# a problem i made up

#### Drummer87

##### New member
MY DEVELOPEMENT OF THE SITUATION

i started makin this problem so i could combine the concepts of vectors and derivatives/tangent lines and integrals. i started with the simple parabola:

y2 = (-(x-3)^2) / 2 + (9/2)

this represents the (x,y) journey taken by an object after taking gravity or some negative acceleration into effect above ground level (x axis) that drops 2 gallons of water per unit^2.

after finding the the tangent lines for the x values where y=0, you get the two lines:

y= 3x

and

y= -3x + 18

as you can see(assuming you have made a graphical representation of this), a triangle is formed, or more like an absolute value function. these equations represent the two vectors which will be givens when the problem is complete.
so it makes sense to me to try and combine these vectors to get a single equation that represents the path of the object WITHOUT the gravity/negative acceleration.
mathematically i dont know how to do this process but i did find a solution which is:

y1 = - |3x - 9| + 9

anywayz, we now have two paths that the object can possibly take:

For y >= 0
y1 = - |3x - 9| + 9

and

For y >= 0
y2 = (-(x-3)^2) / 2 + (9/2)

to find the area under these functions we use integrals. set the limits equal to the x intercept (x=0 and x=6).

the area under y1 = 27 units^2

the area under y2 = 18 units^2

QUESTION: "how much more water could the object drop if it had a force that canceled out the negative acceleration?" or how much more water could the object drop if gravity/negative acceleration were not a factor?

to get the answer we subtract the areas from one another under each curve:

27 units^2 - 18 units^2 = 9 units^2

and since there is 2 gallons per unit^2:

9 * 2 = 18 gallons

SOLUTION: it would be able to drop 18 more gallons of water if acceleration were not an issue.

plz check this, im 90% sure it is correct.

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BRINGING IT ALL TOGETHER TO MAKE AN ACTUAL PROBLEM

now my problem begins ..........

i want to give the solver/student only the two vectors.

y=3x

and

y=-3x + 18

from there they can derive the points P0 = (x0,y0) , P1 = (x1,y1) , P2 = (x2,y2) , and P3 = (x3,y3). then plug these into bezier curve equations. i suppose this is the correct method to take to get the path of the of the object given a negative acceleration. but we probably need to give the student a GIVEN NEGATIVE ACCELERATION...help here please... and then i need to figure out some way to work that acceleration into the process of deriving the new path.

anywayz, solve accordingly with bezier curve to get the PARAMETRIC equations:

x(t) = 6t^3 - 9t^2 + 9t

and

y(t) = -27t^2 + 27t

then the student will somehow convert to a single and simple f(x) function .... i have no earthly clue how to do this conversion, help please! ..... which will be something like:

y2 = (-(x-3)^2) / 2 + (9/2)

so now we have 2 vectors and 1 function! next we need to convert the 2 vectors into a single and simple f(x) function....... again i have no clue how to do this conversion so help me please..... it will look something like:

y1 = - |3x - 9| + 9

.....now that we have 2 simple functions each representing 2 dimensional motion,
we can solve the problem using integrals, which has already been explained in my development of this problem.

if you can help in anyway possible it will be greatly appreciated.
thanks,
Richard W.