Find the exact curve where the following curve is the steepest:

y = 50 / (1+6e^-2t) for t>or = to 0

First take the derivative of the Y w. r. t t

so

dy/dt = 50 (-1) (1 + 6e^ (-2t)) ^ (-2). (6e^ (-2t). (-2)

=600 e^ (-2t) / (1 + 6e^ (-2t)) ^2

to find that where it is maximum we take the drivative again

d²y/dt² = 600 e^ (-2t). (-2). (1 + 6e^ (-2t)) ^ (-2) + 600 e^ (-2t). (-2) (1 + 6e^ (-2t)) ^ (-3). (-12 e^ (-2t)) = [-1200 e^ (-2t) (1 + 6e^ (-2t)) + 1200 e^ (-2t). 12e^ (-2t) ] / (1 + 6e^ (-2t)) ^3 = 1200 e^ (-2t) [6e^ (-2t) - 1] / (1 + 6e^ (-2t)) ^3 which will be 0 when 6e^ (-2t) = 1, so e^ (-2t) = 1/6 = > - 2t = ln (1/6) = - ln 6 = > t = (1/2) ln 6

for value of y we know that

6e^ (-2t) = 1, so y = 50 / (1 + 6e^ (-2t)) = 50 / (1 + 1) = 25. So the coordinates are ((ln 6) / 2, 25).