6.3: Gateway Arc Length
This activity is intended to supplement Calculus, Chapter 5, Lesson 3.
ID: 12439
Time Required: 15 minutes
Activity Overview
Students will investigate the arc length of the Gateway Arch. They will use the Pythagorean Theorem to approximate and use Calculus to find the exact solution. They will also use CAS capabilities, including arcLen( ), to solve a variety of arc length questions.
Topic: Differential Equations
 Arc length approximation, calculus formula, and using CAS
 Find the arc length of parametric equation
Teacher Preparation and Notes

Arc length is a Calculus BC topic. Calculus AB teachers may enjoy using this activity after the \begin{align*}AP^*\end{align*}
AP∗ exam or using with students in your AB class who want to prepare for the BC exam. After completing the activity, students should be more successful with AP questions like multiple choice 03BC15, 98BC21, 88BC#33, and free response 04formB BC1c, 02formB BC1d&3c, 01BC1c, 97BC1e&3b. For four of these six free response questions the graph is given in parametric form. 
The syntax for arcLen is arcLen\begin{align*}(f(x),x,a,b)\end{align*}
(f(x),x,a,b) where \begin{align*}f(x)\end{align*}f(x) is the function, \begin{align*}x\end{align*}x is the variable and the arc length is to be found from \begin{align*}x = a\end{align*}x=a to \begin{align*}x = b\end{align*}x=b . This activity will help students approximate arc length and use calculus to find the exact arc length.
Associated Materials
 Student Worksheet: Gateway Arch Length http://www.ck12.org/flexr/chapter/9730, scroll down to the third activity.
\begin{align*}^*\end{align*}
Part 1 – Arc Length Introduced
The first question investigates the Gateway Arch and the distance that one would travel if they rode the elevator tram to reach the top. The Pythagorean Theorem is used to approximate the distance. The graph on the student worksheet helps students visually understand why those numbers where used in the solution to Exercise 1. The formula for arc length is derived from the Pythagorean Theorem. To help students understand why the integral formula approximates the arc length, compare this method to finding the area under a curve using a Riemann Sum. When the infinitesimal values of \begin{align*}dL\end{align*}
Discussion Questions
 What are the conditions for which the Pythagorean Theorem applies? If students say, “It works for triangles,” press them further. What are characteristics of a triangle? Perhaps they will see then, “Oh yeah, the Pythagorean Theorem only works for right triangles. If you would like, you could go a bit deeper and ask, “What relationship (principle or law) applies for triangles that are not right? Explain it.” Law of Cosines \begin{align*}c^2 = a^2 + b^2  2ab \cos \theta\end{align*}
c2=a2+b2−2abcosθ where \begin{align*}\theta\end{align*}θ is the angle between \begin{align*}a\end{align*}a and \begin{align*}b\end{align*}b . The Pythagorean Theorem is a special case of this where \begin{align*}\theta = 90^{\circ}\end{align*}θ=90∘ . (They may also say Law of Sines.)
 Ask students, “Remind me, what is an integral? What does it mean?” You may need to remind them that the definition was based on the area of rectangles or Riemann sums. Ask again, “What does this mean? What are you doing with Riemann Sums?” Adding infinite infinitesimals.
For Exercise 2, students are to use CAS to find the arc length of the Gateway Arch equation. Have students store the equation modeling the arch before entering the arc length formula. Students compare this solution with their length from Exercise 1.
Arc length for parametric equations is introduced and students are to solve this arc length by hand. For Exercise 4, students use CAS to find the arc length for the function \begin{align*}y = \sqrt{4  x^2}\end{align*}
The arcLen command is also introduced.
Student Solutions
1. The distance is at least \begin{align*}704 = \sqrt{315^2 + 630 ^2}\end{align*}
2. \begin{align*}\int \limits_{0}^{300} \sqrt{1 + \left(\frac{d}{dx}(f1(x))\right)^2} dx = 739.449\end{align*}
This is reasonable since it is a little larger than the straight line found in Exercise 1.
3. \begin{align*} & \int \limits_{a}^{b} \sqrt{ \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt} \right)^2} dt \\ & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{\left( \frac{d}{dt} \ 2 \cos(t) \right)^2 + \left( \frac{d}{dt} \ 2 \sin (t) \right)^2} \ dt \\ & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{(2 \sin (t))^2 + (2 \cos (t))^2} \ dt \\ & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{4(\sin^2(t) + \cos^2(t))} \ dt = 2t _{0}^{\frac{\pi}{2}} = \pi\end{align*}
4. \begin{align*}\int \limits_{0}^{2} \sqrt{1 + \left(\frac{d}{dx} \left(\sqrt{4  x^2} \right) \right)^2} dx \approx 3.14159\end{align*}
5. The solution will be around 10.
\begin{align*}\int \limits_{0}^{3} \sqrt{1 + \left(\frac{d}{dx} (x^2  9) \right)^2} dx \approx 9.747\end{align*}
6. The arc length will be more than 5, because \begin{align*}5 = \sqrt{3^2 + 4^2}\end{align*}
\begin{align*}\int \limits_{0}^{3} \sqrt{1 + \left(\frac{d}{dx} \left(x^2 + \frac{5}{3}x + 4 \right) \right)^2} dx \approx 6.492\end{align*}
This answer is larger than 5 which is expected.
Part 2 – Additional Practice
Students are expected to know the arc length formula and answer multiplechoice questions without a calculator. Question 2 is a parametric arc length question.
Student Solutions
1. D) \begin{align*}\int \limits_{a}^{b} \sqrt{\frac{x^2  5}{x^2  4}} \ dx\end{align*}
2. A) \begin{align*}\int \limits_{0}^{\pi} \sqrt{ \cos^2t + 1} \ dt\end{align*}
3. E)\begin{align*}\int \limits_{a}^{b} \sqrt{1 + \sec^4 x} \ dx\end{align*}
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