Answer on Calculus Question for Abhilash
S(t) = At^p e^-kt
t > 0
The the second derivative :
d2S/dt2 = d/dt (dS/dt) = d/dt [e-kt (Aptp-1 - kAtp)] = -ke-kt (Aptp-1 - kAtp) + e-kt (Ap(p-1)tp-2 - kpAtp-1)] =
= e-kt [k2Atp -2 kpAtp-1 + Ap(p-1)tp-2]
For A = 0.02, p = 3, k = 3:
d2S/dt2 = e-3t [0.18t3 -0.36At2 + 0.12t] . The zeros of this second derivative are the inflection points:
e-3t [0.18t3 -0.36At2 + 0.12t] = 0.
e-3t t(3t2 - 6t + 2) = 0.
t1 = 0
3t2 - 6t + 2 = 0;
D = 9-6=3
t2 = (3 - √3)/3 = 1 - 1/√3 = 0.423;
t3 = (3 + √3)/3 = 1 + 1/√3 = 1.577;
Thus the points of inflection are 0, 0.423, 1.577
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