Arc Lengths of Parametric Equations (College Board AP® Calculus BC)

Study Guide

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on

Arc lengths of parametric equations

How do I find the arc length of a curve given parametrically?

Graph depicting a parametric curve from t1 to t2 with arrows showing the length of a section of the curve (a dashed line alongside), and an integral formula for arc length L above.
  • The formula to find the length of the curve (the arc length), L units, from the point at t equals t subscript 1 to the point at t equals t subscript 2 on the parametric curve x equals f open parentheses t close parentheses and y equals g open parentheses t close parentheses is

L equals integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript square root of open parentheses fraction numerator d x over denominator d t end fraction close parentheses squared plus open parentheses fraction numerator d y over denominator d t end fraction close parentheses squared end root space d t

Examiner Tips and Tricks

Questions on parametric arc lengths may ask you to leave your answer as a definite integral.

Worked Example

(a) Find the length of the curve x equals 3 t plus 1 and y equals 2 t to the power of 3 over 2 end exponent from t equals 0 to t equals 8.

Use the formula L equals integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript square root of open parentheses fraction numerator d x over denominator d t end fraction close parentheses squared plus open parentheses fraction numerator d y over denominator d t end fraction close parentheses squared end root space d t

It helps to find fraction numerator d x over denominator d t end fraction and fraction numerator d y over denominator d t end fraction individually

table row cell fraction numerator d x over denominator d t end fraction end cell equals 3 row cell fraction numerator d y over denominator d t end fraction end cell equals cell 3 t to the power of 1 half end exponent end cell end table

Substitute these derivatives, and the limits t subscript 1 equals 0 and t subscript 2 equals 8, into the formula L equals integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript square root of open parentheses fraction numerator d x over denominator d t end fraction close parentheses squared plus open parentheses fraction numerator d y over denominator d t end fraction close parentheses squared end root space d t

L equals integral subscript 0 superscript 8 square root of open parentheses 3 close parentheses squared plus open parentheses 3 t to the power of 1 half end exponent close parentheses squared end root space d t

If calculators are allowed, evaluate this definite integral on your calculator

If calculators are not allowed, continue by simplifying under the square root and taking out a 9

table row L equals cell integral subscript 0 superscript 8 square root of 9 plus 9 t end root space d t end cell row blank equals cell integral subscript 0 superscript 8 square root of 9 square root of 1 plus t end root space d t end cell row blank equals cell 3 integral subscript 0 superscript 8 square root of 1 plus t end root space d t end cell end table

There are many ways to evaluate this definite integral, for example integration by substitution using u equals 1 plus t

Note that fraction numerator d u over denominator d t end fraction equals 1 so d u equals d t and that t equals 0 rightwards double arrow u equals 1, t equals 8 rightwards double arrow u equals 9

table row L equals cell 3 integral subscript 1 superscript 9 square root of u space d u end cell row blank equals cell 3 integral subscript 1 superscript 9 u to the power of 1 half end exponent space d u end cell row blank equals cell 3 open square brackets 2 over 3 u to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 9 end cell row blank equals cell 2 open square brackets u to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 9 end cell row blank equals cell 2 cross times open parentheses square root of 9 close parentheses cubed minus 2 cross times open parentheses square root of 1 close parentheses cubed end cell row blank equals cell 54 minus 2 end cell end table

The length of the curve is 52 units

(b) Show that the length of the curve x equals 1 minus cos space t and y equals t minus sin space t from t equals 0 to t equals pi over 2 can be written as square root of 2 integral subscript 0 superscript pi over 2 end superscript square root of 1 minus cos space t end root space d t

In this question, you do not need to evaluate the definite integral

Start by finding fraction numerator d x over denominator d t end fraction and fraction numerator d y over denominator d t end fraction individually

table row cell fraction numerator d x over denominator d t end fraction end cell equals cell sin space t end cell row cell fraction numerator d y over denominator d t end fraction end cell equals cell 1 minus cos space t end cell end table

Substitute these derivatives, and the limits t subscript 1 equals 0 and t subscript 2 equals pi over 2, into the formula L equals integral subscript t subscript 1 end subscript superscript t subscript 2 end superscript square root of open parentheses fraction numerator d x over denominator d t end fraction close parentheses squared plus open parentheses fraction numerator d y over denominator d t end fraction close parentheses squared end root space d t

L equals integral subscript 0 superscript pi over 2 end superscript square root of open parentheses sin space t close parentheses squared plus open parentheses 1 minus cos space t close parentheses squared end root space d t

This is not how it is given in the question, so expand and simplify under the square root

L equals integral subscript 0 superscript pi over 2 end superscript square root of sin squared t plus 1 minus 2 cos space t plus cos squared t end root space d t

Here, you can use the trigonometric identity sin squared space t plus cos squared space t equals 1 to simplify further

table row L equals cell integral subscript 0 superscript pi over 2 end superscript square root of 1 plus 1 minus 2 cos space t end root space d t end cell row blank equals cell integral subscript 0 superscript pi over 2 end superscript square root of 2 minus 2 cos space t end root space d t end cell end table

This is almost the answer given, but a 2 must be factored out from inside the square root

table row L equals cell integral subscript 0 superscript pi over 2 end superscript square root of 2 open parentheses 1 minus cos space t close parentheses end root space d t end cell row blank equals cell integral subscript 0 superscript pi over 2 end superscript square root of 2 square root of 1 minus cos space t end root space d t end cell row blank equals cell square root of 2 integral subscript 0 superscript pi over 2 end superscript square root of 1 minus cos space t end root space d t end cell end table

square root of 2 integral subscript 0 superscript pi over 2 end superscript square root of 1 minus cos space t end root space d t

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Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.