For parametric curves, the distance traveled along the curve (arc length) and the speed of a moving particle are closely related. The speed is the magnitude of the velocity vector, and the total distance traveled is the integral of speed over time.
🎯 In this section you will learn
- The arc length formula for parametric curves: $L = \int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2} dt$
- How speed is related to the derivatives: $v(t) = \sqrt{(dx/dt)^2 + (dy/dt)^2}$
- How to find the distance traveled by a particle moving along a parametric path
📌 Arc Length Formula
$$ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt $$
The length of the curve from $t=a$ to $t=b$
💡 Derivation
On a small interval $dt$, the particle moves approximately $\sqrt{(dx)^2 + (dy)^2} = \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt$. Summing (integrating) gives the total length.
Find the arc length of $x = 2t$, $y = 3t$ from $t=0$ to $t=2$.
①
Find derivatives
$\frac{dx}{dt} = 2$, $\frac{dy}{dt} = 3$
②
Set up the integral
$L = \int_0^2 \sqrt{2^2 + 3^2} dt = \int_0^2 \sqrt{13} dt$
③
Evaluate
$L = \sqrt{13} \cdot 2 = 2\sqrt{13}$
Find the arc length of $x = 3\cos t$, $y = 3\sin t$ from $t=0$ to $t=2\pi$.
①
Find derivatives
$\frac{dx}{dt} = -3\sin t$, $\frac{dy}{dt} = 3\cos t$
②
Set up the integral
$L = \int_0^{2\pi} \sqrt{(-3\sin t)^2 + (3\cos t)^2} dt = \int_0^{2\pi} \sqrt{9(\sin^2 t + \cos^2 t)} dt = \int_0^{2\pi} 3 dt$
③
Evaluate
$L = 3 \cdot 2\pi = 6\pi$ (circumference of a circle of radius $3$)
Find the arc length of $x = t$, $y = t^2$ from $t=0$ to $t=2$.
①
Find derivatives
$\frac{dx}{dt} = 1$, $\frac{dy}{dt} = 2t$
②
Set up the integral
$L = \int_0^2 \sqrt{1 + (2t)^2} dt = \int_0^2 \sqrt{1 + 4t^2} dt$
③
Evaluate (using formula or trig substitution)
$\int \sqrt{1+4t^2} dt = \frac{t}{2}\sqrt{1+4t^2} + \frac{1}{4}\ln\left|2t + \sqrt{1+4t^2}\right| + C$
$L = \left[\frac{t}{2}\sqrt{1+4t^2} + \frac{1}{4}\ln\left(2t + \sqrt{1+4t^2}\right)\right]_0^2$
$= \frac{2}{2}\sqrt{17} + \frac{1}{4}\ln(4+\sqrt{17}) - 0 = \sqrt{17} + \frac{1}{4}\ln(4+\sqrt{17}) \approx 4.646$
📌 Speed and Distance Traveled
$$ \text{Speed} = v(t) = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} $$
$$ \text{Distance traveled} = \int_a^b v(t) \, dt $$
Same as arc length when the curve is traversed once
A particle moves with $x = 2\cos t$, $y = 3\sin t$. Find its speed at $t = \pi/4$.
①
Find derivatives
$\frac{dx}{dt} = -2\sin t$, $\frac{dy}{dt} = 3\cos t$
②
Speed formula
$v(t) = \sqrt{(-2\sin t)^2 + (3\cos t)^2} = \sqrt{4\sin^2 t + 9\cos^2 t}$
③
Evaluate at $t = \pi/4$
$\sin(\pi/4) = \cos(\pi/4) = \frac{\sqrt{2}}{2}$
$v(\pi/4) = \sqrt{4 \cdot \frac{1}{2} + 9 \cdot \frac{1}{2}} = \sqrt{\frac{4}{2} + \frac{9}{2}} = \sqrt{\frac{13}{2}} = \frac{\sqrt{26}}{2} \approx 2.55$
⚠️ Common Mistakes
- Forgetting the square root: Arc length uses $\sqrt{(dx/dt)^2 + (dy/dt)^2}$, not $(dx/dt)^2 + (dy/dt)^2$.
- Confusing arc length with speed: Speed is the integrand; arc length is the integral.
- Incorrect limits: The limits of integration should be the $t$-values, not $x$ or $y$ values.
🔍 Key Takeaways
- Arc length: $L = \int_a^b \sqrt{(dx/dt)^2 + (dy/dt)^2} dt$.
- Speed: $v(t) = \sqrt{(dx/dt)^2 + (dy/dt)^2}$.
- Distance traveled = $\int_a^b v(t) dt$ = arc length (if the curve is traversed once).
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