DEV project

Problem 1)

Suppose a Ferris wheel has a 15 ft tall ramp and the diameter of the wheel is 80 ft tall. It makes a complete revolution in 20 sec. Write a mathematical equation that describes a relationship between the height of the rider above the bottom of the Ferris wheel and time.



First, we set the period of  20 seconds equal to 2pi/b, then we determined the b value to be pi/10. we knew our amplitude was -40 because the rider boarded on the bottom of the wheel and the diameter is 80. the vertical shift was +44 because the amplitude plus the height of the ramp equals 44. The equation was cosine because it started at the bottom of the Ferris wheel.

Problem 2)

Richar Ressler, a famous astronaut, came up with the idea that the brightness of a moon increases and decreases with itself. the moons radius is 1000 miles and changes by 1 mile a day with each flash of light. if the time between periods of maximum brightness is 3 days have an equation that describes the radius o this moon as a function of time.

We set the period of 3 to pi/2 and got our b value of 2pi/3. The vertical shift was 1000 because that is the radius of the moon. We knew the sine graph would increase and decrease a 1 mile both ways because that is how much each flash of light changes.

Problem 3)

using our identity formulas, we knew that the right side of the problem was equal to 1. The top of the right formula cancelled out to one and the bottom of the equation was an identity that equaled one. So each side of the equation equals 1.

Problem 4)

 We worked on the top of the left side first and got it sin^2X times cos^2X. We put that over the bottom half of the left side, which we got down to 1/sin. to get rid of the fraction underneath a fraction, we had to multiply the reciprocal of the bottom times the top. The left side came to equal sin^2X times cos^2X and same with the right side of the equation.

Problem 5)

The top side of the left has an identity equal to one, so the top part of the fraction is now "cot^2X + 1" which equals csc^2X. After we put that over the bottom part of the fraction it became csc^2X over tan^2X. we then multiplied the reciprocal and the left side became cos^2X/sin^4X. The right side of the problem was cos^2X/sin^2X times 1/sin^2X. Both sides then equaled cos^2X/sin^4X.

-P.S. we worked really hard on this amazing project, Jackson.



lexi: 

This project was kinda fun to me once we started to do it. iIt took alot of time and effort but i do not think it was too hard! me and richar both worked equally through it so not too much work was put on just one of us.



richar: 

i did not really like this project, we had to put in a lot of hours but i think it turned out good! i really hope we get a good grade on it though because we need it to help our grades and that would be awesome!