Learning Outcomes
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Compute arclengths related to two-dimensional parametric/vector equations.
Section 7.3 Parametric/Vector Arclength (CO3)
Subsection 7.3.1 Activities
Example 7.3.1.
In Figure 102, the blue curve is the graph of the parametric equations \(x=t^2\) and \(y=t^3\) for \(1\leq t\leq 2\text{.}\) This curve connects the point \((1,1)\) to the point \((4,8)\text{.}\) The red dashed line is the straight line segment connecting these points.
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Activity 7.3.2.
Let’s first investigate the length of the dashed red line segment in Figure 102.
(a)
Draw a right triangle with the red dashed line segment as its hypotenuse, one leg parallel to the \(x\)-axis, and the other parallel to the \(y\)-axis.
How long are these legs?
(b)
The Pythagorean theorem states that for a right triangle with leg lengths \(a,b\) and hypotenuse length \(c\text{,}\) we have...
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\(a=b=c\text{.}\)
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\(a+b=c\text{.}\)
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\(a^2=b^2=c^2\text{.}\)
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\(a^2+b^2=c^2\text{.}\)
(c)
Using the leg lengths and Pythagorean theorem, how long must the red dashed hypotenuse be?
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\(\sqrt{20}\approx 4.47\text{.}\)
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\(\sqrt{58}\approx 7.62\text{.}\)
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\(\sqrt{67}\approx 8.19\text{.}\)
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\(\sqrt{100}=10\text{.}\)
(d)
Compared with the blue parametric curve connecting the same two points, is the red dashed line segment length an overestimate or underestimate?
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Overestimate: the blue curve is shorter than the red line.
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Underestimate: the blue curve is longer than the red line.
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Exact: the blue curve is exactly as long as the red line.
Fact 7.3.3.
Recall that the linear distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) may be computed by the distance formula
\begin{equation*}
\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\text{.}
\end{equation*}
Note that \(\Delta x=|x_2-x_1|\) and \(\Delta y=|y_2-y_1|\) measure leg lengths of a right triangle whose hypotenuse is the distance we want to measure, so we may rewrite this formula as
\begin{equation*}
\sqrt{(\Delta x)^2+(\Delta y)^2}\text{.}
\end{equation*}
This formula will need to be modified to measure a curved path between two points.
Observation 7.3.4.
By approximating the curve by several (say \(N\)) segments connecting points along the curve, we obtain a better approximation than a single line segment. For example, the illustration shown in Figure 103 gives three segments whose distances sum to about \(7.6315\text{,}\) while the actual length of the curve turns out to be about \(7.6337\text{.}\)
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Activity 7.3.5.
How should we modify the distance formula \(\sqrt{(\Delta x)^2+(\Delta y)^2}\) to measure arclength as illustrated in Figure 103?
(a)
Let \(\Delta L_1,\Delta L_2,\Delta L_3\) describe the lengths of each of the three segments. Which expression describes the total length of these segments?
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\(\displaystyle \Delta L_1\times \Delta L_2\times \Delta L_3\)
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\(\displaystyle \Delta L_1+ 2\Delta L_2+ 3\Delta L_3\)
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\(\displaystyle \sum_{i=1}^{3} \Delta L_i\)
(b)
We can let each \(\Delta L_i=\sqrt{(\Delta x_i)^2+(\Delta y_i)^2}\text{.}\) But we will find it useful to involve the parameter \(t\) as well, or more accurately, the change \(\Delta t_i\) of \(t\) between each point of the subdivision.
Which of these is algebraically the same as the above formula for \(\Delta L_i\text{?}\)
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\(\displaystyle \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\)
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\(\displaystyle \sqrt{\left[\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2\right]\Delta t_i}\)
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\(\displaystyle \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\Delta t_i\)
(c)
Finally, we’ll want to increase \(N\) from \(3\) so that it limits to \(\infty\text{.}\) What can we conclude when that happens?
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Each segment is infinitely small.
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\(\displaystyle \Delta x_i\to 0\)
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\(\displaystyle \frac{\Delta x_i}{\Delta t_i}\to\frac{dx}{dt}\)
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All of the above.
Observation 7.3.6.
Put together, and limiting the subdivisions of the curve \(N\to \infty\text{,}\) we obtain the Riemann sum
\begin{equation*}
\lim_{N\to\infty}\sum_{i=1}^N \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\Delta t_i\text{.}
\end{equation*}
Thus arclength along a parametric curve from \(a\leq t\leq b\) may be calculated by using the corresponding definite integral
\begin{equation*}
\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\text{.}
\end{equation*}
Activity 7.3.7.
Let’s gain confidence in the arclength formula
\begin{equation*}
\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt
\end{equation*}
by checking to make sure it matches the distance formula for line segments.
The parametric equations \(x=3t-1\) and \(y=2-4t\) for \(1\leq t\leq 3\) represent the segment of the line \(y=-\frac{4}{3}x-\frac{2}{3}\) connecting \((2,-2)\) to \((8,-10)\text{.}\)
(a)
Find \(dx/dt\) and \(dy/dt\text{,}\) and substitute them into the formula above along with \(a=1\) and \(b=3\text{.}\)
(b)
Show that the value of this formula is \(10\text{.}\)
(c)
Show that the length of the line segment connecting \((2,-2)\) to \((8,-10)\) is \(10\) by applying the distance formula directly instead.
Activity 7.3.8.
For each of these parametric equations, use
\begin{equation*}
\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt
\end{equation*}
to write a definite integral that computes the given length. (Do not evaluate the integral.)
(a)
(b)
(c)
Activity 7.3.9.
Let’s see how to modify \(\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\) to produce the arclength of the graph of a function \(y=f(x)\text{.}\)
(a)
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\(\displaystyle dx\)
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\(\displaystyle dt\)
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\(\displaystyle 1\)
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\(\displaystyle 0\)
(b)
(c)
Write a modified, simplified formula for \(\int_{t=a}^{t=b} \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt\) with \(t\) replaced with \(x\text{.}\)
Subsection 7.3.2 Videos
Subsection 7.3.3 Exercises
Exercises available at
https://tbil.org/preview/calculus/exercises/#/bank/CO3/
