Section 7.3 Parametric/vector arclength (CO3)
Learning Outcomes
Compute arclengths related to two-dimensional parametric/vector equations.
Example 7.3.1.
In Figure 144, 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.
Activity 7.3.2.
Let's first investigate the length of the dashed red line segment in Figure 144.
(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?
\(3\) and \(7\text{.}\)
\(4\) and \(8\text{.}\)
\(3\) and \(8\text{.}\)
\(4\) and \(7\text{.}\)
(b)
The Pythagorean theorem states that for a right triangle with leg lengths \(a,b\) and hypotenuse length \(c\text{,}\) we have...
\(a=b=c\text{.}\)
\(a+b=c\text{.}\)
\(a^2=b^2=c^2\text{.}\)
\(a^2+b^2=c^2\text{.}\)
(c)
Using the leg lengths and Pythagorean theorem, how long must the red dashed hypotenuse be?
\(\sqrt{20}\approx 4.47\text{.}\)
\(\sqrt{56}\approx 7.48\text{.}\)
\(\sqrt{67}\approx 8.19\text{.}\)
\(\sqrt{100}=10\text{.}\)
(d)
Compared with the blue parametric curve connecting the same two points, is the red dashed line segement length an overestimate or underestimate?
Overestimate: the blue curve is shorter than the red line.
Underestimate: the blue curve is longer than the red line.
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
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
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 \(N\)) segements connecting points along the curve, we obtain a better approximation than a single line segment, as shown in Figure 145.
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 145?
(a)
Let \(\Delta L_1,\Delta L_2,\Delta L_3\) describe the lengths of each of the three segements. Which expression describes the total length of these segments?
\(\displaystyle \Delta L_1\times \Delta L_2\times \Delta L_3\)
\(\displaystyle \Delta L_1+ 2\Delta L_2+ 3\Delta L_3\)
\(\displaystyle \sum_{i=1}^{3} \Delta L_i\)
(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 \(\Delta L_i\text{?}\)
\(\displaystyle \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\)
\(\displaystyle \sqrt{\left[\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2\right]\Delta t_i}\)
\(\displaystyle \sqrt{\left(\frac{\Delta x_i}{\Delta t_i}\right)^2+\left(\frac{\Delta y_i}{\Delta t_i}\right)^2}\Delta t_i\)
(c)
Finally, we'll want to increase \(N\) from \(3\) so that it limits to \(\infty\text{.}\) What can we conclude when that happens?
Each segment is infintely small.
\(\displaystyle \Delta x_i\to 0\)
\(\displaystyle \Delta t_i\to 0\)
\(\displaystyle \frac{\Delta x_i}{\Delta t_i}\to\frac{dx}{dt}\)
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
Thus arclength along a parametric curve from \(a\leq t\leq b\) may be calculated by using the corresponding definite integral
Activity 7.3.7.
Let's gain confidence in the arclength formula
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
to write a definite integral that computes the given length. (Do not evaluate the integral.)
(a)
The portion of \(x=\sin 3t, y=\cos 3t\) where \(0\leq t\leq \pi/6\text{.}\)
(b)
The portion of \(x=e^t, y=\ln t\) where \(1\leq t\leq e\text{.}\)
(c)
The portion of \(x=t+1, y=t^2\) between the points \((3,4)\) and \((5,16)\text{.}\)
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)
Let \(x=t\text{.}\) How can \(\frac{dx}{dt}\) be simplified?
\(\displaystyle dx\)
\(\displaystyle dt\)
\(\displaystyle 1\)
\(\displaystyle 0\)
(b)
Given \(x=t\text{,}\) how should \(\frac{dy}{dt}\) and \(dt\) be rewritten?
\(\frac{dy}{dt}=\frac{dy}{dx}\) and \(dt=dx\text{.}\)
\(\frac{dy}{dt}=\frac{dx}{dt}\) and \(dt=dx\text{.}\)
\(\frac{dy}{dt}=\frac{dy}{dx}\) and \(dt=1\text{.}\)
\(\frac{dy}{dt}=\frac{dy}{dt}\) and \(dt=1\text{.}\)
(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 \(x\text{.}\)