Derivadas

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|Math Monthly Project |
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Math can be seen all around us, people are always thinking about how can something related to math be useful to him/her someday in their life, actually math is everywhere and the person can be able to see it, he or she just has to be more attentive. Math can make a person’s life easier, of courseif the concept is clear to the person. In this project, we were given an equation of position to which we had to obtain the first and second derivative, meaning the velocity and acceleration an also, we had to graph it and then establish relationships amongst the three different graphs with the purpose of making these concepts clearer. In the second part of the project, another graph was given tous, this one with a more real life situation in which a car was traveling through a highway, the questions were focused on a –when will the headlights or taillights illuminate a sculpture?- type of question.

FIRST PART

1) One problem will be assigned to each of the teams, is the equation of position of a car.
y=(-0.0006x^5)+(0.02325x^4)-(0.304x^3)+(1.44x^2)-8

2) Use a graphingdevice to graph the equation of position assigned to your team (remember to include all the graphs on your final report). Determine the domain and range for the situation.

Domain: (-∞, ∞)
Range: (-∞,∞).
The domain and range have no restrictions, therefore they are infinite.
3) Determine the equation of velocity of the car (remember to include the procedures) and graph the equation of velocity.In order to obtain the velocity one must find the derivative of the original equation.

V= (-0.0006t^5)+(0.02325t^4)-(0.304t^3)+(1.44t^2)-8

V= 5(-0.0006t^4)+4(0.02325t^3)-3(0.304t^2)+2(1.44t)

V=(-0.003x^4)+(0.093x^3)-0.912x^2+2.88x

This graph shows the equation of velocity (m/s).

Domain: (-∞,∞)

Range: (-∞, 3)

4) Determine the equation of acceleration of the car (remember toinclude the procedures) and graph the equation of acceleration.

To obtain the acceleration of the car, one could get the second derivative of the original equation or more easily, one can obtain the derivative of the equation of velocity that will give you the equation for acceleration.

V = (-0.003x^4)+(0.093x^3)-0.912x^2+2.88x

A = 4(-0.003x^3)+3(0.093x^2)+2(-0.921x)+2.88A = -0.012x^3+.279x^2-1.824x+2.88

This graph shows the equation of acceleration (m/s^2).

Domain: (-∞,∞)
Range: (-∞,∞)

5) Using the three equations and their graphs describe the movement of the car. Clearly state in which interval it moves forward (to the right), backward (to the left) or if it stops and if it moves each time faster or slower.

This graph shows the threedifferent equations obtained previously, those of position, velocity and acceleration.

(-∞, 0)= The car is moving left with a negative velocity, always decreasing its velocity due to the positive acceleration, when approaching to 0 it decreases to a complete stop.

(0, 8)= The car starts moving right with a positive velocity and a positive but descendant acceleration, then between 2.5 and 8 theacceleration becomes negative and the car stops again.

(8, 15)= The car once again increases its velocity and acceleration and when approaching to 15 the acceleration becomes negative and the car stops.

(15, ∞)= The car starts moving left again with an ascendant negative acceleration and velocity.

6) When 3 minutes (180 s) have gone by since it started to move, find the position,velocity and acceleration for that time. Describe the movement of the car for the given time.

Position Y =(-0.0006(3)^5)+(0.02325(3)^4)-(0.304(3)^3)+(1.44(3)^2)-8
Y = -1.889568^-14 + 2.366883^-5-0.75855+18.6624-8
Y = 9.917 m After 180 s or 3 minutes, the position of the car would have advanced almost 10 more meters.

Velocity V = (-0.003(3)^4)+(0.093(3)^3)-0.912(3)^2+2.88(3)...
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