AP Calculus BC — Unit 9.3
Finding Arc Lengths of Curves Given by Parametric Equations
Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)
Table of Contents
Unit 9.3 Key Concepts Packet
1) Parametric Arc Length Formula
Main formula (parametric)
If \(x=x(t)\) and \(y=y(t)\) for \(a\le t\le b\), then
\( \displaystyle L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt \)
- The square root is the magnitude of the velocity vector, so it is always nonnegative.
- Because of the squares, you do not manually insert absolute values inside the square root.
2) Arc Length = Integral of Speed (Vector View)
Vector form (cleanest AP interpretation)
If \( \mathbf{r}(t)=\langle x(t),y(t)\rangle \), then
\( \displaystyle L=\int_a^b \|\mathbf{r}'(t)\|\,dt \),
where \( \|\mathbf{r}'(t)\|=\sqrt{(x'(t))^2+(y'(t))^2} \).
- Think “distance traveled” = area under the speed curve.
- If the motion stops briefly (speed = 0), that portion adds 0 distance (but the integral still works).
3) Choosing Bounds + Retracing + Symmetry
Critical warning
If the curve is traced more than once on \([a,b]\), the integral counts the length each time it is traced.
- Pick an interval that traces the curve exactly once whenever the prompt implies “the curve” (not “the path traveled”).
- Use symmetry when it truly reduces work (e.g., quarter circle × 4) — but only if the tracing is clean and non-overlapping.
4) When Arc Length Simplifies Nicely
- Lines: If \(x'(t)\) and \(y'(t)\) are constants, the speed is constant and the integral is easy.
- Circles: For \(x=R\cos t\), \(y=R\sin t\), speed is \(R\) (constant), so length is \(R(b-a)\).
- Factoring tricks: Often \( \sqrt{(2t)^2+(2t)^2} = 2|t|\sqrt{2} \); on intervals where \(t\ge0\), the absolute value disappears automatically.
5) Calculator Expectations + Common AP Traps
- Do not confuse with Cartesian arc length: \( \int \sqrt{1+(dy/dx)^2}\,dx \). For parametrics, use the \(dt\)-formula.
- If you are asked for a numerical approximation, use a calculator to evaluate the integral.
- Retracing is the #1 reason students get “double” (or “half”) the correct length.
Worked Examples + Notes
Example 1 — Line segment (constant speed)
For \(x=3t\), \(y=4t\) on \([0,2]\), find the arc length.
Step 1: \(x'(t)=3\), \(y'(t)=4\).
Step 2: Speed \(=\sqrt{3^2+4^2}=5\) (constant).
Step 3: \(L=\int_0^2 5\,dt=10\).
Example 2 — Quarter circle (classic AP simplification)
For \(x=\cos t\), \(y=\sin t\) on \([0,\pi/2]\), find the arc length.
Step 1: \(x'(t)=-\sin t\), \(y'(t)=\cos t\).
Step 2: Speed \(=\sqrt{\sin^2 t+\cos^2 t}=1\).
Step 3: \(L=\int_0^{\pi/2} 1\,dt=\pi/2\).
Example 3 — A “factorable” integrand
For \(x=t^2\), \(y=t^2\) on \([0,1]\), find the arc length.
Step 1: \(x'(t)=2t\), \(y'(t)=2t\).
Step 2: Speed \(=\sqrt{(2t)^2+(2t)^2}=2t\sqrt{2}\) (since \(t\ge0\) on this interval).
Step 3: \(L=\int_0^1 2t\sqrt{2}\,dt=\sqrt{2}\).
Example 4 — Exact value with substitution
For \(x=t^2\), \(y=t^3\) on \([0,1]\), find the arc length exactly.
Step 1: \(x'(t)=2t\), \(y'(t)=3t^2\).
Step 2: Speed \(=\sqrt{4t^2+9t^4}=t\sqrt{4+9t^2}\) (since \(t\ge0\)).
Step 3: \(L=\int_0^1 t\sqrt{4+9t^2}\,dt\). Let \(u=4+9t^2\), so \(du=18t\,dt\).
Step 4: \(L=\frac{1}{18}\int_{4}^{13} u^{1/2}\,du=\frac{1}{18}\cdot\frac{2}{3}\left[u^{3/2}\right]_{4}^{13}=\frac{1}{27}\left(13\sqrt{13}-8\right)\).
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Unit 9.3 Multiple-Choice Practice (36 Questions)
Answer Key