The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A ,

a 2 + u 2 d u = u 2 a 2 + u 2 + a 2 2 ln | u + a 2 + u 2 | + C .

We set a = 140 and u = −32 v + 2 . This gives d u = −32 d v , so d v = 1 32 d u . Therefore

140 2 + ( −32 v + 2 ) 2 d v = 1 32 a 2 + u 2 d u = 1 32 [ ( −32 v + 2 ) 2 140 2 + ( −32 v + 2 ) 2 + 140 2 2 ln | ( −32 v + 2 ) + 140 2 + ( −32 v + 2 ) 2 | ] + C

and

s ( t ) = 1 32 [ ( −32 t + 2 ) 2 140 2 + ( −32 t + 2 ) 2 + 140 2 2 ln | ( −32 t + 2 ) + 140 2 + ( −32 t + 2 ) 2 | ] + 1 32 [ 140 2 + 2 2 + 140 2 2 ln | 2 + 140 2 + 2 2 | ] = ( t 2 1 32 ) 1024 t 2 128 t + 19604 1225 4 ln | ( −32 t + 2 ) + 1024 t 2 128 t + 19604 | + 19604 32 + 1225 4 ln ( 2 + 19604 ) .

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

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Source:  OpenStax, Calculus volume 2. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11965/1.2
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