The Rate of Change Essay Sample

The Rate of Change Pages
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During this investigation we will achieve to discover how math can be applied to our everyday lives by using what we have learned in our math class to find out data.

y=.0125×2
y=.0125×2

Finding the original equation:
y=ax2vertex= (0,0)point= (20,5)
35=a(55)2Domain= (-∞,∞)Range= [0, ∞)
35/25= a
a=.0125
1) Considering the map assigned to your team determine the position of the car when the headlights illuminate the sculpture.

Position= (-65,52.8)
By using the program graphmatica to be more accurate, we found out that in the parabola representing the road (y=.0125×2) the point where the car’s headlights illuminate the sculpture, which the sculpture is in the coordinate (-20,-20), is in the position of (-65, 52.8) because the car is moving to the right and this means that is going to the positive side of the graph.

2) Determine the equation of the line followed by the light of the car when the headlights hit the sculpture

Tangent line=
y=.0125×2
y’=.025x
y’=.025(-65)
y’= -1.625x
y=-1.625x-52.81

To find out the equation of the tangent line we first found the point where the car illuminated the sculpture which is (-65, 52.81), then we found the derivative of the original equation y=.0125×2 and the derivative is y’=.025x . After finding the derivative we needed to find the slope of the tangent line so we used the equation of the derivative and replaced the x with the x of the position and after solving it we got the slope of y’= -1.625x, and to finish the equation we added the y of the position to finally get y=-1.625x-52.81

3) Now determine the position of the car when the tail lights hit the sculpture.

Position (30,11.25)

By using the program graphmatica to get more accurate results, we found out that in the parabola representing the road (y=.0125×2), the point where the car’s tail lights hit the sculpture, which is found in the coordinate (-20,-20), is in the position of (30, 11.25). The car is moving to the right and this means that is going to the positive side of the graph. 4) Determine the equation of the line followed by the light of the car when the tail lights hit the sculpture.

Tangent Line
y=.0125×2 y=.0125x^2 y-11.25=0.75(x-30) y=.0125(30)2 y= 0.025x y=0.75x-22.5+11.25 y= 11.25 y=0.025(30) y=0.75x+33.75 m=0.75

To find out the equation of the tangent line we first found the point where
the car illuminated the sculpture which is (30,11.25), then we found the derivative of the original equation y=.0125×2 which is is y’=.025x . After finding the derivative we needed to find the slope of the tangent line so we used the equation of the derivative and replaced the x of that equation with the x-coordinate of the position (30,11.25) and after solving it we got the slope of m=0.75, and to finish the equation of the line which is: y-y0=m(x-x0) , we substitute the y and x values with the slope and we solve to get the final equation which is: y=0.75x+33.75

5) Find the point of intersection of the two lines, where do the two lines meet?

6) There is another sculpture located on the coordinate (20, 40) on the map assigned to your team, when will the lights (head or tail) of the car hit the sculpture?

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