1) Early cultures explained rainbows through myths symbolizing both hope and fear. Today, rainbows can be explained using calculus and the properties of light. 2) Using Fermat's principle that light follows the fastest path, calculus can derive the law of reflection - the angle of incidence equals the angle of reflection. It can also derive Snell's law of refraction - the ratio of sines of the incidence and refraction angles is a constant. 3) For a rainbow to form, light enters a raindrop and is reflected or refracted. The angle of maximum rainbow deflection occurs when the derivative of the deflection angle equals 0, relating the incidence and refraction angles through

1. The Calculus of Rainbows By Angela Han, Rabeea Abbas, Laura Boardman, and Cathy Ma

3. Ancint Greece- the goddess Iris used the rainbow as a sign of both warning and hope

4. Africa- rainbow was a large snake coming out to graze after the storm, again a symbol of both hope and fear

5. Australian aboriginals- Rainbow serpent Is the Creator that can give fertility by creating rain but also let loose blindness and disease

6. Native Americans- a bridge anchored in this world and leading to the next

9. Suppose a source of light rays at point P and passes through point Q after reflection off the mirror. At what point R does the ray reflect off the mirror?

10. Using Fermat's principle we know the ray follows a path that minimizes the time necessary to travel from P to Q. Since the speed of light is constant, we can use the triangle in the figure to get the following expression for the path as a function of x:

11. To find the minimum path length we find the derivative L’(x):

12. Then we set this expression equal to 0:

13. x/(√(p² + x²)) = (d – x)/(√q² + (d – x)²)

14. Referring to the figure, we see that:

15. h

16. Therefore L’(x) = 0 when cosa = cosb, this is the minimum path length

17. So…. The Law of Reflection:

18. For reflection, the angle of incidence (a) equals the angle of reflection (b)P Q p q a b x (d-x) R d

19. Refraction With reflection, we assumed that the light rays maintained a constant speed, as the rays traveled through a fixed medium What happens if part of the light’s path is in water and part is in air? P = source of light rays in the air Q = point in water R = point at which a light ray crosses the air/water interface α = angle of incidence β = angle of refraction The Law of Refraction states that (sinα)/(sinβ) is a constant (k)

20. The Law of Refraction The total time the light takes to travel from P to Q is displayed as ca= speed in air cw = speed in water To find the minimum of T(x), we need to first derive the function Then we set T’(x) equal to 0, which gives: This is a constant ratio, known as the index of refraction. For water this value is approximately 1.33.

22. The Law of Refraction states that the light ray that strikes the drop will be bent toward the perpendicular line.

23. Generally, as light travels form one medium to another, art of the light is reflected, as the other part continues to another medium, where it’s refracted.

24. It is reasonable to think of a water drop as a spherical shape:

26. For example, if A is on the upper half of the raindrop, the ray will exit on the lower half

27. If the ray enters the drop along the circle’s diameter, the angle of incidence and refraction will both be zero; the total deflection will be 180 degrees

28. However, as the ray from Point A in figure 2 cross the drop, the deflection angle changes

30. Solving a problem involving Calculus This problem is trying to derive Snell’s Law of Refraction using Lagrange Multipliers, which are useful for calculating minima or maxima for functions with subject to constraints (a condition that a solution to an optimization problem must satisfy.) Lagrange multipliers are denoted by a new variable (λ), and the function is denoted by

31. Problem Continued