As of Monday morning I am done with school for the semester. This has been a taxing few months, and I am ready for a break. Hopefully the time off will give me an opportunity to blog more often, as it has clearly been lacking. To celebrate the end of the semester I decided to compile a blog post that would include some aspect from each of the three classes I took: Calculus, Physics, and Mineralogy.

Here is a problem I created to acknowledge the three in harmony. Test me to see if I solved it correctly, and if not, please feel free to solve it yourselves and share your answer. (It took me awhile to come up with a problem that included all three, and while it makes sense in my head, it may not translate well to people outside of it. Thanks for your understanding)

**If a calcite crystal is thrown vertically down from a cliff 50 meters high with an initial velocity of 5 m/s, and a horizontally projected lamp is placed at a height of 20 meters and 8 meters from the base of the cliff. What speed is the calcite travelling when it passes its critical angle (the angle at which light cannot refract in a medium)?**

The first step was to determine at what height the critical angle of calcite would be reached, and I did so using Snell’s Law (n_{2}/n_{1} = sinϴ_{1}/sinϴ_{2}) and the understanding that the critical angle for all minerals is when sinϴ_{2} = 1. Also needed were the refractive indices of calcite which are found easily enough in any of my textbooks or at handbookofmineralogy.org.

n_{1} = 1.658 and n_{2} = 1.486.

From the math above the critical angle of calcite was found to be 63.7° by solving for the sin^{-1} of (n_{1}/n_{2}). Now that we have the angle we can determine the height of the calcite by using a little trigonometry. But, before we do that it is best to point out that the calcite will pass the angle twice in its path, and therefore we will be solving for two different velocities.

With the help of a handy triangle and SOHCAHTOA I found the height by solving 8 * tan (63.7) which happens to be ~16 meters. Therefore the calcite passes the critical angle at 20 m (height of the lamp) ± 16 meters, so, 4 and 36 m.

By jumbling a bunch of physics and calculus formulas together I surmised the path of the calcite to follow the function:

f(t) = 4.9t^{2} + 5t – 50

By setting this function equal to the two determined heights and solving for t with the quadratic formula I found the time it took for the calcite to reach the critical height after being thrown. (Also by understanding time cannot be negative).

t = 2.8 and 3.7 seconds

Here is where the calculus really kicked in as finding the derivative of the position function gives the velocity function.

f(t) = 4.9t^{2} + 5t -50

f ‘(t) = 9.8t + 5

Dropping the two values for time into the velocity function allowed me to solve for how fast the calcite was traveling at those particular moments.

f ‘(2.8) = 9.8(2.8) +5

f ‘(2.8) = 32.4 m/s

f ‘(3.7) = 9.8(3.7) + 5

f ‘(3.7) = 41.3 m/s

So, if my calculations are correct that calcite was really flying at 32.4 m/s and 41.3 m/s!

Let me know what you think. Does my math hold up? Is the problem viable? Chime in!

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