According to the Adams-Moulton scheme, the derivative of a function T at time-step t is given by _________(a) \(\frac{3 T(t)+4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)(b) \(\frac{3 T(t)-4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)(c) \(\frac{3 T(t)+4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)(d) \(\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)I got this question during an interview.Origin of the question is Transient Flows topic in section Transient Flows of Computational Fluid Dynamics

According to the Adams-Moulton scheme, the derivative of a function T

1.	According to the Adams-Moulton scheme, the derivative of a function T at time-step t is given by _________(a) \(\frac{3 T(t)+4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)(b) \(\frac{3 T(t)-4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)(c) \(\frac{3 T(t)+4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)(d) \(\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)I got this question during an interview.Origin of the question is Transient Flows topic in section Transient Flows of Computational Fluid Dynamics
Answer» Correct answer is (d) \(\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\) Easiest explanation: To FIND the derivative, the Adams-Moulton METHOD USES the previous and the second previous steps. The mathematical expression is \(\frac{\PARTIAL T(t)}{\partial t}=\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\).

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