NAME COLLEGE ALGEBRA LAB All work must be shown for credit T

NAME COLLEGE ALGEBRA LAB All work must be shown for credit. The figure below shows an arched truss 80ft. long with heights as indicated and the vertical stringers are 10ft. apart. Both the top and the bottom of the arch are arcs of parabolas. 16 16 80\' 1. Find the equation of the top parabola. 2. Find the equation of the bottom parabola. ya(40416 o= a (0-40)-HC 0 100 Find to the nearest foot the sum of the lengths of the vertical stringers. 3. 10

Solution

using base as x axis , bisecting(dividing into 2 equal parts) line as y axis

parabolas has zeroes at x =-40 ,x=40

general form is y=k(x+40)(x-40)

1)

for top parabola:

y_t(0)=16+16

y_t(0)=32

=>k(0+40)(0-40)=32

=>k=-0.02

so equation of top parabola is y_t(x)=-0.02(x+40)(x-40)

so equation of top parabola is y_t(x)=-0.02(x²-1600)

2)

for bottom parabola:

y_b(0)=16

=>k(0+40)(0-40)=16

=>k=-0.01

so equation of bottom parabola is y_b(x)=-0.01(x+40)(x-40)

so equation of bottom parabola is y_b(x)=-0.01(x²-1600)

3)

sum of lengths of vertical strings =[y_t(-30)+y_t(-20)+y_t(-10)+y_t(0)+y_t(10)+y_t(20)+y_t(30)]-[y_b(-30)+y_b(-20)+y_b(-10)+y_b(0)+y_b(10)+y_b(20)+y_b(30)]

sum of lengths of vertical strings =84 feet