Part of my DEV project

1. Q1: The Velodrome Solutions

2. When racer 1 has double racer 2’s speed their functions will be equal. But racer 1 is now picking up speed. So his function is now… j (x) = 10sin((pi/5)(x))+40 +x Because every x twentieth of the track he picks up x speed so your adding the x twentieths of the track to his speed And racer 1 will be trying to double racer 2’s speed so racer 2’s function that has double the speed and half the range is… k(x) =5 sin((pi/5)(x))+ 70 (5 before the sine would stay the same because it would be doubled and halved) . Since their functions will be equal we can make an equation out of this… j (x) = k(x) 10sin((pi/5)(x))+40 = 5sin((pi/5)(x))+ 70

3. We then try to solve for x… j (x) = g(x) 10sin((pi/5)(x))+40 +x = 5sin((pi/5)(x))+ 70 10sin((pi/5)(x))+40 +x - ( 5sin((pi/5)(x))+ 70) = 0 5 sin((pi/5)(x)) + x – 30 = 0 5 sin((pi/5)(x)) + x = 30 5 sin((pi/5)(x)) = 30 – x Any way that you manipulate it you are going to get an x within the argument of sine. But from here it is solvable.

4. 5 sin((pi/5)(x)) = 30 – x Take the reciprocal of both sides 1/5 sin((pi/5)(x)) = 1/(30 – x) The outputs of 1/5 sin((pi/5)(x)) are + when 0<x<5 + a10; a E Real The outputs of 1/5 sin((pi/5)(x)) are - when 5<x<10 + a10; a E Real Domain: {x|x cannot equal a * pi; aEI; xER} Range: {y|y<=-0.2, y=>0.2;yER} From what we know about reciprocals 1/(30 – x) will have invariant points at y = –1 and 1. Setting it equal to those outputs… 1/(30 – x) = 1 1/(30 – x) = -1 x = 29 x = 31 We have invariant points at (29,1) and (31,-1)

7. The reciprocal function 1/(30-x) has only 1 input that it cannot use and that input (x = 30) must be where the other graphs meet, since they intersect and reciprocal graphs didn’t. If you input it into the question for x it is true. 5sin((pi/5)(30)) = 30 – 30 5sin(6pi) = 0 5(0) = 0 0=0 30 is how many twentieths of the track it will take. So it will take him one and a half laps at this ever increasing pace to double his opponents speed.

8. 30 – x (blue) and 5sin((pi/5)(x)) (green) Intersection at 30 supports our answer.

9. 10sin((pi/5)(x))+70(black) and 10sin((pi/5)(x))+40+x (red) Intersection at 30 supports our answer.

10. There is a shorter way to do this though. At one point we had 5 sin((pi/5)(x)) = 30 – x. Now if you recognize that if x was any multiple of 10 that you would get sin(multiples of 2pi), which no matter what is 0 and 5 times 0 is 0. 30 is a multiple of 10 and would give you zero on the other side of the equation which would make both sides of the equation equal and 30 a valid answer.