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Solution since the initial dose is y the drug

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Solution: Since the initial dose is y 0 , the drug concentration at any time t ≥o is found by the equation y=y 0 e -kt , the solution of the equation ky dt dy - = At t=T the second dose of y 0 is taken, which increases the drug level to y(T)=y 0 +y 0 e -kT = y 0 (1+e -kT ) 104
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The drug level immediately begins to decay. To find its mathematical expression we solve the initial-value problem: ky dt dy - = y(T)=y 0 (1+e -kT ) Solving this initial value problem we get y=y 0 (1+e -kT )e -k(t-T) This equation gives the drug level for t>T. The third dose of y 0 is to be taken at t=2T and the drug just before this dose is taken is given by kT e kT e o y T T k e kT e o y y - - + = - - - + = ) 1 ( ) 2 ( 1 The dosage y 0 taken at t=2T raises the drug level to y(2T) = y 0 + y 0 (1+e -kT )e -kT = y 0 (1+e -kT +e -2kt ) Continuing in this way, we find after (n+1)th dose is taken that the drug level is y(nT)=y 0 (1+e -kT +e -2kT +…..+e -nkT ) We notice that the drug level after (n+1) th dose is the sum of the first n terms of a geometric series, with first term as y o and the common ratio e -kT . This sum can be written as kT e kT n e o y nT y - - + - - = 1 ) ) 1 ( 1 ( ) ( As n becomes large, the drug level approaches a steady state value, say y s given by y s = lim y(nT) n →∞ 105
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= kT e 1 o y - - The steady state value y s is called the saturation level of the drug. 106
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4.11 A Pursuit Problem Figure 4.4 A dog chasing a rabbit is shown in Figure 4.4. The rabbit starts at the position (0,0) and runs at a constant speed v R along the y-axis. The dog starts chase at the position (1.0) and runs at a constant speed v D so that its line of sight is always directed at the rabbit. If v D >v R , the dog will catch the rabbit; otherwise the rabbit gets away. Finding the function representing the pursuit curve gives the path the dog follows. Since the dog always runs directly at the rabbit during the pursuit, the slope of the line of sight between the dog and the rabbit at any time t is given by x R y y R x x R y y m - = - - = If we assume that the line of sight is tangent to the pursuit curve y=f(x), then m= dx dy and therefore x R y y dx dy - = (4.21) is the mathematical model of the “Pursuit Problem”. The solution of (4.21) will give the path taken by the dog. 107
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The position of the dog at any time t>0 is (x,y), and the y coordinate of the rabbit at the corresponding time is y R =0+v R t =v R t, so x t v y dx dy R - = or t v y dx dy x R - = Implicitly differentiating this expression with respect to x yields - = + dx dt v ' y ' y " xy R where dx dy y dx y d y = = ' , 2 2 " This may be written as dx dt v " xy R - = (4.22) Finally, we note that the speed of the dog can be written as 2 2 dt dy dt dx dt ds D v + - = = 2 1 + - = dx dy dt dx Solving this for dx dt , we have 2 ) ' y ( 1 D v 1 dx dt + - = Substituting this result into Equation (4.22) yields 108
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2 ) ' y ( 1 D v 1 R v " xy + = Put y'=w, then this equation takes the form 2 w 1 D v 1 R v ' xw + = or x dx D v R v 2 w 1 dw = + Integrating both sides we get w and by integrating w we get y. The constant of integration can be found by using the initial conditions y(1)=0 and y' (1)=0.
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