Lesson Word problems on engineering constructions of parabolic shapes

Word problems on engineering constructions of parabolic shapes

Problem 1

There is no need to explain that the parabola is opened down in this problem.


So, let assume that the origin of the coordinate system is established at one edge of the tunnel.
Then the other edge is at x= 20 meters.


So, our parabola has two intersections with x-axis: at x= 0 and x= 20.


Thus the parabola has two zeroes, 0 and 20. Therefore, we can write it in the form

    y = a*(x-0)*(x-20) = a*x*(x-20),


where "a" is some real coefficient, now unknown.


To find "a", use the fact that  y(4) = 16 meters, which is given.  So

    a*4*(4-20) = 16,   or

    4a*(-16) = 16, 

    -4a      =  1,

      a      = -1/4 = -0.25.


Thus, the parabola is

      y = -0.25x*(x-20).


Its vertex is half-way between 0 and 20, i.e. at x= 10.


Therefore, the highest point of the parabola is  y(10) = -0.25*10*(10-20) = -0.25*10*(-10) = 25.


ANSWER.  The height of the tunnel is 25 meters over the ground level.

Problem 2

Let x-axis be horizontal at the ground level, with the origin under the upper 
point of the bridge; y-axis vertical.


Then the parabola has x-intercepts at x= -10 ft and x= 10 ft.


So, the parabola has the form  y = a*(x-(-10))*(x-10) = a*(x+10)*(x-10) = a*(x^2-100).


Coefficient "a" is some real negative number (since the parabola is opened downward).
We do not know this number. It is the only unknown in this problem,
and our goal is to find this single unknown.


To find "a", use the fact that at x= 1 ft we have from the problem y= 8 ft.
It gives you this equation

    8 = a*(1^2 - 100),  or  8 = -99a,  which implies  a =  = -0.08080808...


Thus the quadratic function is y = ,

and it has the maximum at x= 0  (at the axis of symmetry).


Thus the maximum height of the bridge is   =  = 8.08080808... ft, 
or, after rounding, about 8.08 ft.    ANSWER

Problem 3

Place the origin of the coordinate system half way between the towers,
at the level of the bridge.


Using vertex form, write equation of a parabola 

    y = ax^2 + 30.


Find the unknown coefficient "a" from the condition 

    a*200^2 + 30 = 150  ft,


which says that the cable is suspended at the top of a tower  (for each cable and each tower).


From this equation, find "a"

    40000a = 150 - 30 = 120

    a =  =  = 0.003.


So, the parabola equation is

    y = 0.003*x^2 + 30.


They want you find "y" at x = 200-50 = 150 ft.

Substitute x= 150 into equation of the parabola and get

    y = 0.003*150^2+30 = 97.5  ft.


ANSWER.  50 ft away from either tower, the cable is 97.5 ft over the bridge.

Problem 4

Place the origin of the coordinate system (x,y) where the first jet begins.


Then the jet (the quadratic function) has x-intercepts at x= 0 and x= 12+5 = 17 feet.


So, we can write the quadratic function in the form

    y = a(x-0)*(x-17) = ax*(x-17).


In this form, we have only one unknown, which is the real coefficient "a".


We will find it from the condition  y(12) = 5,  which says that the jet passes 
through the hole located at x= 12 and y= 5.


So, for "a" we have this equation

    5 = a*12*(12-17) = a*12*(-5) = -60a.


It gives  a =  = .


Thus, the quadratic function representing the path of the jet is

    y = .     ANSWER