+ Math Module 3 Distances

+ Navigating If we want to navigate our robot we have two choices. We could sense and recognize some type of landmark to tell us the robot’s location. We could approximate the robot’s location based on the turns it has made and the distances it has driven – this is called navigating by dead reckoning. The GPS system in a car usually gets its position from the GSP satellites but this can be unreliable. Mountains, weather, tall buildings, and tunnels can block the signal. Knowing how fast the car is moving (from the speedometer) and in what direction (from an internal compass) allows navigating by dead reckoning.

+ Distance We know how to drive our robot straight and how to turn 90 degrees to the right and left. We can move vertically and horizontally. This allows us to navigate mazes and identify our position using dead reckoning. We are often not so interested in the total distance travelled but in the straight line distance. Distance from our starting point. Distance to our target. We can represent this using right triangles.

+ Straight Line Distance

+ Vertical and Horizontal Distance Let us label the starting point with the coordinates (x 0, y 0 ). Suppose we drive our robot to point (x 1, y 1 ). The does it really matter how we get there? The net vertical distance travelled will be (y 1 – y 0 ). The net horizontal distance travelled will be (x 1 – x 0 ). We know two legs of the right triangle. How do we find the third longer side – the hypotenuse? This is the straight line distance.

+ Vertical and Horizontal Distance

+ Pythagorean Theorem We need a formula to relate the lengths of two legs of a right triangle to the length of the hypotenuse. If we have a right triangle with legs a and b and hypotenuse c, the we can use the famous Pythagorean Theorem a 2 + b 2 = c 2

+ Proving the Pythagorean Theorem There are many proofs of the Pythagorean Theorem. Probably the easiest to understand is to take a right triangle and use 4 copies of the triangle to construct a square. We will do this two different ways but get the same size square both times.

+ Proving the Pythagorean Theorem The area of the shaded middle square is c 2. The area of the big triangle is (a + b) 2 = 4 (area of triangle) + c 2.

+ Proving the Pythagorean Theorem The area of the two shaded squares are a 2 and b 2. The area of the big triangle is (a + b) 2 = 4 (area of triangle) + a 2 + b 2.

+ Proving the Pythagorean Theorem We constructed the same large square two different ways, computing its area both times. The area is 4 (area of triangle) + c 2 = 4 (area of triangle) + a 2 + b 2 c 2 = a 2 + b 2

+ Verifying Pythagorean Theorem for a 3, 4, 5 Triangle If the legs of a right triangle have sides 3 and 4 then the hypotenuse must have length 5 because 5 2 = 25 and = = 25 In this special case we can verify the Pythagorean theorem by using unit squares (or unit cubes). We construct a square along each side of the right triangle. The areas of these squares will be 3 2 = 9, 4 2 = 16, and 5 2 = 25.

+ Verifying Pythagorean Theorem for a 3, 4, 5 Triangle

+ Distance Formula Back to the straight line distance problem. We have a right triangle where the legs are a = x 1 – x 0 b = y 1 – y 0 c = straight line distance We know a 2 + b 2 = c 2 So (x 1 – x 2 ) 2 + (y 1 – y 0 ) 2 = c 2

+ Distance Formula Solving for the straight line distance c we get We will consider distances to always be positive so we can discard the negative square root.

+ Distance as a Function of Time Let’s look at the distance of the robot as a function of time. Suppose our robot is driving along the x axis, makes a 90 degree turn, and then drives directly to the target. We will assume the robot is driving at one unit per second and we will ignore the time it takes to make the 90 degree turn. What is the robots distance as a function of time?

+ Distance as a Function of Time

+ Travelling the x Axis As the robot travels along the x axis, we really don’t need the distance formula because only the x coordinate is changing. The distance from the starting point is increasing by one unit per second. This will go on for 8 seconds. The graph of distance VS time will look like the following.

+ Travelling the x Axis

+ Travelling in the y Direction At this point the robot turns and the graph changes. Over the next 6 seconds the robot moves directly toward the target. It starts a distance of 8 units from the start. The distance of the target from the start is Over 6 seconds the robot moves from 8 to 10 units from the start. The graph now looks like the following.

+ Travelling the x Axis

+ Distance To the Target We might also be concerned with the distance the robot still must travel to get to the target. The robot starts 10 units from the target. Over 8 seconds it moves to be 6 units from the target. It then turns 90 degrees and moves directly toward the target. The graph looks like the following.

+ Distance To the Target

+ Equidistant We can see that the robot is equidistant from the start and the target where the two graphs cross (somewhere between 6 and 7 seconds). The segments of the two graphs where they cross are lines. d = t and d = -1/2 t + 10 Both equations will be true when the lines cross. Since in d = t the first equation, we can replace the t in the second line with d. d = -1/2 d /2 d = 10 d = 20/3 = 6.67 units Again since d = t this means the robot will be equidistant from the start and the target a 6.67 seconds.