UPI Study › Courses › Calculus 2 › Free lessons › Lesson 4.4 Free lesson
Arc Length of a Curve
Calculus 2 · Chapter 4: Applications of IntegrationSection 4.4 · lesson 19 of 5411 min · free
← Volumes: The Cylindrical Shell MethodNext lesson: Surface Area of a Solid of Revolution →
In this lesson
Measuring something that bends.
- Approximate a curve by chords and take the limit
- Derive the arc length element ds
- Integrate in x or in y, whichever matches the curve
- Recognise why most real arc lengths must be computed numerically
Transcript
Everything said in the video, section by section (1,649 words). Read along, or search it.
MAT-265 - Calculus 2 · Section 4.4
Arc Length of a Curve
A highway engineer laying asphalt along a curving mountain road needs the true length of the path. Not the straight-line distance between its endpoints. The actual distance the asphalt has to cover. A manufacturer cutting decorative metal trim shaped like a parabola must know exactly how much material the curve consumes. A geneticist estimating the length of a coiled strand, or a cartographer measuring a winding river, faces the same question. How do you measure the length of something that bends? A ruler works for straight segments. But a curve has no straight pieces to lay a ruler against. So the resolution is to build straight pieces ourselves. Approximate the curve by many tiny straight segments. Measure each one with the distance formula. Add them up. And let the segments shrink. Which is precisely the logic of the integral.
The Question
How Long Is the Road You Are Driving?
Before we start, a note on why this section matters more than its length suggests. The derivation below produces a compact differential — the arc length element, written d s. And that single object becomes the foundation for a great deal that follows. Not just arc length. Also surface area of revolution in the next section. Work along a path. And the length of parametric and polar curves in Chapter Nine. So master this one differential, and several later sections become almost free. That's why those sections cite this derivation rather than repeating it.
The Idea
Approximate the Curve by Straight Segments
Consider a smooth curve y equals f of x, from x equals a to x equals b. Divide the interval into small pieces, and mark the corresponding points on the curve. Now connect consecutive points with straight line segments. These are called chords. Together they produce a polygonal path that hugs the curve. The total length of that polygonal path approximates the length of the curve. And the approximation improves as the segments become shorter and more numerous. Look at the picture. The green curve is the truth; the red path is our approximation. Already close, with only seven chords. Now zoom in on one chord. Each chord is the hypotenuse of a small right triangle. Its horizontal leg is the change in x, delta x. Its vertical leg is the change in y, delta y. So by the Pythagorean theorem, the length of one chord is root delta x squared, plus delta y squared. The total polygonal length is the sum of all such chords. Taking the limit as the chords shrink to zero width converts that sum into an integral. But first, the chord length has to be rewritten into a form ready for integration.
The Derivation
From a Chord to the Element ds
Here's the pivotal algebraic step. And it's worth understanding rather than memorizing. Start from the chord length, root delta x squared plus delta y squared. Factor delta x squared out from under the radical. You get root of delta x squared, times the quantity one plus delta y squared over delta x squared. Pull the delta x squared out of the root, and it becomes a factor of delta x on the outside. So delta s equals root of one plus, delta y over delta x, squared — all times delta x. Why is that the pivotal step? Because it isolates the ratio delta y over delta x. And that ratio is the slope of the chord. As delta x approaches zero, that slope approaches the derivative, d y d x. And the discrete delta s becomes the infinitesimal arc length element, d s. So d s equals root of one plus d y d x squared, d x. Or equivalently, root of one plus d x d y squared, d y. Or in fully symmetric form: d s equals root of d x squared plus d y squared. Those aren't three different facts. They're the same element, organized for integration in different variables.
The Formulas
Integrate the Element
Integrating the arc length element over the appropriate interval gives the two formulas. For a curve y equals f of x with continuous derivative on a to b, the length is the integral of root one plus f prime of x squared, d x. For a curve x equals g of y with continuous derivative on c to d, it's the integral of root one plus g prime of y squared, d y. The requirement that the derivative be continuous — that the curve is smooth — is what guarantees the integral exists. And the choice between the two formulas is dictated by how the curve is most naturally described. Exactly as the choice between d x and d y slicing was in section four point one. Let's do one. Find the arc length of y equals two thirds x to the three halves, from zero to three. This function is designed so the radical simplifies. Watch. The derivative is two thirds times three halves x to the one half. Which is just x to the one half. So d y d x squared is simply x. And the integrand becomes root one plus x. A perfect square under the radical is not needed here — it's already elementary. Substitute u equals one plus x, and you get the integral of root u from one to four. That's two thirds times eight minus one. Fourteen thirds. About four point six seven units.
Worked Example 2
When the Curve Is Given as x = g(y)
Now one given the other way round. As x in terms of y. Find the arc length of x equals one third, times y squared plus two, to the three halves. From y equals zero to y equals two. The curve is given as x equals g of y. So use the second formula, and differentiate with respect to y. Apply the chain rule. You get one third, times three halves, times y squared plus two to the one half, times two y. Simplify: y times y squared plus two, to the one half. Now square it. You get y squared times y squared plus two. Which expands to y to the fourth plus two y squared. Add one, and look for a perfect square. One plus y to the fourth plus two y squared. That's y squared plus one, all squared. So the radical collapses — root of y squared plus one squared is just y squared plus one, since it's nonnegative on our interval. The integral becomes the integral of y squared plus one, from zero to two. Which is eight thirds plus two. Fourteen thirds again. And notice what made that work. The function was deliberately designed so that one plus d x d y squared is a perfect square. That design is the hallmark of arc length problems that can be solved in closed form.
An Honest Note
How Rare That Clean Outcome Is
Now let me be honest with you about how rare that clean outcome actually is. The two examples above were constructed so the radical collapses. The overwhelming majority of curves you meet in practice produce an integrand with no elementary antiderivative at all. Go back to the highway engineer from the opening. Suppose the road follows the profile y equals x squared over some stretch. As innocent a curve as exists. The resulting arc length integral is the integral of root one plus four x squared. That cannot be finished with the power rule. It demands a trigonometric substitution at best. And for most real survey data — where the curve is known only as a table of measured points — there's no formula to substitute into at all. The length is computed numerically. Modern engineering software evaluates arc length by methods such as Simpson's rule, applied directly to root one plus f prime squared. Which is conceptually nothing more than the polygonal approximation from our figure — carried out by a computer, with a very large number of very short chords. So the exact-antiderivative examples in this section are best understood as special cases that build intuition. While the chord-summing idea behind them is what actually scales to practice. Here's the thing to think about. The arc length integrand contains a square root of a sum. And square roots of sums almost never have elementary antiderivatives. Why does that suggest most real arc lengths get computed numerically rather than symbolically?
Key Takeaways
What to Carry Forward
The most frequent errors here are mechanical rather than conceptual. First. Forgetting the plus one under the radical. Writing root of d y d x squared collapses to the absolute value of d y d x — and measures something meaningless. That one comes from the horizontal leg, delta x, of the triangle. It is never optional. Second. When a curve is given as x equals g of y, it's tempting to compute d y d x anyway. The correct move is to differentiate in the variable matching the formula. Use d x d y for the y-integral. A reliable habit: write down the symmetric element, d s equals root d x squared plus d y squared, first. Then decide which variable to factor out, based on how the curve is described. And finally, always confirm the derivative is continuous on the interval. A corner or a vertical tangent can invalidate the formula, or require splitting the integral. Five takeaways. Arc length comes from approximating a curve by chords, measuring each with the Pythagorean theorem, and taking the limit. The arc length element is root one plus d y d x squared, d x. Equivalently, root d x squared plus d y squared. And it gets reused for surface area, and for parametric and polar lengths. For y equals f of x, integrate in x. For x equals g of y, integrate in y. Workable problems are engineered so the expression under the radical becomes a perfect square. Most real curves require numerical integration. And never drop the plus one — write the symmetric d s first to avoid that error entirely. Next up is section four point five — Surface Area of Revolution, which multiplies this same d s by a circumference.
About Calculus 2 at UPI Study
Calculus 2 (Math 265) is a self-paced online course of 54 video lessons in 10 chapters, from integration techniques to series, parametric curves and vectors. It carries a NCCRS credit recommendation for 4 semester credits.
Every lesson video and transcript is free on pages like this one. Enrolling adds a practice quiz after each lesson, the graded work and final exam that earn the credit, lifetime access, and an official transcript on completion. UPI Study has helped over 130,000 students with college credits. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.
Enroll now → Course details Download the syllabus Every course with PRO →
Why take it here
Why students take Calculus 2 with UPI Study
Where the credit goes
Universities that work with NCCRS
Calculus 2 carries a NCCRS credit recommendation. These universities work with NCCRS; some are UPI Study's signed partners. ACE and NCCRS have over 2,100 cooperating colleges that will consider accepting credits; the final decision is with the universities.
Thomas Edison State UniversitySigned partner
University of the PeopleSigned partner
Charter Oak State CollegeSigned partner
Strayer UniversitySigned partner
Columbia Southern UniversitySigned partner
Indiana TechSigned partner
Penn StateWorks with NCCRS
Southern New Hampshire UniversityWorks with NCCRS
SUNYWorks with NCCRS
+many more cooperating universitiesFind my college →Find my college → Not listed? Ask Athena in the corner of this page, or book a 15-minute call.
FAQ
Questions students ask us
About Arc Length of a Curve
What is the arc length formula in calculus?
Where does the arc length formula come from?
How do you find the arc length of y=\tfrac{2}{3}x^{3/2} from x=0 to x=3?
Why are most arc lengths computed numerically instead of exactly?
About the Calculus 2 course
How many college credits is Calculus 2?
How much does Calculus 2 cost and what is included?
Which universities accept UPI Study credits?
How long does Calculus 2 take?
How are the exams done?
How do my credits reach my college?
Ready for the full course? Earn 4 credits in Calculus 2.
54 video lessons, a quiz after each one, and the graded work that earns the credit. Recommended for credit by NCCRS.
4 credits · NCCRSEnroll →