Arc length of a parametric curve — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Derivation of the arc length formula for smooth parametric curves, evaluating definite integrals for arc length, accounting for curve orientation, solving numerical integration problems, and applied motion problems involving path length.

You should already know: Derivatives of parametric functions, definite integral evaluation techniques, chain rule for differentiation.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Arc length of a parametric curve?

Arc length is the total distance along a curve between two endpoints, distinct from the straight-line distance between those points. For parametric curves, where $x$ and $y$ are both defined as functions of a third parameter $t$ ($x=x(t),y=y(t)$ for $a\le t\le b$), arc length gives the total length of the path traced out as $t$ moves from $a$ to $b$. This topic is explicitly required by the AP Calculus BC Course and Exam Description (CED), accounting for approximately 1-3% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is commonly paired with parametric motion problems, where $t$ represents time, and arc length equals total distance traveled by a particle over an interval. Unlike Cartesian arc length, the parametric formula works for curves that are not functions of $x$ or $y$, including loops, cycloids, and projectile paths, making it far more general for common use cases.

2. Derivation and Arc Length Formula for Smooth Parametric Curves

To derive the parametric arc length formula, we use the same Riemann sum approach used for Cartesian arc length, adapted for two parametric functions. First, split the parameter interval $[a,b]$ into $n$ subintervals of width $\Delta t=\frac{b-a}{n}$. At each parameter value $t_i$, we have a corresponding point $(x(t_i),y(t_i))$ on the curve. By the Pythagorean theorem, the straight-line distance between consecutive points $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$ is $\sqrt{(\Delta x_i{)}^2+(\Delta y_i{)}^2}$. By the Mean Value Theorem, there exists a $t_i^{\ast }$ in $[t_i,t_{i+1}]$ such that $\Delta x_i=x^{\prime }(t_i^{\ast })\Delta t$ and $\Delta y_i=y^{\prime }(t_i^{\ast })\Delta t$. Substituting gives: $$\sqrt{(x^{\prime }(t_i^{\ast })\Delta t{)}^2+(y^{\prime }(t_i^{\ast })\Delta t{)}^2}=\sqrt{(x^{\prime }(t_i^{\ast }){)}^2+(y^{\prime }(t_i^{\ast }){)}^2}\Delta t$$ Taking the limit as $n\to \infty$ converts the Riemann sum to a definite integral, giving the final formula: $$L={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$ This formula only applies to smooth parametric curves, meaning $x^{\prime }(t)$ and $y^{\prime }(t)$ are continuous on $[a,b]$ and never both zero at the same point (to avoid cusps or sharp corners that break the curve).

Worked Example

Find the arc length of the parametric curve $x(t)=3\cos t$, $y(t)=3\sin t$ for $0\le t\le \frac{\pi }{2}$.

1. Compute derivatives: $\frac{dx}{dt}=-3\sin t$, $\frac{dy}{dt}=3\cos t$.
2. Square and add the derivatives: ${\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=9{\sin }^{2}t+9{\cos }^{2}t=9({\sin }^{2}t+{\cos }^{2}t)=9$.
3. Substitute into the arc length formula: $L={\int }_0^{\frac{\pi }{2}}\sqrt{9}dt={\int }_0^{\frac{\pi }{2}}3dt$.
4. Evaluate the integral: $3t{|}_0^{\frac{\pi }{2}}=\frac{3\pi }{2}-0=\frac{3\pi }{2}$. This result matches the expected length of a quarter-circle of radius 3, confirming the calculation is correct.

Exam tip: Always check for points where both derivatives are zero before integrating. If such a point exists inside your interval, split the integral at that point to avoid undefined values.

3. Arc Length and Parametric Curve Orientation

Orientation is the direction a parametric curve is traced as $t$ increases, and it is important to understand how orientation affects arc length calculations. Arc length is a measure of total distance, so it is always non-negative, regardless of the direction the curve is traced. In the derivation of the formula, $\Delta t$ is always positive, so the lower limit of integration must always be the smaller $t$-value, and the upper limit the larger $t$-value. If you reverse the limits, you will get the negative of the correct arc length, because ${\int }_b^{a}f(t)dt=-{\int }_a^{b}f(t)dt$, and the integrand $\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}$ is always non-negative. This is especially relevant for particle motion: even if a particle backtracks along its path, the formula automatically adds the length of every segment, regardless of direction, so you do not need to adjust bounds when direction changes.

Worked Example

What is the arc length of the line segment from $(0,0)$ to $(4,3)$ traced parametrically as $x(t)=4-4t$, $y(t)=3-3t$ for $0\le t\le 1$?

1. Compute derivatives: $\frac{dx}{dt}=-4$, $\frac{dy}{dt}=-3$.
2. Square and add derivatives: $(-4{)}^2+(-3{)}^2=16+9=25$.
3. Set up the integral with lower limit $t=0$ and upper limit $t=1$ (the given interval, with smaller $t$ first): $L={\int }_0^1\sqrt{25}dt={\int }_0^{1}5dt$.
4. Evaluate: $5t{|}_0^1=5$, which matches the straight-line distance between $(0,0)$ and $(4,3)$ ($\sqrt{4^2+3^2}=5$), even though the curve is traced from $(4,3)$ to $(0,0)$ as $t$ increases. The length is still positive and correct.

Exam tip: For FRQ questions asking for total distance traveled by a parametric particle, you do not need to find when the particle changes direction. This is only required for displacement; the arc length formula automatically accounts for backtracking.

4. Numerical Integration for Non-Elementary Arc Length Integrals

Most parametric arc length integrals do not simplify to functions with elementary antiderivatives, so AP exams regularly test your ability to set up the integral correctly and evaluate it numerically using a graphing calculator. For these problems, the majority of points are awarded for the correct set up, even if your final decimal value is slightly off, so prioritizing writing the correct integral is key. The AP exam standard requires rounding to 3 decimal places for numerical answers unless explicitly stated otherwise.

Worked Example

Set up and evaluate the arc length of the parametric curve $x(t)=\sin (t^2)$, $y(t)=\cos (t^3)$ for $0\le t\le 1$, rounded to 3 decimal places.

1. Compute derivatives using the chain rule: $\frac{dx}{dt}=2t\cos (t^2)$, $\frac{dy}{dt}=-3t^2\sin (t^3)$.
2. Square and add derivatives: ${\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2=4t^2{\cos }^2(t^2)+9t^4{\sin }^2(t^3)$. This expression cannot be simplified to an elementary integrable function.
3. Set up the arc length integral: $L={\int }_0^1\sqrt{4t^2{\cos }^2(t^2)+9t^4{\sin }^2(t^3)}dt$.
4. Use a graphing calculator's numerical integration function to evaluate, giving $L\approx 1.838$, rounded to three decimal places.

Exam tip: Always write the full integral on your FRQ response before evaluating with a calculator. AP graders award at least half the points for the correct set up, even if your final rounded value is incorrect.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Using the Cartesian arc length formula $\int \sqrt{1+(dy/dx{)}^2}dx$ for a non-function parametric curve (e.g., a full circle). Why: Students confuse parametric and Cartesian arc length, and try to convert to Cartesian form incorrectly, leading to wrong integrands for curves that cross themselves. Correct move: Always use the parametric arc length formula for parametric curves, regardless of whether you can convert to Cartesian form.
• Wrong move: Reversing lower and upper bounds of integration, leaving a negative arc length as the final answer. Why: Students mix up orientation and sign rules, forgetting arc length is always positive. Correct move: Always set the lower limit equal to the smaller $t$-value, upper limit equal to the larger $t$-value, so the result is automatically non-negative.
• Wrong move: Forgetting to square both derivatives inside the square root, or writing $(x^{\prime }+y^{\prime }{)}^2$ instead of $(x^{\prime }{)}^2+(y^{\prime }{)}^2$. Why: Students rush through algebra steps for problems with multi-term derivatives. Correct move: Write each squared derivative separately, expand, and simplify before combining into the integrand.
• Wrong move: When finding total distance traveled, adjusting bounds to subtract backtracked path length. Why: Students confuse total distance (arc length) with net displacement. Correct move: For total distance, always use the full interval from start time to end time in the integral.
• Wrong move: On numerical integration problems, only writing the final decimal answer without writing the integral set up. Why: Students forget AP awards points for set up even if the calculator result is wrong. Correct move: Always write the complete integral with correct integrand and bounds before using your calculator.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following is the correct integral for the arc length of the parametric curve $x(t)=\ln t$, $y(t)=t^2$ for $1\le t\le e$?

A) ${\int }_1^e\sqrt{\frac{1}{t}+4t^2}dt$ B) ${\int }_1^e\sqrt{{\ln }^{2}t+t^4}dt$ C) ${\int }_1^e\sqrt{\frac{1}{t^2}+4t^2}dt$ D) ${\int }_1^e\sqrt{\frac{1}{t^2}+2t}dt$

Worked Solution: First compute derivatives: $\frac{dx}{dt}=\frac{1}{t}$ and $\frac{dy}{dt}=2t$. Next, square each derivative: ${\left(\frac{dx}{dt}\right)}^2=\frac{1}{t^2}$ and ${\left(\frac{dy}{dt}\right)}^2=4t^2$. Substitute into the parametric arc length formula to get the correct integral, which matches option C. Option A is a common error where $dx/dt$ is not squared, option B incorrectly squares $x(t)$ and $y(t)$ instead of their derivatives, and option D incorrectly squares the derivative of $y(t)$ as $2t$ instead of $(2t{)}^2=4t^2$. Correct answer: C.

Question 2 (Free Response)

A particle moves along a parametric path for $0\le t\le 4$, with position given by $x(t)=2t+1$, $y(t)=t^2-t$. (a) Write an integral expression for the total distance traveled by the particle over the interval $0\le t\le 4$. (b) Simplify the integrand and find the exact value of the total distance traveled. (c) Find the straight-line displacement distance between the particle's starting and ending positions, and compare it to the total distance traveled.

Worked Solution: (a) Compute derivatives: $\frac{dx}{dt}=2$, $\frac{dy}{dt}=2t-1$. The total distance (arc length) integral is: $$L={\int }_0^4\sqrt{2^2+(2t-1{)}^2}dt={\int }_0^4\sqrt{4+(2t-1{)}^2}dt$$ (b) Use substitution $u=2t-1$, $du=2dt$, so $dt=\frac{du}{2}$, with bounds $u=-1$ (t=0) to $u=7$ (t=4): $$L=\frac{1}{2}{\int }_{-1}^7\sqrt{u^2+4}du=\frac{1}{2}{\left[\frac{u}{2}\sqrt{u^2+4}+2\ln \left(u+\sqrt{u^2+4}\right)\right]}_{-1}^7$$ Evaluating gives the exact value: $$L=\frac{7\sqrt{53}+\sqrt{5}}{4}+\ln \left(\frac{7+\sqrt{53}}{\sqrt{5}-1}\right)\approx 16.135$$ (c) Starting position: $(x(0),y(0))=(1,0)$, ending position: $(x(4),y(4))=(9,12)$. Displacement distance is $\sqrt{(9-1{)}^2+(12-0{)}^2}=\sqrt{208}=4\sqrt{13}\approx 14.422$. The total distance traveled along the curved path (~16.135) is longer than the straight-line displacement distance between endpoints, as expected.

Question 3 (Application / Real-World Style)

A projectile is launched from ground level, and its position at time $t$ (seconds, $0\le t\le 11$) is given by the parametric equations $x(t)=40t$ meters, $y(t)=50t-4.9t^2$ meters. Find the total length of the projectile's trajectory from launch until it hits the ground, rounded to the nearest tenth of a meter. What does this value represent in context?

Worked Solution: First, find impact time by setting $y(t)=0$: $t(50-4.9t)=0$, so $t=0$ (launch) and $t=\frac{50}{4.9}\approx 10.204$ seconds (impact). Compute derivatives: $\frac{dx}{dt}=40$, $\frac{dy}{dt}=50-9.8t$. Set up the arc length integral: $$L={\int }_0^{10.204}\sqrt{40^2+(50-9.8t{)}^2}dt$$ Evaluate numerically with a graphing calculator to get $L\approx 497.7$ meters. In context, this is the total distance the projectile travels along its curved flight path from launch to impact, which is ~90 meters longer than the straight-line horizontal range between launch and impact points.

7. Quick Reference Cheatsheet

8. What's Next

Mastering arc length of parametric curves is a critical prerequisite for upcoming topics in this unit: arc length of polar curves and surface area of revolution for parametric and polar curves. Without correctly applying the parametric arc length formula, you will not be able to derive or evaluate these more advanced integrals, which are also tested on the AP Calculus BC exam. This topic builds on your knowledge of definite integrals and parametric differentiation, and connects directly to vector-valued functions, where the integrand of the arc length formula is exactly the speed of a moving particle (the magnitude of the velocity vector). Understanding parametric arc length also lays the foundation for line integrals in multivariable calculus, which you will encounter in college-level calculus after AP Calculus BC.

• Arc length of a polar curve
• Surface area of revolution for parametric curves
• Speed and motion of parametric particles
• Derivatives of parametric functions