# Download A Source Book in Mathematics: volume 1 by David Eugene Smith PDF By David Eugene Smith

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Hence Integrating (2) we obtain Using (7), (8) and (3) as before we obtain Table 2. The last row will be explained shortly. r We would like to somehow compare Tables 1 and 2 to see if they make similar predictions concerning /. Since / depends heavily on C in Table 1 and on K in Table 2, it is not clear how to do this. We adopt the following scheme. Start with a column of Table 1. Let K be the average value of 6 + 0 as the bus moves from its initial position to the point at which /(0) is a maximum.

The wheel positions can be computed if the axle positions are known. The necessity of the unknown scalar factor y(t) is what makes this problem interesting. The solution of (1) is Integrating by parts and setting p,(0) = />/i0, / = 1, 2, we obtain Let £(0 = exp (/0' 7(5) ds). ,,0 + fof(s)p'\(s) ds, so that q'(t) = £ ( t ) p \ ( t ) . ;(/), (4) becomes Unfortunately, we are unable to explicitly solve this equation. We are able to get a simple differential equation for £(/) when the motion of the front of the bus is described as a solution of the second order linear differential equation, where a, b, c are constant and d = (*/,, d2) is a fixed vector.

Whenever the initial conditions and the amount of fuel available are compatible with a soft landing, the optimal control law consists of a final stage of braking with maximum thrust, generally preceded by an initial stage of free fall. An analogous treatment can be applied to the minimum time problem, with performance index / = \',10 dt, f = 1, as we shall now show. From the Hamiltonian (4), with /= 1, one sees that condition (6) can only be satisfied by control laws where u = 0 or « = a, whenever the switching function is respectively positive or negative.