From a Curve Segment to an Integral
A parametric curve describes points by means of a parameter. As the parameter changes, the coordinate functions \(x(t)\) and \(y(t)\) determine the point \((x(t),y(t))\). To find the length of one specified segment, you need two kinds of information: the parameter values at the segment’s endpoints, and the rates at which both coordinates change over that interval.
The distinction matters. A point such as \((2,5)\) is a pair of coordinates; it is not automatically the parameter value \(t=2\). The parameter value must be found from the equations \(x(t)=2\) and \(y(t)=5\), together, when that point is an endpoint of the requested segment.
The Arc Length Setup
Suppose a curve is given on a parameter interval by \(x=x(t)\) and \(y=y(t)\), where both coordinate functions are continuously differentiable. On an interval \(a\leq t\leq b\), the arc length is
The expression under the square root combines the two coordinate rates to give the speed along the curve. The integral adds that speed over the chosen parameter interval. The bounds are parameter values, not \(x\)- or \(y\)-coordinates.
For length, use increasing bounds \(a\leq b\). If the segment is described from the parameter value \(b\) back to \(a\), the geometric length is still calculated over \([a,b]\). Integrating from \(b\) to \(a\) would give the negative of the length because the integral’s limits would be reversed.
Why the Coordinate Derivatives Give Length
Over a short parameter change \(\Delta t\), the changes in the coordinates are approximately \(x'(t)\Delta t\) and \(y'(t)\Delta t\). The Pythagorean theorem gives the corresponding short straight-line distance as approximately
For increasing parameter steps, \(|\Delta t|=\Delta t\). Adding these short distances across the interval and taking the limit gives the integral of the speed. This explains both the square root in the integrand and why the parameter interval controls which part of the curve is measured.
Worked Example: A Polynomial Curve Between Two Points
Worked Example: A Polynomial Curve Between Two Points
Let \(x=t^2+1\) and \(y=2t\). Set up the arc length integral for the segment from \((2,2)\) to \((10,6)\).
Step 1: Find the parameter at each endpoint. For \((2,2)\), the equation \(y=2t=2\) gives \(t=1\). Check the other coordinate: \(x=1^2+1=2\). For \((10,6)\), \(y=2t=6\) gives \(t=3\), and \(x=3^2+1=10\). Thus the requested segment is traced for \(1\leq t\leq 3\).
Step 2: Differentiate both coordinates.
Step 3: Substitute into the formula.
Check: At \(t=1\), the parametrization gives \((1^2+1,2\cdot1)=(2,2)\). At \(t=3\), it gives \((3^2+1,2\cdot3)=(10,6)\). Differentiating \(t^2+1\) and \(2t\) gives \(2t\) and \(2\), as used. The integral is set up for the requested segment; its bounds are parameter values \(1\) and \(3\), not the endpoint coordinates.
Choosing Bounds for a Trigonometric Parametrization
Endpoint coordinates do not always reveal parameter values by inspection. With trigonometric coordinate functions, it is especially important to use both coordinate equations and to check which interval traces the intended segment. Trigonometric functions repeat values, so one point on the curve can correspond to multiple parameter values.
Worked Example: A Trigonometric Curve on a Specified Interval
Let \(x=2\cos t\) and \(y=\sin(2t)\). Set up the arc length for the segment traced from \(t=0\) to \(t=\frac{\pi}{2}\), and identify its endpoints.
Step 1: Evaluate the coordinates at the interval endpoints. At \(t=0\),
so the starting point is \((2,0)\). At \(t=\frac{\pi}{2}\),
so the ending point is \((0,0)\). The specified parameter interval is \(\left[0,\frac{\pi}{2}\right]\).
Step 2: Differentiate both coordinates.
The second derivative uses the chain rule: the derivative of \(\sin(2t)\) is \(\cos(2t)\cdot 2\).
Step 3: Form the arc length integral.
Check: The endpoint substitutions give \((2,0)\) and \((0,0)\), as stated. Differentiating \(2\cos t\) gives \(-2\sin t\); differentiating \(\sin(2t)\) gives \(2\cos(2t)\). Both coordinate derivatives appear squared in the integrand.
When the Segment Is Described in Reverse
A problem may name the endpoint encountered first when \(t\) is decreasing. The endpoints still determine the same geometric segment, but the integral for length should have its smaller parameter value first. This is because speed is nonnegative, while reversing the integral’s limits changes its sign.
Worked Example: Reversing the Endpoint Order
Let \(x=1-t\) and \(y=t^2\). Set up the length of the segment joining the point reached at \(t=2\) to the point reached at \(t=0\).
Step 1: Identify both endpoint coordinates. At \(t=2\),
At \(t=0\),
The named traversal goes from \(t=2\) down to \(t=0\), but the increasing interval covering this segment is \(0\leq t\leq 2\).
Step 2: Differentiate each coordinate.
Step 3: Use increasing bounds.
Check: Substitution at \(t=2\) gives \((-1,4)\), and substitution at \(t=0\) gives \((1,0)\). The derivatives are \(-1\) and \(2t\). The integrand is nonnegative, and the bounds \(0\) to \(2\) give a nonnegative length. Writing \(\int_2^0\sqrt{1+4t^2}\,dt\) instead would give the negative of this length.
Endpoint Coordinates May Not Identify a Unique Parameter
A curve can pass through the same point at different parameter values. Therefore, finding one solution to the endpoint equations is not enough if the curve revisits that point. You must choose the parameter values that trace the requested segment. The interval between those values determines which portion of the parametrized motion is measured.
For instance, in the trigonometric example, the endpoint \((2,0)\) occurs at \(t=0\), but periodic trigonometric functions can produce that same coordinate pair again at other parameter values. If a problem specifies a particular arc or a parameter interval, use that information to select the intended occurrence. If it gives only endpoint coordinates and the parametrization revisits them, the segment may not be uniquely determined without additional information.
Also remember that the integral measures the distance traced during the interval. If the parametrization travels over part of the curve more than once during that interval, each pass contributes to the total. The integral does not automatically remove repeated tracing.
A Reliable Setup Process
Use the named endpoints, the stated parameter interval, and any indicated direction or branch to identify the intended portion of the parametrization.
Substitute candidate parameter values into both coordinate functions. Confirm that the resulting ordered pair is the endpoint, and use the values corresponding to the requested segment.
For a length integral, use \(a\leq b\). Reversing the direction of traversal does not make geometric length negative.
Compute \(x'(t)\) and \(y'(t)\) with respect to the parameter, including any chain-rule factors.
Write \(\int_a^b\sqrt{[x'(t)]^2+[y'(t)]^2}\,dt\). Verify the endpoint coordinates, derivative calculations, and bounds before proceeding.
Common Mistake: Using Coordinates as Bounds
Suppose \(x=t^2+1\), \(y=2t\), and the requested segment runs from \((2,2)\) to \((10,6)\). A common incorrect setup is to use the \(x\)-coordinates as parameter bounds and write an integral from \(2\) to \(10\). That treats coordinate values as though they were parameter values.
The correction is to solve for the parameter at each endpoint. The point \((2,2)\) corresponds to \(t=1\), and \((10,6)\) corresponds to \(t=3\). Thus the correct setup is
Another tempting error is to differentiate only \(x(t)\). A parametric curve changes in two coordinate directions, so its speed depends on both \(x'(t)\) and \(y'(t)\). Omitting either coordinate gives the wrong integrand unless that coordinate’s derivative is identically zero on the interval.
Why Careful Setup Matters
Arc length questions often require more than substituting into a formula. The essential decisions are which parameter values trace the requested piece and whether the coordinate derivatives have been computed correctly. A correct integrand with incorrect bounds measures the wrong segment; correct bounds with an omitted derivative also give an incorrect length.
On the AP exam, show the parameter-endpoint check when it is not immediate, write both derivatives explicitly, and leave the integral in setup form when the task asks only to set it up. This makes clear that the bounds belong to the parameter and that both coordinate changes contribute to distance.
Check Your Understanding
For each question, identify the parameter interval or arc length setup as requested. Keep parameter values distinct from coordinate values.
- For \(x=t^2\) and \(y=t+1\), which parameter values correspond to the endpoints \((1,2)\) and \((9,4)\)?
- For \(x=3\cos t\) and \(y=2\sin t\), set up the arc length integral over \(0\leq t\leq\frac{\pi}{2}\).
- For \(x=t\) and \(y=t^3\), set up the length of the segment traversed from \(t=2\) down to \(t=-1\), using increasing bounds.
- Why can endpoint coordinates alone fail to identify a unique segment of a parametrized curve?
- What sign error results from integrating a nonnegative speed function with the larger parameter value as the lower limit?
