Topic 9.3 Notes – Finding Arc Lengths of Curves Given by Parametric Equations

1. Arc Length of a Parametric Curve

Suppose a curve is defined by

$$x = x(t), \quad y = y(t), \quad t \in [a,b].$$

The arc length of this curve from $t=a$ to $t=b$ is

$$S = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt$$

That square root expression is the key. It represents the speed of the particle moving along the curve.

Where this comes from

If you zoom in on a tiny piece of the curve:

• Horizontal change ≈ $dx$
• Vertical change ≈ $dy$

By the Pythagorean Theorem,

$$ds = \sqrt{(dx)^2 + (dy)^2}$$

Now divide by $dt$:

$$ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt$$

So arc length is just:

$$\text{Arc length} = \int \text{speed} \, dt$$

That connection to accumulation of change is straight Unit 6 thinking. We are adding up tiny distance pieces over an interval.

If you remember the Cartesian formula
$\int_a^b \sqrt{1 + (f'(x))^2}\, dx$,
this is the parametric version where both coordinates are changing.

2. How to Compute Arc Length

When this shows up on a quiz or FRQ, the process is very consistent.

Step-by-step

1. Find the derivatives

   $$\frac{dx}{dt}, \quad \frac{dy}{dt}$$

2. Build the speed

   $$\sqrt{(x')^2 + (y')^2}$$

3. Set up the definite integral

   $$\int_a^b \sqrt{(x')^2 + (y')^2}\, dt$$

4. Simplify before integrating
   Especially with trig expressions.

Quick Example

Let

$$x(t) = 3\cos t, \quad y(t) = 3\sin t, \quad t \in [0,\pi].$$

Derivatives:

$$x' = -3\sin t, \quad y' = 3\cos t$$

Inside the radical:

$$9\sin^2 t + 9\cos^2 t = 9$$

Speed = $3$

Arc length:

$$\int_0^\pi 3\, dt = 3\pi$$

This makes sense. That’s half a circle of radius 3, so length should be $3\pi$.

On non-calculator sections, they often design it so a trig identity collapses everything nicely like this.

3. Geometric and Physical Meaning

It helps to picture what’s happening.

Upper semicircle traced by $x=3\cos t,\; y=3\sin t$

As $t$ increases, the particle moves along the curve.
The formula measures how much ground it covers.

Two interpretations show up in problems:

• Geometric: “Find the length of the curve.”
• Physical: “Find the total distance traveled.”

Those are the same computation. Distance traveled is the integral of speed.

Special Situations

1. One coordinate constant

If $x(t)=5$, then $x'=0$.

The formula becomes:

$$S = \int_a^b |y'(t)|\, dt$$

You’re just moving vertically.

2. When the integral isn’t nice

Sometimes
$$\sqrt{(x')^2 + (y')^2}$$
doesn’t simplify.

• On calculator-active parts, you may evaluate numerically.
• On FRQs, they may leave the answer as a definite integral.

Do not force an antiderivative that doesn’t exist in elementary form.

Common Errors I See

• Forgetting to square both derivatives.
• Dropping the square root.
• Using $x$-bounds instead of $t$-bounds.
• Forgetting trig identities like $\sin^2 t + \cos^2 t = 1$.
• Mixing this up with $\int |v(t)| dt$ in 1D motion. Here speed already accounts for both components.