(t^2+1)^100
USE CHAIN RULE
Outside first (using power rule)
100*(t^2+1)^99 * derivative of the inside
100(t^2+1)^99 * d(t^2+1)
100(t^2+1)^99 * 2t
200t(t^2+1)^99
50,000 = 5 · 10,000 = 5 · 10⁴
Check the picture below.
since the focus point is above the directrix, that simply means the parabola is a vertical one, and therefore the square variable is the "x".
keeping in mind that, there's a distance "p" from the vertex to either the focus point or the directrix, that puts the vertex half-way between those fellows, in this case at 0, right between 1 and -1, as you see in the picture, and the parabola looks like so.
since the parabola is opening upwards, the value for "p" is positive, thus
P = 2(L + W)
P = 22
L = 2W + 2
22 = 2(2W + 2 + W)
22 = 2(3W + 2)
22 = 6W + 4
22 - 4 = 6W
18 = 6W
18/6 = W
3 = W.....this is the width
L = 2W + 2
L = 2(3) + 2
L = 6 + 2
L = 8 meters <=== this is the length