Genie and Hitch

Suppose Genie walks her dog Hitch on a trail where Hitch can run around without a leash. Genie wants to make sure Hitch gets a certain number of miles of running per day, but she doesn't want to walk that far herself. How long does Genie have to walk to make sure Hitch gets enough running?

The first step to solving most problems is to figure out what you know, and what you are trying to find out. Writting things down is always a good idea. Trying to keep it all in your head just makes things harder.

For Genie to figure out how far she needs to walk, she's going to have to make some estimates or measurements. She wants to know how fast Hitch is running so that she'll know how far he runs fur the duration of the walk. Let's say Genie estimates Hitch's running speed to be twice as fast as her walking speed. So if Genie walks at X mph, Hitch runs at 2X mph. In time T, Genie walks T*X miles, and Hitch runs T*2X miles. In comparing those distances, we see that Hitch runs twice as far (2XT) as Genie walks (XT). That means, Genie only needs to walk half the distance she wants Hitch to cover.

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S.A.Tricky

Here's an S.A.T. problem that stumped my students and me (for a little while).

Problem:

a - b = 10
a² - b² = 50

Find b.

Solution:

My instincts were to look for a connection between (a - b)² and a² - b², but it's really much simpler than that. Solve for a in terms of b in the first equation, and substitute into the second equation.

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Edit: The original post contained a mistake. (10 + b)² = 100 + 20b + b², not 100 + 2b + b².

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Atari Space

My friend Justin was reminiscing about some Atari-aged video games: Subspace/Continuum. The problem they inspired reminds me more of Asteroids.

Problem:

A turret in a video game can rotate but not change location. It shoots bullets that travel with a constant velocity V_B. An asteroid travels at a constant velocity V_A. Where should the turret aim to hit the asteroid?

Solution:

You'll never find where you're going if you don't know where you started. Let's put the turret at the center of a Cartesian plane. Add a point A for the asteroid at (x₀, y₀), and add a point for the eventual collision at (x₁, y₁). Label the distances traveled by the asteroid and the bullet D_A and D_B. Recall that little ditty: "Distance = Rate × Time." We can use this whenever we are dealing with constant velocities. I've already labeled arrows R_A and R_B to represent the rates in the diagram.(I'm listing the equation numbers (abbreviated "eqn") on the left. When I use those equations later, I'll show which equation number I used on the right.) Note: the bullet and the asteroid travel for the same amount of time once the bullet is fired, so T has the same value in each equation.

We have two equations and three unknowns, so we'll need to find some more relationships. I can find more information about D_A and D_B by breaking the distances into their horizontal and vertical components. I've added these components to the diagram below.Let β be the angle that the bullet's trajectory makes with the x-axis, and call the angle that the ship flies at with respect to the x-axis α.With all of these variables in play, let's take a moment to recap what is known and what we need to find.Now, we need to trade our variables down to get rid of some of those unknowns. I'll start with x₁. Now do the same with the sines to get rid of y₁. Now that we've gotten rid of x₁ and y₁, solve for T in equation 11 and substitute into equation 15... ...and simplify. Now we're down to one unknown: β. To find β, solve for sin(β), square both sides, and substitute 1 - cos²(β) for sin²(β). Then, use the quadratic equation to solve for the possible values of cos(β), and convert those values into the angles we want. Once we have β, we use equation 18 to find T and equations 4 and 5 to find x₁ and y₁.

I'm not going to attempt to finish this solution unless someone asks me to. I'm too likely to make a mistake somewhere in the tangled mass of algebra that is to follow, and Justin has access to math software that can do the manipulations for him. Comment if you have any questions.

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