Parabolas

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I recently came across a problem that required knowledge of parabola formulas. That surprised me. Although parabola problems show up relatively frequently, they usually require little more than logic. However, I thought that provided a nice opportunity to refresh on all things parabola.

A parabola is defined mathematically by this formula: y= ax^2 + bx + c. We see parabolas in nature most often when we look at projectiles, like a cannonball shot out of a cannon or a jump shot out of the hand of Steph Curry. Parabolas are generally u-shaped and are symmetrical about the vertex, which is either the highest or lowest point of the parabola, depending on the orientation.

Whether our parabola is cupped upward or downward is determined by the sign of the “a” term in the formula we saw above. When a is positive, the parabola will have a vertex at the bottom and open upward. When a is negative, the parabola will have a vertex at the top and open downward.

However, there is more than one way to define a parabola mathematically. We can also solve a parabola if we have the vertex and another point on the parabola. We do that by using the similar formula y = a(x-h)^2 + k. The coordinates of the vertex are (h,k).

So using that information, find the equation of a parabola with vertex (-2,1) containing the point (1, 19). The first thing we need to do is solve for a by inserting our points into the formula. We get:

19 = a(1-(-2)^2 + 1

19= a(3^2) + 1

19= 9a + 1

Now, we put our a into the formula with our vertex (h,k), but instead of using the x and y from a specific

point we’re going to solve for the generic x and y.

y= 2(x-(-2)^2 + 1

y= 2(x+2)^2 +1

Now expand:

y=2(x+2)(x+2) +1

y=2x^2 + 8x + 9

Now we’ve solved for the equation of this parabola and we could mathematically figure out all of the

points on this curve.

I hope that’s been a good refresher on parabolas!

 Change in Christian’s Pocket

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Christian has 13 coins in his pocket. The total value of the change is $1.37. The change may be composed of pennies, nickels, dimes and quarters. How many dimes could be in Christian’s pocket?

A) 0

B) 3

C) 5

D) 7

E) 9

There are several ways to approach this problem. Many of them are wrong. That’s because the core problem (13 coins, $1.37) can be solves several different ways. If you approached this and just decided to solve it you might find two or three solutions before you found one that was in the answer choices. That’s the first important step. When you’re given a problem without a clear, obvious and quick solution look to the answer choices.

Once, we’ve decided that we’re going to work through the answer choices, there is still some uncertainty. Look at answer choice B for instance. Even if we know that we have 3 dimes, that still doesn’t sort out how many pennies, nickels and quarters we have. That brings us to tip number two: when you’re given complex information, look for opportunities to simplify. The best example of that in this case is the pennies. Because our total needs to end in a 7, and all the other coins can only get us multiples of 5, we know that we must have either 2 or 7 pennies in order to make this work. That makes quick work of checking the possibilities for answer choice B. If we have 3 dimes and 2 pennies, can we make $1.05 out of 8 coins? Well, quarters can only get us to multiples of 25, so we must have at least one nickel so we can simplify again. Can we make $1.00 composed of quarters and nickels? Nope. Doesn’t work. Test 3 dimes and 7 pennies and you’ll very quickly see that you can’t make $1.00 out of 3 of the coins we’re given.

The final tip is that challenging problems don’t need to be tough. If you went into this and said I’m not sure how to solve this, but I can work from the answer choices and I can simplify problems, you might have started by testing answer A.

If I don’t have any dimes, I might have exactly 2 pennies. If that’s the case I have 11 coins to make $1.35. I need at least 2 nickels, so now I have 9 coins to make $1.25. 4 quarters and 5 nickels would work. Done. Answer choice A is correct. (You can also make this work with 0 dimes and 7 pennies).

When you don’t know where to start, look at the answer choices and look to simplify.

7 pennies, 1 nickel, 0 dimes, 5 quarters

 Now and Venn

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One of the books I’m currently reading is Priceless: The Myth of Fair Value (and How to Take Advantage of It) by William Poundstone. It’s an interesting look as some of the psychology behind how we think about prices and perceive value. Poundstone recounts one experiment done by Kahneman and Tversky called “Linda the Bank Teller” that I’d like you to think about.

“Linda is 31 years old, outspoken, and very bright. She majored in philosophy. As a student, she was deeply concerned with issues of discrimination and social justice, and also participated in anti-nuclear demonstrations.”

Which of the following is more likely to be true?

Linda is a bank teller.

Linda is a bank teller and is active in the feminist movement.

Astonishingly, 85% of the college students tested responded that the second statement is more likely to be true than the first, which is absolutely ridiculous. The point that Kahneman and Tversky were making is that we rely on heuristics or “rules of thumb” more often than is logically sound. The point I’d like to make is about drawing diagrams.

The information you’re given in this situation, although different in format from what you’re going to see on a standardized test is nonetheless similar. You’re given a few different pieces of information that fit together in some way that may not be intuitively obvious. One of the best ways to deal with that situation is to draw a graphical representation of what you’ve been told. When given two overlapping sets, you should draw a Venn diagram.

Now, let’s consider the two situations. First, the chance that Linda is a bank teller is represented by the region in red.

The chance that Linda is a bank teller and is active in the feminist movement is represented by the region in blue.

As you can see, the second set is a subset of the first. A drawing with only the two regions we’re concerned with would look something like this:

Linda can’t be a feminist and bank teller without being a bank teller. In this graphic form it’s pretty obvious, and yet 85% of the students missed this fact the first time around. This is a great example of the benefits of converting words into graphical forms so that you can make sure you have a good understanding of what you’re doing before you go forward with a problem.

 Augie’s Socks: Part II

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In part one, I introduced this set of facts:

Augie is blind. Thanks to his organized systems he is able to live alone. However, one day he finds that his laundry service has forgotten to pair his socks together. Augie has 12 black socks, 8 white socks, 6 blue socks and 4 red socks. Augie puts two random socks on.

Now, let’s ask a probability question.

What is the probability that Augie has put on a matching pair of socks?

This question tests both OR and AND probabilities. Let’s lay it out systematically.

There are four ways Augie could be wearing a matching pair of socks (black, black), (white, white), (blue, blue) or (red,red).

Since any one of these situations satisfies our condition of matching socks we want to add their individual probabilities together.

P(matching black) + P(matching white) + P(matching blue) + P(matching red) = P(matching pair of socks)

Now, we need to find those four probabilities individually. In order to pull a matching pair of black socks. In order to do that the first sock we pull out must be black. The odds of that are 12/30 (number of black socks/number of total socks) (12/30 reduces to 2/5 but we’ll save reducing for the end). But in order to pull a pair of black socks we need to pull a black sock first AND pull a black sock second. In order to pull the second black sock we’ll have to pull one of the remaining 11 black socks out of the remaining 29. Since we need both events to occur in order to get a black pair, we’ll multiply the individual probabilities to get the probability of getting a black pair:

12/30 * 11/29 = 132/870

Following the same process for the other colors gets you:

P(matching white) = 8/30 * 7/29 = 56/870

P(matching blue) = 6/30 * 5/29 = 30/870

P(matching red) = 4/30 * 3/29 = 12/870

Adding the probabilities gets a total of 230/870 = 23/87 or about 26.4%.

In this case you need to remember that when there are two individual events that must BOTH occur, you multiply their probabilities together. When there are multiple desired outcomes that are EACH satisfactory you add their probabilities.