Asked by Anonymous
A window is designed to be the area between a parabolic section and a straight base. The height is 0.640 m and the base is 1.60 m. What is the area of the window?
Answers
Answered by
Reiny
place the region on an x-y grid so that the centre of the base is at the origin.
so the vertex is (0,.64)
equation is:
y = a(x-0)^2 + .64
y = ax^2 + .64
but (.8,0) lies on it
0 = a(.64) + .64
a = -.64/.64 = -1
y = -x^2 + .64
area between curve and x-axis
= 2∫(-x^2 + .64) dx from 0 to .8
I went from 0 to .8 and doubled it because of the nice symmetry
= 2 [(-1/3)x^3 + .64x ] from 0 to .8
= 2( (-.512/3 + .64(.8) - 0)
= 2( -512/3000 + 64/80)
= 2(236/375)
= 472/375 = appr 1.259 m^2
so the vertex is (0,.64)
equation is:
y = a(x-0)^2 + .64
y = ax^2 + .64
but (.8,0) lies on it
0 = a(.64) + .64
a = -.64/.64 = -1
y = -x^2 + .64
area between curve and x-axis
= 2∫(-x^2 + .64) dx from 0 to .8
I went from 0 to .8 and doubled it because of the nice symmetry
= 2 [(-1/3)x^3 + .64x ] from 0 to .8
= 2( (-.512/3 + .64(.8) - 0)
= 2( -512/3000 + 64/80)
= 2(236/375)
= 472/375 = appr 1.259 m^2
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